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Limits and continuity in a normed space

⌚ ~75 min ▢ 9 blocks ✓ 20 exercises ➣ Prerequisites : Limits and continuity, Topology of a normed space

The chapter Topology of a normed space described single normed spaces from the inside --- their open and closed sets, the closure of a subset, density. This chapter studies maps between normed spaces and asks the two questions on which all of analysis rests: does $f(x)$ approach a value as $x$ approaches a point $a$, and does that value equal $f(a)$? The first is the notion of limit, the second the notion of continuity.
The chapter has three sections. Section~1 builds the limit of a map, its sequential characterisation --- the bridge to the theory of sequences --- and the rules for computing it. Section~2 turns to continuity: its operations, its reading through the preimages of open and closed sets, and the finer Lipschitz and uniform-continuity grades. Section~3 treats the remarkably simple case of linear and multilinear maps, where continuity reduces to a single inequality $\norme{u(x)} \le C\,\norme{x}$ and is measured by the operator norm.
Standing notation. Throughout, $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$, and $E$, $F$, $G$ denote normed $\mathbb{K}$-vector spaces; a norm is written $\norme{\cdot}$, or $\norme{\cdot}_E$ and $\norme{\cdot}_F$ when two must be told apart. A map is studied on a subset $A$ of $E$, and the point $a$ at which a limit is taken is an adherent point of $A$ --- written $a \in \overline{A}$, the closure being that of Topology of a normed space. Open balls $B(a,r)$, closed balls $B_f(a,r)$, the distance $d(x,y) = \norme{x-y}$, convergent sequences and their operations are those of Normed vector spaces; open sets, closed sets, neighbourhoods, the closure $\overline{A}$, density and the relative topology are those of Topology of a normed space. This chapter is the analytic layer feeding Compactness, connectedness, finite dimension and every later chapter that differentiates or integrates.

I Limits

I.1 Limit of a map

In Limits and continuity the limit of a real function was written $\norme{x-a}$ small $\Rightarrow$ $\norme{f(x)-\ell}$ small, with absolute values. Replacing each absolute value by the norm of the space it lives in gives, word for word, the limit of a map between normed spaces: $f$ tends to $b$ at $a$ when $f$ sends every point close enough to $a$ to a point as close as wished to $b$. The point $a$ is taken adherent to the domain $A$, so that every ball around $a$ meets $A$.

Definition — Limit of a map at a point

Let $A$ be a non-empty subset of $E$, let $f \colon A \to F$, let $a \in \overline{A}$ and $b \in F$. We say that $f$ admits the limit $b$ at $a$ if $$ \forall \varepsilon > 0,\ \exists \alpha > 0,\ \forall x \in A,\quad \norme{x-a} \le \alpha \implies \norme{f(x)-b} < \varepsilon. $$

The definition says: the part of $A$ inside the closed ball $B_f(a,\alpha)$ is sent inside the open ball $B(b,\varepsilon)$.

Proposition — Uniqueness of the limit

Let $f \colon A \to F$ and $a \in \overline{A}$. If $f$ admits a limit at $a$, that limit is unique. One then writes it $\displaystyle\lim_{a} f$ or $\displaystyle\lim_{x \to a} f(x)$, and the convergence is denoted $f(x) \xrightarrow[x \to a]{} b$.

Proof

Suppose $f$ admits two limits $b$ and $b'$ at $a$. Fix $\varepsilon > 0$. The definition gives $\alpha > 0$ and $\alpha' > 0$ such that $\norme{x-a} \le \alpha \Rightarrow \norme{f(x)-b} < \varepsilon$ and $\norme{x-a} \le \alpha' \Rightarrow \norme{f(x)-b'} < \varepsilon$. Since $a \in \overline{A}$, the ball $B\bigl(a,\min(\alpha,\alpha')\bigr)$ meets $A$: there exists at least one $x \in A$ with $\norme{x-a} \le \min(\alpha,\alpha')$. For such an $x$, $$ \begin{aligned} \norme{b-b'} &= \norme{\bigl(b-f(x)\bigr) + \bigl(f(x)-b'\bigr)} && \\ &\le \norme{f(x)-b} + \norme{f(x)-b'} && \text{(triangle inequality)}\\ &< \varepsilon + \varepsilon = 2\varepsilon. && \text{(the two implications above)} \end{aligned} $$ Thus $\norme{b-b'} < 2\varepsilon$ for every $\varepsilon > 0$, hence $\norme{b-b'} = 0$ and $b = b'$.

Proposition — Reformulation by balls and by neighbourhoods

Let $f \colon A \to F$, $a \in \overline{A}$ and $b \in F$. The following are equivalent:

[(i)] $\displaystyle\lim_{a} f = b$;
[(ii)] for every $\varepsilon > 0$ there is $\alpha > 0$ with $f\bigl(A \cap B_f(a,\alpha)\bigr) \subset B(b,\varepsilon)$;
[(iii)] for every neighbourhood $W$ of $b$ there is a neighbourhood $V$ of $a$ with $f(A \cap V) \subset W$.

Proof

(i) $\iff$ (ii). The membership $x \in A \cap B_f(a,\alpha)$ is exactly « $x \in A$ and $\norme{x-a} \le \alpha$ », and $f(x) \in B(b,\varepsilon)$ is exactly $\norme{f(x)-b} < \varepsilon$. So « $f\bigl(A \cap B_f(a,\alpha)\bigr) \subset B(b,\varepsilon)$ » is a rewriting of « $\forall x \in A,\ \norme{x-a} \le \alpha \Rightarrow \norme{f(x)-b} < \varepsilon$ », and (ii) is the definition of the limit set down with set inclusions.
(ii) $\Rightarrow$ (iii). Let $W$ be a neighbourhood of $b$: it contains a ball $B(b,\varepsilon)$. By (ii) there is $\alpha > 0$ with $f\bigl(A \cap B_f(a,\alpha)\bigr) \subset B(b,\varepsilon) \subset W$. The ball $V = B(a,\alpha)$ is a neighbourhood of $a$, and $A \cap V \subset A \cap B_f(a,\alpha)$, so $f(A \cap V) \subset W$.
(iii) $\Rightarrow$ (ii). Let $\varepsilon > 0$. The ball $W = B(b,\varepsilon)$ is a neighbourhood of $b$; (iii) gives a neighbourhood $V$ of $a$ with $f(A \cap V) \subset W$, and $V$ contains a ball $B(a,\alpha)$. Then $f\bigl(A \cap B_f(a,\alpha/2)\bigr) \subset f(A \cap V) \subset B(b,\varepsilon)$.

The limit here is taken over all $x \in A$, including $x = a$ when $a$ belongs to $A$: it is an unpunctured limit. A consequence is that if $a \in A$ and $\lim_a f$ exists, then necessarily $\lim_a f = f(a)$ --- this is exactly continuity, the subject of Section~2. The punctured limit familiar from one-variable analysis, where the value at $a$ is ignored, is recovered here as the limit of the restriction of $f$ to $A \setminus \{a\}$.

Definition — Limits involving infinity

The definition of a limit extends, by the same template, to infinite base points and infinite values, each case carrying its own adherent hypothesis.

If $A$ is unbounded, $f \colon A \to F$ tends to $b$ as $\norme{x} \to +\infty$ if $\forall \varepsilon > 0,\ \exists M \in \mathbb{R},\ \forall x \in A,\ \norme{x} \ge M \Rightarrow \norme{f(x)-b} < \varepsilon$.
If $A \subset \mathbb{R}$ is unbounded above (resp. below), the limit at $+\infty$ (resp. $-\infty$) is defined likewise, with $x \ge M$ (resp. $x \le M$).
For a real-valued $f \colon A \to \mathbb{R}$ and $a \in \overline{A \setminus \{a\}}$, $f$ tends to $+\infty$ at $a$ if $\forall M \in \mathbb{R},\ \exists \alpha > 0,\ \forall x \in A \setminus \{a\},\ \norme{x-a} \le \alpha \Rightarrow f(x) \ge M$ --- here the punctured domain $A \setminus \{a\}$ is mandatory, since a finite value $f(a)$ would forbid $f(x) \to +\infty$. The limit $-\infty$ is symmetric.

