CommeUnJeu · L2 MP
Normed vector spaces
In the first year, length was measured by the absolute value \(|\cdot|\) on \(\mathbb{R}\), by the modulus \(|\cdot|\) on \(\mathbb{C}\), and by the norm \(\sqrt{\langle x,x\rangle}\) attached to an inner product. Each of these is a way of assigning a non-negative number --- a « length » --- to a vector. This chapter isolates the three properties they all share and turns them into a single definition: a norm. A norm can then live on any vector space --- on \(\mathbb{K}^n\), on a space of polynomials, of matrices, of continuous functions alike.
The chapter has three sections. Section~1 defines a norm, the distance and the balls it produces, and the usual norms \(\norme{\cdot}_1\), \(\norme{\cdot}_2\), \(\norme{\cdot}_\infty\). Section~2 uses the distance to define convergent sequences in a normed vector space. Section~3 compares two norms on the same space: the central question is when they describe the same convergence --- the relation of equivalent norms.
Standing notation. Throughout, \(\mathbb{K}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\); \(E\) denotes a \(\mathbb{K}\)-vector space, with zero vector \(0_E\); a norm on \(E\) is written \(\norme{\cdot}\). The vector-space notions (basis, dimension, finite dimension, product space) are those of Finite-dimensional vector spaces, and the convergence notions for real sequences are those of Real sequences. The function-space norms of \S1.2 also recall, at their point of use, named results from Pre-Hilbert real spaces (Cauchy--Schwarz), Limits and continuity (the extreme value theorem) and Integration on a segment. This chapter is the vocabulary layer of the topology block: Topology of a normed space, Limits and continuity in a normed space and Compactness, connectedness, finite dimension all build directly on it.
The chapter has three sections. Section~1 defines a norm, the distance and the balls it produces, and the usual norms \(\norme{\cdot}_1\), \(\norme{\cdot}_2\), \(\norme{\cdot}_\infty\). Section~2 uses the distance to define convergent sequences in a normed vector space. Section~3 compares two norms on the same space: the central question is when they describe the same convergence --- the relation of equivalent norms.
Standing notation. Throughout, \(\mathbb{K}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\); \(E\) denotes a \(\mathbb{K}\)-vector space, with zero vector \(0_E\); a norm on \(E\) is written \(\norme{\cdot}\). The vector-space notions (basis, dimension, finite dimension, product space) are those of Finite-dimensional vector spaces, and the convergence notions for real sequences are those of Real sequences. The function-space norms of \S1.2 also recall, at their point of use, named results from Pre-Hilbert real spaces (Cauchy--Schwarz), Limits and continuity (the extreme value theorem) and Integration on a segment. This chapter is the vocabulary layer of the topology block: Topology of a normed space, Limits and continuity in a normed space and Compactness, connectedness, finite dimension all build directly on it.
I
Norms and normed vector spaces
I.1
Norm and normed vector space
A norm abstracts the absolute value. We keep exactly the three properties of \(|\cdot|\) that make it a measure of length: it vanishes only at the origin, it scales with scalars, and it satisfies the triangle inequality. Any map with these three properties deserves to be called a length.
Definition — Norm and normed vector space
Let \(E\) be a \(\mathbb{K}\)-vector space. A norm on \(E\) is a map \(N \colon E \to \mathbb{R}\) such that, for all \(x,y \in E\) and all \(\lambda \in \mathbb{K}\): - separation: \(N(x) = 0 \iff x = 0_E\);
- homogeneity: \(N(\lambda x) = |\lambda|\, N(x)\);
- triangle inequality: \(N(x+y) \le N(x) + N(y)\).
Non-negativity is not a fourth axiom: it is a consequence of the three above. Indeed, applying the triangle inequality and homogeneity, $$ 0 = N(0_E) = N\bigl(x + (-x)\bigr) \le N(x) + N(-x) = 2\, N(x), $$ so \(N(x) \ge 0\) for every \(x\). A norm therefore automatically takes its values in \(\mathbb{R}^+\).
Moreover, for any \(x \ne 0_E\), the vector \(x / \norme{x}\) has norm \(1\) by homogeneity --- it is the unit vector along \(x\).
Moreover, for any \(x \ne 0_E\), the vector \(x / \norme{x}\) has norm \(1\) by homogeneity --- it is the unit vector along \(x\).
Example — The modulus is the model norm
On the \(\mathbb{K}\)-vector space \(\mathbb{K}\) itself, the modulus \(|\cdot| \colon \mathbb{K} \to \mathbb{R}\) is a norm: \(|x| = 0 \iff x = 0\), \(|\lambda x| = |\lambda|\,|x|\), and \(|x+y| \le |x| + |y|\). So \((\mathbb{K}, |\cdot|)\) is a normed vector space --- the model on which the general definition is built. Proposition — Reverse triangle inequality
Let \((E, \norme{\cdot})\) be a normed vector space. For all \(x,y \in E\), $$ \textcolor{colorprop}{\bigl|\, \norme{x} - \norme{y} \,\bigr| \le \norme{x - y}}, \qquad \text{and likewise} \qquad \bigl|\, \norme{x} - \norme{y} \,\bigr| \le \norme{x + y}. $$ The first form is the one tied to the distance \(d(x,y) = \norme{x-y}\) of \S1.3.
Write \(x = (x - y) + y\) and apply the triangle inequality: $$ \begin{aligned} \norme{x} = \norme{(x-y) + y} &\le \norme{x-y} + \norme{y} && \text{(triangle inequality)} \end{aligned} $$ hence \(\norme{x} - \norme{y} \le \norme{x-y}\). Exchanging \(x\) and \(y\), and using \(\norme{y-x} = \norme{-(x-y)} = \norme{x-y}\) by homogeneity, gives \(\norme{y} - \norme{x} \le \norme{x-y}\). The two inequalities together yield \(\bigl|\norme{x} - \norme{y}\bigr| \le \norme{x-y}\). Replacing \(y\) by \(-y\) gives the \(\norme{x+y}\) form.
The triangle inequality is geometric: in the triangle with vertices \(0_E\), \(x\) and \(x+y\), the side \(\norme{x+y}\) is no longer than the sum of the other two sides. 
Recall, from Pre-Hilbert real spaces, the Cauchy--Schwarz inequality. For a real inner product space \((E, \langle\cdot,\cdot\rangle)\) it reads \(|\langle x,y\rangle| \le \sqrt{\langle x,x\rangle}\,\sqrt{\langle y,y\rangle}\); for finite families of reals it reads \(\bigl(\sum a_i b_i\bigr)^2 \le \bigl(\sum a_i^2\bigr)\bigl(\sum b_i^2\bigr)\); and for continuous functions it reads \(\bigl(\int_a^b |fg|\bigr)^2 \le \bigl(\int_a^b |f|^2\bigr)\bigl(\int_a^b |g|^2\bigr)\). It is recalled here, without proof, because it is the tool behind every triangle inequality for the norm \(\norme{\cdot}_2\) in this chapter.
Proposition — The norm of a real inner product space
Let \((E, \langle\cdot,\cdot\rangle)\) be a real inner product space (the inner product is real-valued). The map \(x \mapsto \sqrt{\langle x,x\rangle}\) is a norm on \(E\), called the prehilbertian norm associated with the inner product.
Write \(\norme{x} = \sqrt{\langle x,x\rangle}\), which is defined since \(\langle x,x\rangle \ge 0\).
- Separation. \(\norme{x} = 0 \iff \langle x,x\rangle = 0 \iff x = 0_E\), since the inner product is positive-definite.
- Homogeneity. For \(\lambda \in \mathbb{R}\), \(\norme{\lambda x} = \sqrt{\langle \lambda x, \lambda x\rangle} = \sqrt{\lambda^2 \langle x,x\rangle} = |\lambda|\,\norme{x}\).
- Triangle inequality. Expanding by bilinearity, $$ \begin{aligned} \norme{x+y}^2 = \langle x+y, x+y\rangle &= \norme{x}^2 + 2\langle x,y\rangle + \norme{y}^2 && \text{(bilinearity)}\\ &\le \norme{x}^2 + 2\,\norme{x}\,\norme{y} + \norme{y}^2 && \text{(Cauchy--Schwarz)}\\ &= \bigl(\norme{x} + \norme{y}\bigr)^2. \end{aligned} $$ Taking square roots gives \(\norme{x+y} \le \norme{x} + \norme{y}\).