Mixed cases combine the two templates explicitly: for instance, for $A \subset \mathbb{R}$ unbounded above, $f \colon A \to \mathbb{R}$ tends to $+\infty$ as $x \to +\infty$ if $\forall M \in \mathbb{R},\ \exists B \in \mathbb{R},\ \forall x \in A,\ x \ge B \Rightarrow f(x) \ge M$. Uniqueness of the limit and the sequential characterisation extend to all these cases by the same arguments --- a ball around an infinite endpoint being replaced by the corresponding tail, and the convergent test sequences by sequences tending to that endpoint; the operation rules require, exactly as in one-variable analysis, that no indeterminate form ($+\infty + (-\infty)$, $0 \cdot \infty$, $\tfrac{\infty}{\infty}$, $\tfrac{0}{0}$) arise.

Example — A limit by the $\varepsilon$-$\alpha$ definition

Equip $\mathbb{R}^2$ with the norm $\norme{\cdot}_\infty$ (a usual norm of Normed vector spaces) and let $f \colon \mathbb{R}^2 \to \mathbb{R}$, $f(x,y) = 2x - 3y$. Show that $\displaystyle\lim_{(x,y) \to (1,1)} f(x,y) = -1$.

Answer

Write $u = (x,y)$ and $a = (1,1)$, so $f(a) = -1$. For every $u \in \mathbb{R}^2$, $$ \begin{aligned} |f(u) - (-1)| &= |2x - 3y + 1| = |2(x-1) - 3(y-1)| && \\ &\le 2\,|x-1| + 3\,|y-1| && \text{(triangle inequality)}\\ &\le 5\,\norme{u-a}_\infty, && \text{(}|x-1|,|y-1| \le \norme{u-a}_\infty\text{)} \end{aligned} $$ since $\norme{u-a}_\infty = \max(|x-1|,|y-1|)$. Fix $\varepsilon > 0$ and set $\alpha = \varepsilon/10 > 0$. Then $\norme{u-a}_\infty \le \alpha$ gives $|f(u)+1| \le 5\alpha = \varepsilon/2 < \varepsilon$ --- which establishes $\lim_{u \to a} f(u) = -1$.

Example — A limit as the norm goes to infinity

On a non-zero normed space $E$ --- so $E$ is unbounded --- consider $f \colon E \to \mathbb{R}$, $f(x) = \dfrac{1}{1+\norme{x}}$. As $\norme{x} \to +\infty$, the denominator $1+\norme{x} \to +\infty$, so $f(x) \to 0$: for $\varepsilon > 0$, taking $M = 1/\varepsilon$ gives $\norme{x} \ge M \Rightarrow 0 < f(x) \le \frac{1}{1+M} < \varepsilon$. This is the first extension of the previous Definition.

Method — Show that a map admits a given limit

To prove $\lim_a f = b$ directly from the definition:

fix $\varepsilon > 0$ and aim to produce $\alpha > 0$;
bound $\norme{f(x)-b}$ above by an explicit increasing function of $\norme{x-a}$ --- typically of the form $C\,\norme{x-a}$, obtained through the triangle inequality;
read off $\alpha$ from that bound, so that $\norme{x-a} \le \alpha$ forces $\norme{f(x)-b} < \varepsilon$.

The sequential characterisation of the next subsection often replaces this work; the direct route is the fallback when no rule applies.

Skills to practice

Proving a limit from the definition

I.2 Sequential characterisation and limits in a product

The next result transports the limit of a map onto sequences: $f$ tends to $b$ at $a$ exactly when $f$ carries every sequence converging to $a$ to a sequence converging to $b$. It is the bridge that makes the whole theory of convergent sequences of Normed vector spaces available for maps --- and it is the standard tool both to compute a limit and to prove that none exists.

Proposition — Sequential characterisation of the limit

Let $f \colon A \to F$, $a \in \overline{A}$ and $b \in F$. Then $\displaystyle\lim_{a} f = b$ if and only if, for every sequence $(u_n)$ of elements of $A$ with $u_n \to a$, the sequence $\bigl(f(u_n)\bigr)$ converges to $b$.

Proof

$(\Rightarrow)$ Suppose $\lim_a f = b$, and let $(u_n)$ be a sequence of $A$ with $u_n \to a$. Fix $\varepsilon > 0$; the limit gives $\alpha > 0$ with $\norme{x-a} \le \alpha \Rightarrow \norme{f(x)-b} < \varepsilon$ for $x \in A$. Since $u_n \to a$, there is a rank $N$ from which $\norme{u_n-a} \le \alpha$; then $\norme{f(u_n)-b} < \varepsilon$ for $n \ge N$. So $f(u_n) \to b$.
$(\Leftarrow)$ We prove the contrapositive: assume $\lim_a f = b$ fails, and build a sequence breaking the right-hand side. The failure of the definition reads $$ \exists \varepsilon > 0,\ \forall \alpha > 0,\ \exists x \in A,\quad \norme{x-a} \le \alpha \ \text{ and }\ \norme{f(x)-b} \ge \varepsilon. $$ Apply this with $\alpha = \tfrac{1}{n+1}$: for each $n$ there is $u_n \in A$ with $\norme{u_n-a} \le \tfrac{1}{n+1}$ and $\norme{f(u_n)-b} \ge \varepsilon$. Then $u_n \to a$, yet $\bigl(f(u_n)\bigr)$ does not converge to $b$ --- so the right-hand side fails too.

The characterisation reads off a picture: $f$ carries a sequence closing in on $a$ to a sequence closing in on $b$.

Proposition — Limit of a map valued in a finite product

Let $F_1,\dots,F_p$ be normed spaces, the product $F_1 \times \dots \times F_p$ carrying the product norm $\norme{(y_1,\dots,y_p)} = \max_{k} \norme{y_k}$. Let $f \colon A \to F_1 \times \dots \times F_p$ have components $f = (f_1,\dots,f_p)$, and let $a \in \overline{A}$. Then $f$ admits a limit at $a$ if and only if each component $f_k$ admits a limit at $a$, and in that case $$ \lim_{a} f = \Bigl(\lim_{a} f_1,\ \dots,\ \lim_{a} f_p\Bigr). $$

Proof

By the sequential characterisation, $f$ admits the limit $b = (b_1,\dots,b_p)$ at $a$ exactly when $f(u_n) \to b$ for every sequence $(u_n)$ of $A$ with $u_n \to a$. Now $f(u_n) = \bigl(f_1(u_n),\dots,f_p(u_n)\bigr)$, and a sequence of a finite product converges to $(b_1,\dots,b_p)$ for the product norm if and only if each coordinate sequence $\bigl(f_k(u_n)\bigr)$ converges to $b_k$ --- this is the convergence in a finite product recalled from Normed vector spaces. So $f(u_n) \to b$ for every such $(u_n)$ if and only if $f_k(u_n) \to b_k$ for every such $(u_n)$ and every $k$, that is --- again by the sequential characterisation --- if and only if each $f_k$ admits the limit $b_k$ at $a$.

Example — Disproving a limit with two sequences

Let $f \colon \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}$, $f(x,y) = \dfrac{xy}{x^2+y^2}$, and $a = (0,0) \in \overline{\mathbb{R}^2 \setminus \{(0,0)\}}$. Along the sequence $u_n = \bigl(\tfrac1n,0\bigr) \to a$ one has $f(u_n) = 0 \to 0$; along $v_n = \bigl(\tfrac1n,\tfrac1n\bigr) \to a$ one has $f(v_n) = \tfrac12 \to \tfrac12$. Two sequences converging to $a$ produce images with different limits, so by the sequential characterisation $f$ admits no limit at $(0,0)$.

Example — A map valued in a product

The map $f \colon \mathbb{R} \to \mathbb{R}^2$, $f(t) = (t^2,\ 3t-1)$, has components $f_1(t) = t^2$ and $f_2(t) = 3t-1$. Each has a limit at $t = 2$, namely $4$ and $5$ --- elementary one-variable limits, recalled from earlier study --- so $f$ has a limit at $2$ and $\lim_{t \to 2} f(t) = (4,5)$. A limit in a product is computed coordinate by coordinate.

Method — Prove or disprove a limit with sequences

To prove $\lim_a f = b$, show $f(u_n) \to b$ for an arbitrary sequence $(u_n)$ of $A$ with $u_n \to a$ --- a single chosen sequence never proves a limit, it only suggests a candidate.
To disprove a proposed limit, exhibit two sequences $u_n \to a$ and $v_n \to a$ whose images $\bigl(f(u_n)\bigr)$ and $\bigl(f(v_n)\bigr)$ have different limits --- or one sequence $u_n \to a$ for which $\bigl(f(u_n)\bigr)$ diverges.

Skills to practice

Applying the sequential characterisation

I.3 Operations on limits

The sequential characterisation lets every rule for limits of sequences be transferred, without a single $\varepsilon$, to limits of maps. A linear combination, a product by a scalar-valued map, a quotient and a composite are handled in turn --- and from then on a limit is computed by decomposition, never returning to the definition.

Proposition — Algebraic operations on limits

Let $a \in \overline{A}$, let $f,g \colon A \to F$ and $\varphi \colon A \to \mathbb{K}$, and suppose $\lim_a f$, $\lim_a g$ and $\lim_a \varphi$ all exist.

For all $\lambda,\mu \in \mathbb{K}$, $\lim_a (\lambda f + \mu g) = \lambda \lim_a f + \mu \lim_a g$.
$\lim_a (\varphi f) = \bigl(\lim_a \varphi\bigr)\bigl(\lim_a f\bigr)$.
If $\lim_a \varphi \ne 0$, then $\varphi$ is non-zero on $A \cap B(a,\alpha)$ for some $\alpha > 0$, and on that domain $\lim_a \tfrac{f}{\varphi} = \tfrac{\lim_a f}{\lim_a \varphi}$.

Proof

Write $b = \lim_a f$, $c = \lim_a g$, $\ell = \lim_a \varphi$. Let $(u_n)$ be an arbitrary sequence of $A$ with $u_n \to a$. By the sequential characterisation, $f(u_n) \to b$, $g(u_n) \to c$ and $\varphi(u_n) \to \ell$.

Linear combination. By the operations on convergent sequences (recalled from Normed vector spaces), $\lambda f(u_n) + \mu g(u_n) \to \lambda b + \mu c$. As this holds for every such $(u_n)$, the sequential characterisation gives $\lim_a (\lambda f + \mu g) = \lambda b + \mu c$.
Product by a scalar-valued map. Likewise the product of the convergent scalar sequence $\bigl(\varphi(u_n)\bigr)$ by the convergent sequence $\bigl(f(u_n)\bigr)$ converges: $\varphi(u_n) f(u_n) \to \ell\,b$, hence $\lim_a (\varphi f) = \ell\,b$.
Quotient. Suppose $\ell \ne 0$. Applying the limit of $\varphi$ with $\varepsilon = \tfrac{|\ell|}{2}$ gives $\alpha > 0$ such that $|\varphi(x)-\ell| < \tfrac{|\ell|}{2}$ for $x \in A \cap B(a,\alpha)$; then $|\varphi(x)| \ge |\ell| - \tfrac{|\ell|}{2} = \tfrac{|\ell|}{2} > 0$, so $\varphi$ does not vanish on $A \cap B(a,\alpha)$, and $a$ remains adherent to that set. On it, for $u_n \to a$ the quotient of convergent sequences gives $\tfrac{f(u_n)}{\varphi(u_n)} \to \tfrac{b}{\ell}$, hence $\lim_a \tfrac{f}{\varphi} = \tfrac{b}{\ell}$.

Proposition — Limit of a composite

Let $f \colon A \to F$ with $f(A) \subset B$, where $B$ is a subset of $F$, and let $g \colon B \to G$. Let $a \in \overline{A}$. If $f$ admits a limit $b$ at $a$, then $b \in \overline{B}$ automatically. If, moreover, $g$ admits a limit $\ell$ at $b$, then $g \circ f$ admits the limit $\ell$ at $a$.

Proof

Since $a \in \overline{A}$, there is a sequence $(u_n)$ of $A$ with $u_n \to a$; then $f(u_n) \to b$ by the sequential characterisation, and $\bigl(f(u_n)\bigr)$ is a sequence of $f(A) \subset B$, so $b$ is a limit of a sequence of $B$, hence $b \in \overline{B}$ --- the limit $\lim_b g$ is therefore meaningful. Now take any sequence $(u_n)$ of $A$ with $u_n \to a$. Then $f(u_n) \to b$, and $\bigl(f(u_n)\bigr)$ is a sequence of $B$ converging to $b$; by the sequential characterisation applied to $g$, $g\bigl(f(u_n)\bigr) \to \ell$. As this holds for every such $(u_n)$, the sequential characterisation gives $\lim_a (g \circ f) = \ell$.

Example — A limit computed by operations

Compute $\displaystyle\lim_{(x,y) \to (1,2)} \frac{x^2 + y}{x + y}$ for the map defined on $A = \{(x,y) \in \mathbb{R}^2 : x + y \ne 0\}$, with $\mathbb{R}^2$ normed by $\norme{\cdot}_\infty$.

Answer

The point $a = (1,2)$ is adherent to $A$ (it lies in $A$, since $1 + 2 = 3 \ne 0$). The two coordinate maps $(x,y) \mapsto x$ and $(x,y) \mapsto y$ have limits $1$ and $2$ at $a$ --- each is one bound away, $|x-1| \le \norme{(x,y)-a}_\infty$ and likewise $|y-2| \le \norme{(x,y)-a}_\infty$. By the algebraic operations:

the numerator $x^2 + y = x \cdot x + y$ has limit $1 \cdot 1 + 2 = 3$ (product, then sum);
the denominator $x + y$ has limit $1 + 2 = 3 \ne 0$;
the quotient therefore has limit $\dfrac{3}{3} = 1$.

Hence $\displaystyle\lim_{(x,y) \to (1,2)} \frac{x^2 + y}{x + y} = 1$.

Example — A limit of a composite

Let $f \colon \mathbb{R} \to \mathbb{R}$, $f(t) = 1 + t^2$, and $g \colon [1,+\infty[ \to \mathbb{R}$, $g(s) = \sqrt{s}$, with $f(\mathbb{R}) \subset [1,+\infty[$. At $a = 0$, $f$ tends to $b = 1$; since $1 \in \overline{[1,+\infty[}$ and $g$ tends to $\ell = 1$ at $1$, the composite $g \circ f \colon t \mapsto \sqrt{1+t^2}$ tends to $1$ at $0$. A limit of a composite is read off the two limits in turn.

Method — Compute a limit using the operations

To compute $\lim_a f$ when $f$ is built from simpler maps:

decompose $f$ into sums, scalar products, quotients and composites of maps whose limits are known --- typically the coordinate maps and elementary one-variable functions;
apply the algebraic-operations and composite Propositions, checking, before any quotient, that the denominator's limit is non-zero;
fall back on the direct $\varepsilon$-$\alpha$ definition only for the atomic pieces.

Skills to practice

Computing a limit by operations

II Continuity

II.1 Continuity at a point and on a set

Continuity at a point $a$ is the limit notion with no surprise: $f$ is continuous at $a$ when its limit at $a$ exists and equals the value $f(a)$. The map $f$ is then continuous on its whole domain when it is continuous at every point. The operations on limits carry over at once, and a new result appears --- two continuous maps that agree on a dense subset agree everywhere.

Definition — Continuity at a point and on a set

Let $f \colon A \to F$ and $a \in A$.

$f$ is continuous at $a$ if $\displaystyle\lim_{a} f = f(a)$, that is $$ \forall \varepsilon > 0,\ \exists \alpha > 0,\ \forall x \in A,\quad \norme{x-a} \le \alpha \implies \norme{f(x)-f(a)} < \varepsilon. $$
$f$ is continuous on $A$ if it is continuous at every point of $A$. The set of continuous maps from $A$ to $F$ is written $\mathcal{C}(A,F)$.

Proposition — Sequential characterisation of continuity

Let $f \colon A \to F$ and $a \in A$. Then $f$ is continuous at $a$ if and only if, for every sequence $(u_n)$ of $A$ with $u_n \to a$, one has $f(u_n) \to f(a)$.