Example — The canonical Euclidean norm
On \(\mathbb{R}^n\) with the canonical inner product \(\langle x,y\rangle = \sum_{i=1}^n x_i y_i\), the prehilbertian norm is \(\norme{x}_2 = \sqrt{\sum_{i=1}^n x_i^2}\), the canonical Euclidean norm. By the previous Proposition, it is a norm on \(\mathbb{R}^n\) --- no separate verification of the triangle inequality is needed. Method — Show that a map is a norm
To prove that a map \(N \colon E \to \mathbb{R}\) is a norm, check the three axioms in order: - separation: prove \(N(x) = 0 \Rightarrow x = 0_E\) (the converse \(N(0_E) = 0\) is usually immediate from homogeneity with \(\lambda = 0\));
- homogeneity: compute \(N(\lambda x)\) and factor out \(|\lambda|\);
- triangle inequality: bound \(N(x+y)\) --- this is the only step that may need a tool, typically the triangle inequality of \(|\cdot|\) applied termwise, or the Cauchy--Schwarz inequality for an \(\norme{\cdot}_2\)-type norm.
Skills to practice
- Verifying that a map is a norm
I.2
The usual norms
Three norms appear over and over: \(\norme{\cdot}_1\), \(\norme{\cdot}_2\) and \(\norme{\cdot}_\infty\). We define them first on \(\mathbb{K}^n\), then on a space of continuous functions, and finally on a product of normed vector spaces. Every triangle inequality for an \(\norme{\cdot}_2\) rests on the Cauchy--Schwarz inequality recalled in \S1.1.
Proposition — The usual norms on \(\mathbb{K}^n\)
Let \(n \ge 1\). For \(x = (x_1, \dots, x_n) \in \mathbb{K}^n\), set $$ \norme{x}_1 = \sum_{i=1}^n |x_i|, \qquad \norme{x}_2 = \sqrt{\sum_{i=1}^n |x_i|^2}, \qquad \norme{x}_\infty = \max_{1 \le i \le n} |x_i|. $$ These three maps are norms on \(\mathbb{K}^n\).
Each map is non-negative and real-valued. Let \(x,y \in \mathbb{K}^n\) and \(\lambda \in \mathbb{K}\).
- The norm \(\norme{\cdot}_1\). Separation: \(\norme{x}_1 = \sum |x_i| = 0\) forces every \(|x_i| = 0\), so \(x = 0\). Homogeneity: \(\norme{\lambda x}_1 = \sum |\lambda x_i| = |\lambda| \sum |x_i| = |\lambda|\,\norme{x}_1\). Triangle inequality: from \(|x_i + y_i| \le |x_i| + |y_i|\) for each \(i\), summing gives \(\norme{x+y}_1 \le \norme{x}_1 + \norme{y}_1\).
- The norm \(\norme{\cdot}_\infty\). Separation: \(\norme{x}_\infty = \max |x_i| = 0\) forces every \(|x_i| = 0\), so \(x = 0\). Homogeneity: \(\norme{\lambda x}_\infty = \max |\lambda x_i| = |\lambda| \max |x_i| = |\lambda|\,\norme{x}_\infty\). Triangle inequality: for each \(i\), \(|x_i + y_i| \le |x_i| + |y_i| \le \norme{x}_\infty + \norme{y}_\infty\); taking the maximum over \(i\) gives \(\norme{x+y}_\infty \le \norme{x}_\infty + \norme{y}_\infty\).
- The norm \(\norme{\cdot}_2\). Separation and homogeneity are as for \(\norme{\cdot}_1\). For the triangle inequality, expand and use the Cauchy--Schwarz inequality on the families \((|x_i|)\) and \((|y_i|)\) of non-negative reals: $$ \begin{aligned} \norme{x+y}_2^2 = \sum_{i=1}^n |x_i + y_i|^2 &\le \sum_{i=1}^n \bigl(|x_i| + |y_i|\bigr)^2 && \text{(triangle inequality of \(|\cdot|\))}\\ &= \norme{x}_2^2 + 2 \sum_{i=1}^n |x_i|\,|y_i| + \norme{y}_2^2 && \text{(expansion)}\\ &\le \norme{x}_2^2 + 2\,\norme{x}_2\,\norme{y}_2 + \norme{y}_2^2 && \text{(Cauchy--Schwarz)}\\ &= \bigl(\norme{x}_2 + \norme{y}_2\bigr)^2. \end{aligned} $$ Taking square roots gives \(\norme{x+y}_2 \le \norme{x}_2 + \norme{y}_2\). This argument is identical for \(\mathbb{K} = \mathbb{R}\) and \(\mathbb{K} = \mathbb{C}\), since it only uses the moduli \(|x_i|\), \(|y_i|\). For \(\mathbb{K} = \mathbb{R}\), \(\norme{\cdot}_2\) is the prehilbertian norm of the canonical inner product.
Example — The three norms of one vector
Take \(x = (3, -4, 12) \in \mathbb{R}^3\). Then $$ \norme{x}_1 = 3 + 4 + 12 = 19, \qquad \norme{x}_2 = \sqrt{9 + 16 + 144} = \sqrt{169} = 13, \qquad \norme{x}_\infty = \max(3, 4, 12) = 12. $$ The same vector receives three different lengths, with \(\norme{x}_\infty \le \norme{x}_2 \le \norme{x}_1\) --- an ordering that will be explained in \S3. Proposition — The usual norms on a space of continuous functions
Let \(a < b\) be two reals. For \(f \in \mathcal{C}([a,b], \mathbb{K})\), set $$ \norme{f}_\infty = \sup_{[a,b]} |f|, \qquad \norme{f}_1 = \int_a^b |f|, \qquad \norme{f}_2 = \sqrt{\int_a^b |f|^2}. $$ These three maps are norms on \(\mathcal{C}([a,b], \mathbb{K})\); the first is the uniform norm, the other two are referred to simply as the norm \(\norme{\cdot}_1\) and the norm \(\norme{\cdot}_2\).
For \(f \in \mathcal{C}([a,b], \mathbb{K})\), the functions \(|f|\) and \(|f|^2\) are continuous and real-valued on the segment \([a,b]\) (composition of \(f\) with the continuous map \(z \mapsto |z|\)); the extreme value theorem and the zero-integral lemma apply to them in both the real and the complex case.
- The uniform norm \(\norme{\cdot}_\infty\). Well-defined: \(|f|\) is continuous on a segment, hence bounded (extreme value theorem, recalled from Limits and continuity), so \(\sup_{[a,b]} |f|\) exists. Separation: \(\norme{f}_\infty = 0\) forces \(|f| \equiv 0\), so \(f = 0\). Homogeneity: \(\sup |\lambda f| = |\lambda| \sup |f|\). Triangle inequality: from \(|f+g| \le |f| + |g|\) pointwise, \(\norme{f+g}_\infty \le \norme{f}_\infty + \norme{g}_\infty\).
- The norm \(\norme{\cdot}_1\). Separation: if \(\int_a^b |f| = 0\) then, \(|f|\) being continuous and non-negative on \([a,b]\), the zero-integral lemma (recalled from Integration on a segment) gives \(|f| \equiv 0\), so \(f = 0\). Homogeneity and triangle inequality follow from the linearity and the monotonicity of the integral: \(\int |\lambda f| = |\lambda| \int |f|\), and \(|f+g| \le |f| + |g|\) gives \(\int |f+g| \le \int |f| + \int |g|\).