Proof

Continuity at $a$ is the equality $\lim_a f = f(a)$. Apply the sequential characterisation of the limit with $b = f(a)$: $\lim_a f = f(a)$ holds if and only if $f(u_n) \to f(a)$ for every sequence $(u_n)$ of $A$ with $u_n \to a$. That is exactly the stated equivalence.

Continuity at $a$ reads on the same kind of picture as the limit, now with the target point pinned to $f(a)$.

Proposition — Operations on continuous maps

Let $a \in A$.

A linear combination $\lambda f + \mu g$ of two maps $f,g \colon A \to F$ continuous at $a$ is continuous at $a$; so $\mathcal{C}(A,F)$ is a $\mathbb{K}$-vector space.
The product $\varphi f$ of a map $f \colon A \to F$ continuous at $a$ by a scalar-valued map $\varphi \colon A \to \mathbb{K}$ continuous at $a$ is continuous at $a$; and the product of two scalar-valued continuous maps is continuous, so $\mathcal{C}(A,\mathbb{K})$ is a $\mathbb{K}$-algebra.
If $f \colon A \to F$ and $\varphi \colon A \to \mathbb{K}$ are both continuous at $a$ and $\varphi(a) \ne 0$, then $\tfrac{f}{\varphi}$ is defined near $a$ and continuous at $a$.
If $f \colon A \to F$ is continuous at $a$ with $f(A) \subset B$, and $g \colon B \to G$ is continuous at $f(a)$, then $g \circ f$ is continuous at $a$.

Proof

Continuity at $a$ is the equality $\lim_a = (\text{value at }a)$, so each statement is the corresponding rule of \S1.3 read with the limits equal to the values.

For the linear combination: $\lim_a f = f(a)$ and $\lim_a g = g(a)$ give, by the algebraic-operations Proposition, $\lim_a (\lambda f + \mu g) = \lambda f(a) + \mu g(a) = (\lambda f + \mu g)(a)$.
For the products: $\lim_a (\varphi f) = \varphi(a) f(a) = (\varphi f)(a)$, and the scalar product is the case $F = \mathbb{K}$.
For the quotient: $f$ and $\varphi$ continuous at $a$ give $\lim_a f = f(a)$ and $\lim_a \varphi = \varphi(a) \ne 0$; the quotient rule of \S1.3 then makes $\varphi$ non-zero on some $A \cap B(a,\alpha)$ and gives $\lim_a \tfrac{f}{\varphi} = \tfrac{f(a)}{\varphi(a)}$.
For the composite: $f$ continuous at $a$ and $g$ continuous at $f(a)$ are $\lim_a f = f(a)$ and $\lim_{f(a)} g = g(f(a))$; the composite Proposition of \S1.3 gives $\lim_a (g \circ f) = g(f(a)) = (g \circ f)(a)$.

Proposition — Coincidence on a dense subset

Let $f,g \colon A \to F$ be two maps continuous on $A$, and let $D$ be a subset dense in $A$. If $f$ and $g$ coincide on $D$, then $f = g$ on the whole of $A$.

Proof

Let $x \in A$. Since $D$ is dense in $A$, the sequential form of density (recalled from Topology of a normed space) gives a sequence $(u_n)$ of elements of $D$ with $u_n \to x$. As $f$ and $g$ are continuous at $x$, the sequential characterisation of continuity gives $f(u_n) \to f(x)$ and $g(u_n) \to g(x)$. But $u_n \in D$, where $f$ and $g$ coincide, so $f(u_n) = g(u_n)$ for every $n$; the two sequences $\bigl(f(u_n)\bigr)$ and $\bigl(g(u_n)\bigr)$ are equal, hence have the same limit, and by uniqueness of the limit $f(x) = g(x)$. As $x \in A$ was arbitrary, $f = g$ on $A$.

Example — The coordinate projections are continuous

On a finite product $F_1 \times \dots \times F_p$ with the product norm, each coordinate projection $\pi_k \colon (y_1,\dots,y_p) \mapsto y_k$ is continuous: a sequence converging in the product converges in each coordinate (the finite-product Proposition of \S1.2), which is exactly $\pi_k(u_n) \to \pi_k(\ell)$. The continuity of the norm $x \mapsto \norme{x}$ is treated separately in \S2.3, where it is shown to be $1$-Lipschitz.

Example — A continuity check by operations

Show that $f \colon \mathbb{R}^2 \to \mathbb{R}$, $f(x,y) = \dfrac{x^2 - y}{1 + x^2 + y^2}$, is continuous on $\mathbb{R}^2$ (normed by $\norme{\cdot}_\infty$).

Answer

Fix any point $a = (x_0,y_0) \in \mathbb{R}^2$ and argue by operations.

The coordinate projections $(x,y) \mapsto x$ and $(x,y) \mapsto y$ are continuous at $a$ (previous Example).
By the operations Proposition, the numerator $x^2 - y$ --- a product and a difference of continuous maps --- is continuous at $a$, and the denominator $1 + x^2 + y^2$ likewise.
The denominator never vanishes: $1 + x^2 + y^2 \ge 1 > 0$. So the quotient $f$ is continuous at $a$.

Since $a$ was arbitrary, $f$ is continuous on $\mathbb{R}^2$.

Example — A continuous map vanishing on the rationals

A map $f \colon \mathbb{R} \to F$ continuous on $\mathbb{R}$ that vanishes at every rational point is identically zero. Indeed $\mathbb{Q}$ is dense in $\mathbb{R}$ (recalled from Topology of a normed space); $f$ and the zero map are both continuous and coincide on $\mathbb{Q}$, so they coincide on all of $\mathbb{R}$. A continuous map is pinned down by its values on a dense subset.

Method — Show that a map is continuous

Three routes establish that $f$ is continuous at a point $a$ --- or on $A$:

by operations --- decompose $f$ into sums, products, quotients and composites of maps already known continuous (coordinate projections, elementary functions), checking non-vanishing denominators;
by the sequential characterisation --- take an arbitrary $u_n \to a$ and show $f(u_n) \to f(a)$;
by the direct $\varepsilon$-$\alpha$ argument --- the fallback for an atomic map, bounding $\norme{f(x)-f(a)}$ by an explicit function of $\norme{x-a}$.

Skills to practice

Proving a map continuous

II.2 Topological characterisation of continuity

Continuity can be expressed with no $\varepsilon$ at all, through the open and closed sets of Topology of a normed space: a map is continuous exactly when the preimage of every open set is open, and the preimage of every closed set is closed --- understood, for a map defined on a subset $A$, with the relative open and closed sets of $A$.

Theorem — Topological characterisation of continuity

Let $f \colon A \to F$ with $A \subset E$. The following are equivalent:

[(i)] $f$ is continuous on $A$;
[(ii)] for every open set $O$ of $F$, the preimage $f^{-1}(O)$ is a relative open set of $A$;
[(iii)] for every closed set $C$ of $F$, the preimage $f^{-1}(C)$ is a relative closed set of $A$.

Proof

(i) $\Rightarrow$ (ii). Let $O$ be open in $F$ and $x \in f^{-1}(O)$, so $f(x) \in O$. As $O$ is open there is $\varepsilon > 0$ with $B(f(x),\varepsilon) \subset O$. By continuity of $f$ at $x$ there is $\alpha > 0$ with: $y \in A$ and $\norme{y-x} \le \alpha$ imply $\norme{f(y)-f(x)} < \varepsilon$, i.e. $f(y) \in B(f(x),\varepsilon) \subset O$, i.e. $y \in f^{-1}(O)$. So $A \cap B(x,\alpha) \subset f^{-1}(O)$. As this holds at every $x \in f^{-1}(O)$, the characterisation of relative open sets (from Topology of a normed space) makes $f^{-1}(O)$ a relative open set of $A$.
(ii) $\Rightarrow$ (i). Let $x \in A$ and $\varepsilon > 0$. The ball $B(f(x),\varepsilon)$ is open in $F$, so by (ii) $f^{-1}\bigl(B(f(x),\varepsilon)\bigr)$ is a relative open set of $A$ containing $x$; the characterisation of relative open sets gives $\alpha > 0$ with $A \cap B(x,\alpha) \subset f^{-1}\bigl(B(f(x),\varepsilon)\bigr)$. Then $y \in A$, $\norme{y-x} \le \alpha/2$ imply $\norme{f(y)-f(x)} < \varepsilon$: $f$ is continuous at $x$.
(ii) $\iff$ (iii). For any subset $X$ of $F$, $f^{-1}(F \setminus X) = A \setminus f^{-1}(X)$, the complement on the preimage side being taken inside the domain $A$ (since $f^{-1}(X) \subset A$). A subset $C$ of $F$ is closed exactly when $F \setminus C$ is open; and $f^{-1}(C)$ is a relative closed set of $A$ exactly when $A \setminus f^{-1}(C) = f^{-1}(F \setminus C)$ is a relative open set of $A$. So (iii) holds for every closed $C$ if and only if (ii) holds for every open $F \setminus C$ --- that is, (ii) and (iii) are equivalent.

Proposition — The case of a map defined on the whole space

Let $f \colon E \to F$ be continuous on $E$. Then the preimage of every open set of $F$ is an open set of $E$, and the preimage of every closed set of $F$ is a closed set of $E$.

Proof

Apply the Theorem with $A = E$. A relative open set of $E$ is a set of the form $E \cap O = O$ with $O$ open in $E$ --- that is, simply an open set of $E$; likewise a relative closed set of $E$ is a closed set of $E$. So the relative statements (ii) and (iii) become: the preimage of an open set is open, the preimage of a closed set is closed.

For a continuous real-valued map $g \colon E \to \mathbb{R}$, the preimage of an open interval is an open set of $E$ --- the practical way to recognise an open set described by a strict inequality.

Example — Open and closed sets cut out by a continuous map

Let $g \colon E \to \mathbb{R}$ be continuous on $E$. The set $\{x \in E : g(x) > 0\} = g^{-1}(\,]0,+\infty[\,)$ is open, as the preimage of the open set $]0,+\infty[$. The set $\{x \in E : g(x) = 0\} = g^{-1}(\{0\})$ is closed, as the preimage of the closed set $\{0\}$, and likewise $\{g(x) \ge 0\} = g^{-1}([0,+\infty[)$ is closed.

Example — An open half-plane

Show that the half-plane $H = \{(x,y) \in \mathbb{R}^2 : 2x - y > 1\}$ is an open set of $\mathbb{R}^2$.

Answer

Introduce $g \colon \mathbb{R}^2 \to \mathbb{R}$, $g(x,y) = 2x - y - 1$. The map $g$ is continuous on $\mathbb{R}^2$ --- a linear combination of the two continuous coordinate projections, minus a constant. Now $$ H = \{(x,y) : 2x - y > 1\} = \{(x,y) : g(x,y) > 0\} = g^{-1}(\,]0,+\infty[\,). $$ The interval $]0,+\infty[$ is an open set of $\mathbb{R}$, and $g$ is continuous on the whole space $\mathbb{R}^2$, so by the Proposition above $H = g^{-1}(\,]0,+\infty[\,)$ is an open set of $\mathbb{R}^2$.

Example — The image of a closed set need not be closed

The Theorem concerns preimages, not images. The direct image of a closed set by a continuous map need not be closed: the map $\arctan \colon \mathbb{R} \to \mathbb{R}$ is continuous, the set $[0,+\infty[$ is closed in $\mathbb{R}$, yet its image $\arctan([0,+\infty[) = [0,\tfrac{\pi}{2}[$ is not closed. Continuity controls preimages, not images.

Method — Show that a set is open or closed by a preimage

To prove that a subset $S$ of $E$ is open (resp. closed):

write $S$ as $\{x : g(x) \in X\} = g^{-1}(X)$ for a well-chosen continuous map $g \colon E \to F$ and a subset $X$ of $F$;
recognise $X$ as an open set of $F$ (resp. a closed set) --- a strict inequality points to an open $X$, a non-strict inequality or an equality to a closed $X$;
conclude by the Proposition: $g$ continuous on $E$ sends the open (resp. closed) $X$ to an open (resp. closed) preimage.

Skills to practice

Identifying open and closed sets by preimage

II.3 Lipschitz and uniformly continuous maps

Continuity at a point lets the threshold $\alpha$ depend on the point. Two stronger grades demand more: uniform continuity asks one $\alpha$ valid at every point at once, and the Lipschitz property asks a controlled rate $\norme{f(x)-f(y)} \le k\,\norme{x-y}$. This subsection also introduces the distance $d(x,A)$ from a point to a subset --- a map that turns out to be Lipschitz.

Definition — Lipschitz map

Let $f \colon A \to F$ with $A \subset E$, and let $k \ge 0$ be a real number. The map $f$ is $k$-Lipschitz if $$ \forall x,y \in A,\quad \norme{f(x)-f(y)} \le k\,\norme{x-y}. $$ It is Lipschitz if it is $k$-Lipschitz for some $k \ge 0$.

Definition — Uniformly continuous map

A map $f \colon A \to F$ is uniformly continuous on $A$ if $$ \forall \varepsilon > 0,\ \exists \alpha > 0,\ \forall x,y \in A,\quad \norme{x-y} \le \alpha \implies \norme{f(x)-f(y)} < \varepsilon. $$ The same $\alpha$ serves all pairs $x,y$ of the domain; it depends on $\varepsilon$ alone. The universal quantifier over $x$ and $y$ is what separates uniform continuity from ordinary continuity, where $\alpha$ may vary from point to point.

Proposition — The Lipschitz$\virgule$ uniform and continuous hierarchy

Let $f \colon A \to F$. Then $f$ Lipschitz $\implies$ $f$ uniformly continuous $\implies$ $f$ continuous on $A$.

Proof

Lipschitz $\Rightarrow$ uniformly continuous. Suppose $f$ is $k$-Lipschitz. Fix $\varepsilon > 0$. If $k = 0$, then $\norme{f(x)-f(y)} \le 0$ for all $x,y$, so $f$ is constant and any $\alpha > 0$ works. If $k > 0$, set $\alpha = \varepsilon/(2k) > 0$; then $\norme{x-y} \le \alpha$ gives $\norme{f(x)-f(y)} \le k\,\norme{x-y} \le k\alpha = \varepsilon/2 < \varepsilon$. In both cases $f$ is uniformly continuous.
Uniformly continuous $\Rightarrow$ continuous. Suppose $f$ is uniformly continuous, and fix $a \in A$ and $\varepsilon > 0$. The uniform-continuity $\alpha$, read with the second variable frozen at $y = a$, gives: $x \in A$, $\norme{x-a} \le \alpha$ imply $\norme{f(x)-f(a)} < \varepsilon$. That is continuity of $f$ at $a$; as $a$ was arbitrary, $f$ is continuous on $A$.

Definition — Distance from a point to a subset

Let $A$ be a non-empty subset of $E$ and $x \in E$. The distance from $x$ to $A$ is $$ d(x,A) = \inf_{a \in A} \norme{x-a}. $$ The infimum exists: $\{\norme{x-a} : a \in A\}$ is a non-empty subset of $\mathbb{R}$ bounded below by $0$.

Proposition — The norm and the distance to a subset are $1$-Lipschitz

The norm $x \mapsto \norme{x}$ on $E$ is $1$-Lipschitz. For any non-empty subset $A$ of $E$, the map $x \mapsto d(x,A)$ is $1$-Lipschitz. Both are therefore continuous on $E$.

Proof

The norm. The reverse triangle inequality (recalled from Normed vector spaces) gives $\bigl|\,\norme{x}-\norme{y}\,\bigr| \le \norme{x-y}$ for all $x,y$, which is exactly the $1$-Lipschitz inequality for $x \mapsto \norme{x}$.
The distance to $A$. Fix $x,y \in E$. For every $a \in A$, the triangle inequality gives $$ \begin{aligned} d(x,A) &\le \norme{x-a} && \text{(infimum is a lower bound)}\\ &\le \norme{x-y} + \norme{y-a}. && \text{(triangle inequality)} \end{aligned} $$ So $d(x,A) - \norme{x-y} \le \norme{y-a}$ for every $a \in A$; the left side is a lower bound of $\{\norme{y-a} : a \in A\}$, hence is $\le$ its infimum: $d(x,A) - \norme{x-y} \le d(y,A)$, that is $d(x,A) - d(y,A) \le \norme{x-y}$. Exchanging $x$ and $y$ gives $d(y,A) - d(x,A) \le \norme{x-y}$, so $\bigl|\,d(x,A)-d(y,A)\,\bigr| \le \norme{x-y}$.