- The norm \(\norme{\cdot}_2\). Separation: \(\int_a^b |f|^2 = 0\) with \(|f|^2\) continuous and non-negative gives \(|f|^2 \equiv 0\), so \(f = 0\). Homogeneity is immediate. Triangle inequality: from \(|f+g|^2 \le \bigl(|f| + |g|\bigr)^2\) pointwise, integrate and bound the cross term by the integral Cauchy--Schwarz inequality: $$ \begin{aligned} \norme{f+g}_2^2 = \int_a^b |f+g|^2 &\le \int_a^b |f|^2 + 2\int_a^b |f|\,|g| + \int_a^b |g|^2 && \text{(expansion and monotonicity)}\\ &\le \norme{f}_2^2 + 2\,\norme{f}_2\,\norme{g}_2 + \norme{g}_2^2 && \text{(Cauchy--Schwarz)}\\ &= \bigl(\norme{f}_2 + \norme{g}_2\bigr)^2. \end{aligned} $$ Taking square roots gives the triangle inequality. For \(\mathbb{K} = \mathbb{R}\), \(\norme{\cdot}_2\) is the prehilbertian norm of the inner product \(\langle f,g\rangle = \int_a^b fg\).
Example — The three norms of one function
Take \(f \colon t \mapsto t\) on \([0,1]\). Then $$ \norme{f}_\infty = \sup_{[0,1]} |t| = 1, \qquad \norme{f}_1 = \int_0^1 t\, \mathrm{d}t = \tfrac{1}{2}, \qquad \norme{f}_2 = \sqrt{\int_0^1 t^2\, \mathrm{d}t} = \sqrt{\tfrac{1}{3}}. $$ Here too the three numbers differ: \(\norme{f}_1 \le \norme{f}_2 \le \norme{f}_\infty\), the reverse of the ordering \(\norme{x}_\infty \le \norme{x}_2 \le \norme{x}_1\) obtained for the vector of \(\mathbb{R}^3\) above. No single ordering of the three usual norms holds both on \(\mathbb{K}^n\) and on a function space. Proposition — Product norm
Let \(p \ge 1\) and let \((E_1, N_1), \dots, (E_p, N_p)\) be normed vector spaces. The map $$ N \colon (x_1, \dots, x_p) \longmapsto \max_{1 \le i \le p} N_i(x_i) $$ is a norm on the product space \(E_1 \times \cdots \times E_p\), called the product norm.
Let \(x = (x_1, \dots, x_p)\) and \(y = (y_1, \dots, y_p)\) in the product, and \(\lambda \in \mathbb{K}\). The three axioms are checked componentwise.
- Separation. \(N(x) = 0\) means \(\max_i N_i(x_i) = 0\), so \(N_i(x_i) = 0\) for every \(i\), hence \(x_i = 0_{E_i}\) for every \(i\), that is \(x = 0\).
- Homogeneity. \(N(\lambda x) = \max_i N_i(\lambda x_i) = \max_i |\lambda|\, N_i(x_i) = |\lambda| \max_i N_i(x_i) = |\lambda|\, N(x)\).
- Triangle inequality. For each \(i\), \(N_i(x_i + y_i) \le N_i(x_i) + N_i(y_i) \le N(x) + N(y)\); taking the maximum over \(i\) gives \(N(x+y) \le N(x) + N(y)\).
Example — The infinity norm is a product norm
Take \(E_1 = \cdots = E_n = (\mathbb{K}, |\cdot|)\). The product space is \(\mathbb{K}^n\), and the product norm is \((x_1, \dots, x_n) \mapsto \max_i |x_i| = \norme{x}_\infty\). So \(\norme{\cdot}_\infty\) on \(\mathbb{K}^n\) is the product norm of \(n\) copies of the model normed space \((\mathbb{K}, |\cdot|)\). Skills to practice
- Computing the usual norms
I.3
Distance\(\virgule\) balls and bounded sets
A norm produces a distance, the distance produces balls, and the balls let us say what a bounded set is. This subsection runs through that chain. The vocabulary --- ball, sphere, bounded --- is the one every later chapter of the topology block reuses.
Definition — Distance associated with a norm
Let \((E, \norme{\cdot})\) be a normed vector space. The distance associated with the norm is the map $$ d \colon E \times E \longrightarrow \mathbb{R}, \qquad d(x,y) = \norme{x - y}. $$ Proposition — Properties of the distance
The distance \(d\) associated with a norm on \(E\) satisfies, for all \(x,y,z \in E\): $$ d(x,y) \ge 0, \qquad d(x,y) = d(y,x), \qquad d(x,y) = 0 \iff x = y, \qquad d(x,y) \le d(x,z) + d(z,y). $$
Each property follows directly from a norm axiom.