In both cases the map is $1$-Lipschitz, hence continuous by the hierarchy.

Proposition — The distance to a subset detects its closure

Let $A$ be a non-empty subset of $E$ and $x \in E$. Then $d(x,A) = 0$ if and only if $x \in \overline{A}$.

Proof

By the characterisation of the infimum, $d(x,A) = 0$ means: for every $r > 0$ there is $a \in A$ with $\norme{x-a} < r$, that is $a \in B(x,r) \cap A$. So $d(x,A) = 0$ holds exactly when every open ball $B(x,r)$ meets $A$ --- which is precisely the definition of an adherent point of $A$, recalled from Topology of a normed space. Hence $d(x,A) = 0 \iff x \in \overline{A}$.

The $1$-Lipschitz inequality $\bigl|\,d(x,A)-d(y,A)\,\bigr| \le \norme{x-y}$ says the distance to $A$ changes no faster than the displacement.

Example — The hierarchy is strict

Neither implication of the hierarchy reverses.

The map $x \mapsto \sqrt{x}$ on $\mathbb{R}_+$ is uniformly continuous but not Lipschitz. Uniform continuity follows from the classical inequality $|\sqrt{x}-\sqrt{y}| \le \sqrt{|x-y|}$ (valid for $x,y \ge 0$): given $\varepsilon > 0$, any $\alpha$ with $0 < \alpha < \varepsilon^2$ gives $|x-y| \le \alpha \Rightarrow |\sqrt{x}-\sqrt{y}| \le \sqrt{\alpha} < \varepsilon$. It is not Lipschitz: for $x_n = \tfrac1n$ and $y_n = 0$, the rate $\tfrac{|\sqrt{x_n}-\sqrt{y_n}|}{|x_n-y_n|} = \sqrt{n} \to +\infty$, so no constant $k$ can bound it.
The map $x \mapsto \ln x$ on $]0,+\infty[$ is continuous but not uniformly continuous. Suppose, for contradiction, it were uniformly continuous: applied with $\varepsilon = \ln 2$, that would give $\alpha > 0$ such that $|x-y| \le \alpha \Rightarrow |\ln x - \ln y| < \ln 2$ for all $x,y > 0$. But for $n$ large enough that $\tfrac1n \le \alpha$, the points $x_n = \tfrac1n$ and $y_n = \tfrac2n$ satisfy $|x_n-y_n| = \tfrac1n \le \alpha$ while $|\ln x_n - \ln y_n| = \bigl|\ln\tfrac12\bigr| = \ln 2$ --- contradicting the strict inequality. So no $\alpha$ works, and $\ln$ is not uniformly continuous.

Lipschitz is strictly stronger than uniformly continuous, which is strictly stronger than continuous.

Example — A distance to a subset computed

In $\mathbb{R}$ with the absolute value, compute $d(x,A)$ for $A = [0,1]$ and an arbitrary $x \in \mathbb{R}$.

Answer

By definition $d(x,A) = \inf_{a \in [0,1]} |x-a|$. Three cases, by the position of $x$ relative to $[0,1]$:

if $x < 0$, then $|x-a| = a - x$ is smallest at $a = 0$, so $d(x,A) = -x = |x|$;
if $0 \le x \le 1$, then $x \in A$ and $a = x$ gives $|x-a| = 0$, so $d(x,A) = 0$;
if $x > 1$, then $|x-a| = x - a$ is smallest at $a = 1$, so $d(x,A) = x - 1$.

In short $d(x,[0,1]) = \max(0,\ -x,\ x-1)$. Consistently with the previous Proposition, $d(x,A) = 0$ exactly on $[0,1] = \overline{[0,1]}$.

Method — Show that a map is Lipschitz or uniformly continuous

Lipschitz --- bound $\norme{f(x)-f(y)}$ above by $k\,\norme{x-y}$ with a constant $k \ge 0$ independent of $x,y$; the triangle inequality and the reverse triangle inequality are the usual tools.
Uniformly continuous --- exhibit, for each $\varepsilon > 0$, one $\alpha > 0$ valid for all pairs $x,y$; the quickest route is to prove $f$ Lipschitz, since Lipschitz implies uniformly continuous.
to show a map is not uniformly continuous, exhibit two sequences $(x_n)$, $(y_n)$ with $\norme{x_n-y_n} \to 0$ but $\norme{f(x_n)-f(y_n)}$ bounded away from $0$.

Skills to practice

Verifying the Lipschitz and uniform-continuity properties

III Continuous linear and multilinear maps

III.1 Continuity criterion for a linear map

Linear maps enjoy a continuity test far simpler than the general $\varepsilon$-$\alpha$ definition. By linearity, continuity everywhere reduces to continuity at $0_E$, and that in turn is captured by a single inequality $\norme{u(x)} \le C\,\norme{x}$. When $E$ is finite-dimensional the inequality is automatic --- every linear map is then continuous --- as the later chapter Compactness, connectedness, finite dimension establishes; here $E$ is arbitrary, so the criterion has genuine content.

Theorem — Continuity criterion for a linear map

Let $u \colon E \to F$ be a linear map. Then $u$ is continuous on $E$ if and only if there exists a real $C \ge 0$ such that $$ \forall x \in E,\quad \norme{u(x)} \le C\,\norme{x}. $$

Proof

$(\Leftarrow)$ Suppose $\norme{u(x)} \le C\,\norme{x}$ for all $x$. For any $x_0 \in E$ and any $x \in E$, linearity gives $u(x) - u(x_0) = u(x - x_0)$, hence $$ \norme{u(x) - u(x_0)} = \norme{u(x-x_0)} \le C\,\norme{x - x_0}. $$ So $u$ is $C$-Lipschitz, therefore continuous on $E$ by the hierarchy of \S2.3.
$(\Rightarrow)$ Suppose $u$ is continuous; in particular it is continuous at $0_E$, where $u(0_E) = 0_F$ by linearity. Applying the definition with $\varepsilon = 1$, there is $\alpha > 0$ such that $\norme{x} \le \alpha \Rightarrow \norme{u(x)} \le 1$. Let $x \in E$ with $x \ne 0_E$. The vector $\dfrac{\alpha x}{\norme{x}}$ has norm $\alpha$, so $\norme{u\!\left(\dfrac{\alpha x}{\norme{x}}\right)} \le 1$; by linearity and homogeneity, $$ \begin{aligned} \norme{u\!\left(\tfrac{\alpha x}{\norme{x}}\right)} &= \tfrac{\alpha}{\norme{x}}\,\norme{u(x)} && \text{(linearity and homogeneity of the norm)}\\ &\le 1, && \end{aligned} $$ hence $\norme{u(x)} \le \tfrac{1}{\alpha}\,\norme{x}$. This inequality also holds at $x = 0_E$ (both sides are $0$). So $C = \tfrac{1}{\alpha}$ works.

Proposition — Equivalent forms of continuity for a linear map

For a linear map $u \colon E \to F$, the following are equivalent:

[(i)] $u$ is continuous on $E$;
[(ii)] $u$ is continuous at $0_E$;
[(iii)] there is $C \ge 0$ with $\norme{u(x)} \le C\,\norme{x}$ for all $x$;
[(iv)] $u$ is bounded on the closed unit ball $B_f(0_E,1)$;
[(v)] $u$ is Lipschitz.

Proof

The Theorem already gives (i) $\iff$ (iii), and its proof shows (iii) $\Rightarrow$ (v) ($u$ is $C$-Lipschitz) and (v) $\Rightarrow$ (i) (Lipschitz implies continuous). It remains to weave in (ii) and (iv).

(i) $\Rightarrow$ (ii) is immediate: continuity on $E$ includes continuity at $0_E$.
(ii) $\Rightarrow$ (iii): the proof of the Theorem used only continuity at $0_E$ to produce $C = 1/\alpha$.
(iii) $\Rightarrow$ (iv): if $\norme{u(x)} \le C\,\norme{x}$, then for $\norme{x} \le 1$ one has $\norme{u(x)} \le C$, so $u$ is bounded on $B_f(0_E,1)$.
(iv) $\Rightarrow$ (iii): suppose $\norme{u(x)} \le M$ for $\norme{x} \le 1$. For $x \ne 0_E$, the vector $x/\norme{x}$ is in $B_f(0_E,1)$, so $\norme{u(x/\norme{x})} \le M$, that is $\norme{u(x)} \le M\,\norme{x}$ by homogeneity; the inequality also holds at $0_E$.

The implications chain to make (i)--(v) all equivalent.

A continuous linear map keeps the image of the unit ball bounded; a discontinuous one sends it off to infinity along a sequence.

Example — A continuous linear form

On the space $\mathcal{C}([0,1],\mathbb{K})$ of continuous functions, normed by the uniform norm $\norme{f}_\infty = \sup_{t \in [0,1]} |f(t)|$ (a usual function-space norm of Normed vector spaces), consider the evaluation $u \colon f \mapsto f(1)$. Show that $u$ is a continuous linear form.

Answer

The map $u$ is linear: $(\lambda f + \mu g)(1) = \lambda f(1) + \mu g(1)$. For its continuity, bound $|u(f)|$ by the norm of $f$: for every $f \in \mathcal{C}([0,1],\mathbb{K})$, $$ |u(f)| = |f(1)| \le \sup_{t \in [0,1]} |f(t)| = \norme{f}_\infty. $$ So $\norme{u(f)} \le C\,\norme{f}_\infty$ with $C = 1$. By the continuity criterion, the linear form $u$ is continuous.

Example — A discontinuous linear form

The very same evaluation $u \colon f \mapsto f(1)$, taken now on $\mathcal{C}([0,1],\mathbb{K})$ with the integral norm $\norme{f}_1 = \int_0^1 |f(t)|\,\mathrm{d}t$ (also a usual function-space norm of Normed vector spaces), is not continuous. Consider $f_n(t) = t^n$: then $\norme{f_n}_1 = \int_0^1 t^n\,\mathrm{d}t = \tfrac{1}{n+1} \to 0$, whereas $u(f_n) = f_n(1) = 1$ for every $n$. No constant $C$ can satisfy $1 = |u(f_n)| \le C\,\norme{f_n}_1 = \tfrac{C}{n+1}$ for all $n$, so the criterion fails: $u$ is discontinuous. Continuity of a linear map genuinely depends on the chosen norm.

Method — Show that a linear map is continuous or discontinuous

For a linear map $u \colon E \to F$:

to prove $u$ continuous, find a constant $C \ge 0$ and establish $\norme{u(x)} \le C\,\norme{x}$ for every $x \in E$ --- usually by bounding $\norme{u(x)}$ through the triangle inequality and the definition of the chosen norm;
to prove $u$ discontinuous, exhibit a sequence $(x_n)$ of non-zero vectors with $\dfrac{\norme{u(x_n)}}{\norme{x_n}} \to +\infty$ --- then no constant $C$ can bound the ratio, and the criterion fails.

Skills to practice

Applying the linear continuity criterion

III.2 The operator norm

The continuity criterion attaches to a continuous linear map a whole set of admissible constants $C$. Their smallest element measures the map: it is the operator norm. It makes the space $\mathcal{L}_c(E,F)$ of continuous linear maps a normed space in its own right. Throughout this subsection the spaces serving as the domain of an operator norm --- $E$, and $F$ where it is a domain --- are assumed non-zero, so that each unit sphere is non-empty.

Definition — Operator norm

The set of continuous linear maps from $E$ to $F$ is written $\mathcal{L}_c(E,F)$. For $u \in \mathcal{L}_c(E,F)$, the operator norm (or subordinate norm) of $u$, written $\norme{u}_{\mathrm{op}}$, is $$ \norme{u}_{\mathrm{op}} = \sup_{x \ne 0_E} \frac{\norme{u(x)}}{\norme{x}} = \sup_{\norme{x} \le 1} \norme{u(x)} = \sup_{\norme{x} = 1} \norme{u(x)}. $$ The three suprema are equal and finite: finite because $u$ is continuous; equal because homogeneity gives $\norme{u(x)}/\norme{x} = \norme{u(x/\norme{x})}$, so each ratio is the value of $\norme{u(\cdot)}$ at the unit vector $x/\norme{x}$ --- whence the $\sup$ over $x \ne 0_E$ equals the $\sup$ over $\norme{x} = 1$ --- while for a point of the unit ball ($0 < \norme{x} \le 1$) one has $\norme{u(x)} \le \norme{u(x/\norme{x})}$, a value already counted in the sphere supremum, so the $\sup$ over $\norme{x} \le 1$ equals the $\sup$ over $\norme{x} = 1$ too --- an equality of suprema, with no claim that any is attained.

Proposition — The operator norm is a norm

The map $u \mapsto \norme{u}_{\mathrm{op}}$ is a norm on the vector space $\mathcal{L}_c(E,F)$. Moreover, for every $u \in \mathcal{L}_c(E,F)$, $$ \forall x \in E,\quad \norme{u(x)} \le \norme{u}_{\mathrm{op}}\,\norme{x}, $$ and for any real $C \ge 0$, $\norme{u}_{\mathrm{op}} \le C$ if and only if $\norme{u(x)} \le C\,\norme{x}$ for all $x$ --- so $\norme{u}_{\mathrm{op}}$ is exactly the smallest valid constant of the continuity criterion.

Proof

First, $\mathcal{L}_c(E,F)$ is a vector space: the operations on continuous maps make a linear combination of continuous linear maps continuous and linear.

The fundamental inequality and the least-constant property. For $x \ne 0_E$, $\norme{u(x)}/\norme{x} \le \norme{u}_{\mathrm{op}}$ by definition of the supremum, so $\norme{u(x)} \le \norme{u}_{\mathrm{op}}\,\norme{x}$; this also holds at $0_E$. And $\norme{u}_{\mathrm{op}} \le C$ means $C$ is an upper bound of all the ratios $\norme{u(x)}/\norme{x}$, which is exactly $\norme{u(x)} \le C\,\norme{x}$ for all $x \ne 0_E$ --- hence for all $x$. So $\norme{u}_{\mathrm{op}}$, being the supremum, is the least admissible $C$.
Separation. If $\norme{u}_{\mathrm{op}} = 0$, then $\norme{u(x)} \le 0$ for all $x$, so $u(x) = 0_F$ for all $x$, i.e. $u$ is the zero map. Conversely the zero map has operator norm $0$.
Homogeneity. For $\lambda \in \mathbb{K}$, $\norme{(\lambda u)(x)} = |\lambda|\,\norme{u(x)}$, so taking the supremum over $\norme{x} = 1$ gives $\norme{\lambda u}_{\mathrm{op}} = |\lambda|\,\norme{u}_{\mathrm{op}}$.
Triangle inequality. For $\norme{x} = 1$, $\norme{(u+v)(x)} \le \norme{u(x)} + \norme{v(x)} \le \norme{u}_{\mathrm{op}} + \norme{v}_{\mathrm{op}}$; the right side is an upper bound, so $\norme{u+v}_{\mathrm{op}} \le \norme{u}_{\mathrm{op}} + \norme{v}_{\mathrm{op}}$.

The three norm axioms hold, so $\norme{\cdot}_{\mathrm{op}}$ is a norm on $\mathcal{L}_c(E,F)$.

Proposition — Sub-multiplicativity of the operator norm

Let $u \in \mathcal{L}_c(E,F)$ and $v \in \mathcal{L}_c(F,G)$, with $E$ and $F$ non-zero. Then $v \circ u \in \mathcal{L}_c(E,G)$ and $$ \norme{v \circ u}_{\mathrm{op}} \le \norme{v}_{\mathrm{op}}\,\norme{u}_{\mathrm{op}}. $$ In particular, for the identity of a non-zero space, $\norme{\mathrm{id}_E}_{\mathrm{op}} = 1$.