- Non-negativity. \(d(x,y) = \norme{x-y} \ge 0\) since a norm is non-negative.
- Symmetry. \(d(y,x) = \norme{y-x} = \norme{-(x-y)} = |-1|\,\norme{x-y} = \norme{x-y} = d(x,y)\) by homogeneity.
- Separation. \(d(x,y) = 0 \iff \norme{x-y} = 0 \iff x - y = 0_E \iff x = y\).
- Triangle inequality. \(d(x,y) = \norme{(x-z) + (z-y)} \le \norme{x-z} + \norme{z-y} = d(x,z) + d(z,y)\).
In this course, every distance comes from a norm, as above. One can also define abstract « distances » on a set with no vector-space structure; such a set is called a metric space. The notions of metric space and, a fortiori, of topological space, are out of the program's scope. The same holds for the notions of Cauchy sequence and of Banach space. None of them is used here: the chapter works entirely with norms on vector spaces.
Definition — Balls and spheres
Let \((E, \norme{\cdot})\) be a normed vector space, \(a \in E\) and \(r\) a real number. - For \(r > 0\), the open ball of centre \(a\) and radius \(r\) is \(B(a,r) = \{\, x \in E : \norme{x-a} < r \,\}\).
- For \(r \ge 0\), the closed ball of centre \(a\) and radius \(r\) is \(B_f(a,r) = \{\, x \in E : \norme{x-a} \le r \,\}\).
- For \(r \ge 0\), the sphere of centre \(a\) and radius \(r\) is \(S(a,r) = \{\, x \in E : \norme{x-a} = r \,\}\).
Example — The three unit balls of \(\mathbb{R}^2\)
The shape of a ball depends on the norm. In \(\mathbb{R}^2\), the closed unit balls of \(\norme{\cdot}_1\), \(\norme{\cdot}_2\), \(\norme{\cdot}_\infty\) are respectively a diamond, a disk and a square: 
Definition — Convex subset
A subset \(C\) of a \(\mathbb{K}\)-vector space \(E\) is convex if, for all \(x,y \in C\) and all \(t \in [0,1]\), the vector \(t x + (1-t) y\) belongs to \(C\). Geometrically: whenever \(C\) contains two points, it contains the whole segment joining them. Proposition — Balls are convex
In a normed vector space, every ball --- open or closed --- is a convex subset.
Take the closed ball \(B_f(a,r)\), let \(x,y \in B_f(a,r)\) and \(t \in [0,1]\). Since \(t \ge 0\) and \(1-t \ge 0\), homogeneity and the triangle inequality give $$ \begin{aligned} \norme{\bigl(t x + (1-t) y\bigr) - a} = \norme{t(x-a) + (1-t)(y-a)} &\le t\,\norme{x-a} + (1-t)\,\norme{y-a} && \text{(triangle inequality and homogeneity)}\\
&\le t r + (1-t) r = r && \text{(since \(\norme{x-a} \le r\) and \(\norme{y-a} \le r\))}. \end{aligned} $$ So \(t x + (1-t) y \in B_f(a,r)\), and the closed ball is convex. For an open ball \(B(a,r)\) the same computation holds with \(\le\) replaced by \(<\) (each of \(\norme{x-a}\) and \(\norme{y-a}\) is \(< r\), and a convex combination of two numbers \(< r\) is \(< r\)).
A convex set contains every chord between two of its points; a non-convex set has at least one chord that escapes it. 
Definition — Bounded subsets
Let \((E, \norme{\cdot})\) be a normed vector space. - A subset \(A \subset E\) is bounded if it is contained in some closed ball: \(\exists\, a \in E,\ \exists\, r \ge 0,\ A \subset B_f(a,r)\).
- A sequence \((u_n)\) of \(E\) is bounded if the set of its values \(\{\, u_n : n \in \mathbb{N} \,\}\) is bounded.
- A function \(f \colon X \to E\) is bounded if its image \(f(X)\) is bounded.
Proposition — Characterisation of bounded subsets
A subset \(A\) of a normed vector space \(E\) is bounded if and only if $$ \exists\, M \ge 0,\ \forall x \in A,\ \norme{x} \le M, $$ that is, if and only if \(A\) is contained in a closed ball centred at \(0_E\).