Proof

The composite $v \circ u$ is linear, and continuous as a composite of continuous maps, so $v \circ u \in \mathcal{L}_c(E,G)$. For every $x \in E$, applying the fundamental inequality twice, $$ \begin{aligned} \norme{(v \circ u)(x)} &= \norme{v\bigl(u(x)\bigr)} && \\ &\le \norme{v}_{\mathrm{op}}\,\norme{u(x)} && \text{(fundamental inequality for }v\text{)}\\ &\le \norme{v}_{\mathrm{op}}\,\norme{u}_{\mathrm{op}}\,\norme{x}. && \text{(fundamental inequality for }u\text{)} \end{aligned} $$ So $\norme{v}_{\mathrm{op}}\,\norme{u}_{\mathrm{op}}$ is an admissible constant for $v \circ u$; being the least one, $\norme{v \circ u}_{\mathrm{op}} \le \norme{v}_{\mathrm{op}}\,\norme{u}_{\mathrm{op}}$. For the identity: $\norme{\mathrm{id}_E(x)} = \norme{x}$, so every ratio equals $1$ and $\norme{\mathrm{id}_E}_{\mathrm{op}} = \sup_{\norme{x}=1} \norme{x} = 1$ (the unit sphere is non-empty since $E \ne \{0_E\}$).

Definition — Subordinate norm of a matrix

Equip $\mathbb{K}^p$ and $\mathbb{K}^n$ each with one of the usual norms $\norme{\cdot}_1$, $\norme{\cdot}_2$, $\norme{\cdot}_\infty$ of Normed vector spaces. For a matrix $A \in \mathcal{M}_{n,p}(\mathbb{K})$, the subordinate norm $\norme{A}_{\mathrm{op}}$ is the operator norm of the linear map $X \mapsto AX$ from $\mathbb{K}^p$ to $\mathbb{K}^n$. This map is continuous: the bound $\norme{AX} \le C\,\norme{X}$ holds with an explicit $C$ read off the entries of $A$ --- for instance, with $\norme{\cdot}_\infty$ at source and target, $\norme{AX}_\infty \le \bigl(\max_i \sum_j |a_{ij}|\bigr)\,\norme{X}_\infty$ --- a direct entrywise computation, and the same kind of finite entrywise estimate produces such a $C$ for every pairing of the usual norms at source and target, with no appeal to finite-dimensional norm equivalence. The value of $\norme{A}_{\mathrm{op}}$ depends on both chosen norms, the one at the source and the one at the target.

The operator norm is the farthest the unit sphere reaches once mapped by $u$ --- a supremum, not necessarily attained.

Example — An operator norm computed

On $\mathcal{C}([0,1],\mathbb{K})$ with the uniform norm $\norme{\cdot}_\infty$, compute the operator norm of the continuous linear form $u \colon f \mapsto f(1)$.

Answer

Two steps: an upper bound, then a vector reaching it.

Upper bound. For every $f$, $|u(f)| = |f(1)| \le \norme{f}_\infty$, so $C = 1$ is an admissible constant; being the least one, $\norme{u}_{\mathrm{op}} \le 1$.
Equality. Take the constant function $f_0 \equiv 1$. Then $\norme{f_0}_\infty = 1$ and $|u(f_0)| = |f_0(1)| = 1$, so the ratio $|u(f_0)|/\norme{f_0}_\infty = 1$ is attained. Hence $\norme{u}_{\mathrm{op}} \ge 1$.

Combining the two, $\norme{u}_{\mathrm{op}} = 1$.

Example — The identity and a matrix subordinate norm

The identity of any non-zero normed space has operator norm $\norme{\mathrm{id}_E}_{\mathrm{op}} = 1$. For a matrix, take $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ acting on $\mathbb{K}^2$ with $\norme{\cdot}_\infty$: for $X = (x_1,x_2)$, $\norme{AX}_\infty = \max(2|x_1|,3|x_2|) \le 3\,\norme{X}_\infty$, and equality holds for $X = (0,1)$, so $\norme{A}_{\mathrm{op}} = 3$ --- the largest dilation factor.

Method — Compute an operator norm

To determine $\norme{u}_{\mathrm{op}}$ for a continuous linear map $u$:

upper bound --- establish $\norme{u(x)} \le C\,\norme{x}$ for all $x$ with an explicit constant $C$; this gives $\norme{u}_{\mathrm{op}} \le C$;
lower bound --- exhibit a non-zero vector $x_0$ with $\norme{u(x_0)} = C\,\norme{x_0}$, or a sequence $(x_n)$ of non-zero vectors with $\norme{u(x_n)}/\norme{x_n} \to C$; this gives $\norme{u}_{\mathrm{op}} \ge C$;
conclude $\norme{u}_{\mathrm{op}} = C$.

Skills to practice

Computing and bounding operator norms

III.3 Continuous multilinear maps

The simple picture of \S3.1 extends to maps that are linear in each variable separately --- multilinear maps. Continuity is again captured by a single inequality, this time a product bound $\norme{u(x_1,\dots,x_p)} \le C\,\norme{x_1}\cdots\norme{x_p}$.

Definition — Multilinear map

Let $p \ge 1$ be an integer and $E_1,\dots,E_p$, $F$ normed spaces. A map $u \colon E_1 \times \dots \times E_p \to F$ is multilinear (or $p$-linear) if it is linear in each variable separately: for every index $k$ and every fixed choice of the other variables, the partial map $$ x_k \longmapsto u(x_1,\dots,x_{k-1},x_k,x_{k+1},\dots,x_p) $$ is linear from $E_k$ to $F$. For $p = 2$ one says bilinear.

Theorem — Continuity criterion for a multilinear map

Let $u \colon E_1 \times \dots \times E_p \to F$ be a multilinear map, the product $E_1 \times \dots \times E_p$ carrying the product norm $\norme{(x_1,\dots,x_p)} = \max_{k} \norme{x_k}$. Then $u$ is continuous if and only if there exists a real $C \ge 0$ such that $$ \forall (x_1,\dots,x_p) \in E_1 \times \dots \times E_p,\quad \norme{u(x_1,\dots,x_p)} \le C\,\norme{x_1}\,\norme{x_2}\cdots\norme{x_p}. $$

This criterion is admitted: its demonstration is not required by the program. It generalises the linear criterion of \S3.1 --- continuity is again equivalent to a single bound, now multiplicative in the $p$ variables --- and is proved by the same kind of rescaling argument, made more delicate by the several variables.

Example — Matrix multiplication is continuous

Fix a usual norm on $\mathbb{K}^n$ and equip $\mathcal{M}_n(\mathbb{K})$ with the corresponding subordinate norm $\norme{\cdot}_{\mathrm{op}}$ of \S3.2. The product map $\mu \colon (A,B) \mapsto AB$ on $\mathcal{M}_n(\mathbb{K})^2$ is bilinear. Sub-multiplicativity, proved in \S3.2, gives $\norme{AB}_{\mathrm{op}} \le \norme{A}_{\mathrm{op}}\,\norme{B}_{\mathrm{op}}$, that is the product bound with $C = 1$. By the criterion, $\mu$ is continuous --- the concrete norm and the inequality are named, not assumed.

Example — Composition of continuous linear maps is continuous

With $E$, $F$, $G$ non-zero (the operator-norm convention of \S3.2), the composition map $c \colon (u,v) \mapsto v \circ u$ from $\mathcal{L}_c(E,F) \times \mathcal{L}_c(F,G)$ to $\mathcal{L}_c(E,G)$ is bilinear. By the sub-multiplicativity of \S3.2, $\norme{v \circ u}_{\mathrm{op}} \le \norme{v}_{\mathrm{op}}\,\norme{u}_{\mathrm{op}}$ --- the product bound with $C = 1$ --- so $c$ is continuous. Composition of continuous linear maps is itself a continuous operation.

Going further

This chapter built limits and continuity of maps between normed spaces. The chapter Compactness, connectedness, finite dimension singles out the compact and the arcwise-connected sets and proves the great theorems carried by continuity on them --- Bolzano--Weierstrass, Heine's theorem, the extreme value theorem and the generalised intermediate value theorem --- and shows that in finite dimension all norms are equivalent and every linear and multilinear map is automatically continuous, so the criteria of Section~3 are then satisfied for free.

Skills to practice

Applying the multilinear continuity criterion