The condition « \(\norme{x} \le M\) for all \(x \in A\) » means exactly \(A \subset B_f(0_E, M)\), so it is a particular case of « bounded ».
- If \(A \subset B_f(0_E, M)\), then \(A\) is contained in a closed ball, hence bounded.
- Conversely, suppose \(A\) is bounded, say \(A \subset B_f(a,r)\). For \(x \in A\), the triangle inequality gives \(\norme{x} = \norme{(x - a) + a} \le \norme{x-a} + \norme{a} \le r + \norme{a}\). So \(M = r + \norme{a}\) works: the centre can always be moved to \(0_E\) by enlarging the radius.
Example — Bounded and unbounded subsets
The closed unit ball \(B_f(0_E, 1)\) is bounded, by the very definition. A line \(\mathbb{K} u\) with \(u \ne 0_E\) is unbounded: \(\norme{\lambda u} = |\lambda|\, \norme{u} \to +\infty\) as \(|\lambda| \to +\infty\), so no \(M\) bounds \(\norme{\lambda u}\). Whether a given set is bounded can also depend on the choice of norm; \S3.1 shows that two equivalent norms always share the same bounded sets. Method — Determine the unit ball of a norm
To describe or draw the closed unit ball of a norm \(N\) on \(\mathbb{R}^2\) or \(\mathbb{R}^3\): - write the condition \(N(x) \le 1\) explicitly in coordinates;
- identify the region it describes --- a polygon for \(\norme{\cdot}_1\) and \(\norme{\cdot}_\infty\), a disk or an ellipse for an \(\norme{\cdot}_2\)-type norm;
- the unit sphere is the set where \(N(x) = 1\) --- the curve bounding the region in \(\mathbb{R}^2\), the surface bounding it in \(\mathbb{R}^3\).
Skills to practice
- Determining balls and bounded sets
II
Sequences in a normed vector space
II.1
Convergence of a sequence
With a distance available, convergence of a sequence transfers verbatim from Real sequences: a sequence converges to \(\ell\) when its terms eventually lie in every ball around \(\ell\). The key remark is that « \(u_n \to \ell\) » is equivalent to « \(\norme{u_n - \ell} \to 0\) », a statement about a real sequence --- so the whole theory of real sequences becomes available.
Definition — Convergent sequence
Let \((u_n)\) be a sequence of a normed vector space \((E, \norme{\cdot})\), and \(\ell \in E\). The sequence \((u_n)\) converges to \(\ell\) if $$ \forall\, \varepsilon > 0,\ \exists\, N \in \mathbb{N},\ \forall\, n \ge N,\quad \norme{u_n - \ell} < \varepsilon. $$ The sequence is convergent if it converges to some \(\ell \in E\), and divergent otherwise. (« Bounded sequence » was defined in \S1.3.) Proposition — Convergence and the norm of the difference
A sequence \((u_n)\) of \(E\) converges to \(\ell \in E\) if and only if the real sequence \(\bigl(\norme{u_n - \ell}\bigr)\) converges to \(0\).
The defining condition « \(\forall \varepsilon > 0,\ \exists N,\ \forall n \ge N,\ \norme{u_n - \ell} < \varepsilon\) » is, word for word, the \(\varepsilon\)-\(N\) definition of « the real sequence \(\bigl(\norme{u_n - \ell}\bigr)\) converges to \(0\) » (recalled from Real sequences). The two statements are therefore the same.
Proposition — Uniqueness of the limit
If a sequence \((u_n)\) of \(E\) converges, its limit is unique.
Suppose \((u_n)\) converges both to \(\ell\) and to \(\ell'\). For every \(n\), the triangle inequality gives $$ 0 \le \norme{\ell - \ell'} = \norme{(\ell - u_n) + (u_n - \ell')} \le \norme{u_n - \ell} + \norme{u_n - \ell'}. $$ The right-hand side tends to \(0\), so by squeezing \(\norme{\ell - \ell'} = 0\), hence \(\ell = \ell'\).
Proposition — A convergent sequence is bounded
Every convergent sequence of a normed vector space is bounded.
Let \((u_n)\) converge to \(\ell\). Applying the definition with \(\varepsilon = 1\), there is an \(N\) --- which we may take \(\ge 1\), since a larger \(N\) still works --- such that \(\norme{u_n - \ell} < 1\) for \(n \ge N\); hence, for \(n \ge N\), \(\norme{u_n} \le \norme{u_n - \ell} + \norme{\ell} < \norme{\ell} + 1\). Setting $$ M = \max\bigl(\norme{u_0}, \norme{u_1}, \dots, \norme{u_{N-1}},\ \norme{\ell} + 1\bigr), $$ a maximum of finitely many reals, we get \(\norme{u_n} \le M\) for every \(n\). By the characterisation of bounded subsets, \((u_n)\) is bounded.
Proposition — Operations on convergent sequences
Let \((u_n)\) and \((v_n)\) be sequences of \(E\) converging to \(\ell\) and \(\ell'\), and let \(\alpha, \beta \in \mathbb{K}\). Then \((\alpha u_n + \beta v_n)\) converges to \(\alpha \ell + \beta \ell'\). Consequently, the convergent sequences of \(E\) form a vector subspace of \(E^{\mathbb{N}}\), and \((u_n) \mapsto \lim u_n\) is a linear map on that subspace.
For every \(n\), homogeneity and the triangle inequality give $$ \begin{aligned} \norme{(\alpha u_n + \beta v_n) - (\alpha \ell + \beta \ell')} &= \norme{\alpha(u_n - \ell) + \beta(v_n - \ell')} && \text{(regrouping)}\\
&\le |\alpha|\,\norme{u_n - \ell} + |\beta|\,\norme{v_n - \ell'} && \text{(triangle inequality and homogeneity)}. \end{aligned} $$ Both terms on the right tend to \(0\), so the left-hand side tends to \(0\); by the « norm of the difference » Proposition, \((\alpha u_n + \beta v_n)\) converges to \(\alpha \ell + \beta \ell'\). This shows the set of convergent sequences is stable under linear combination --- hence a subspace --- and that taking the limit is linear.
Example — Convergence to zero depends on the norm
On \(\mathcal{C}([0,1], \mathbb{R})\), consider \(f_n \colon t \mapsto t^n\). For the norm \(\norme{\cdot}_1\), $$ \norme{f_n - 0}_1 = \int_0^1 t^n\, \mathrm{d}t = \frac{1}{n+1} \xrightarrow[n \to +\infty]{} 0, $$ so \((f_n)\) converges to the zero function for \(\norme{\cdot}_1\). For the norm \(\norme{\cdot}_\infty\), $$ \norme{f_n - 0}_\infty = \sup_{[0,1]} t^n = 1 \not\to 0, $$ so \((f_n)\) does not converge to the zero function for \(\norme{\cdot}_\infty\). Whether « \((f_n)\) converges to \(0\) » holds depends on the chosen norm.
The graphs of \(t^n\), drawn here for \(n = 1, 2, 4\), display the phenomenon: in the \(\norme{\cdot}_1\) sense the area under the curve shrinks to \(0\), while the highest point of each graph stays at height \(1\). 
Proposition — Convergence in a finite product
Let \((E_1, N_1), \dots, (E_p, N_p)\) be normed vector spaces and equip \(E = E_1 \times \cdots \times E_p\) with the product norm \(N\). A sequence \(\bigl(u_n\bigr)\) of \(E\), with \(u_n = (u_{1,n}, \dots, u_{p,n})\), converges if and only if each component sequence \((u_{i,n})_n\) converges in \(E_i\); in that case $$ \lim_{n \to +\infty} u_n = \Bigl(\lim_{n \to +\infty} u_{1,n},\ \dots,\ \lim_{n \to +\infty} u_{p,n}\Bigr). $$
Let \(\ell = (\ell_1, \dots, \ell_p) \in E\). By definition of the product norm, \(N(u_n - \ell) = \max_{1 \le j \le p} N_j(u_{j,n} - \ell_j)\).
- Direct sense. Suppose \(u_n \to \ell\). For each fixed \(i\), \(N_i(u_{i,n} - \ell_i) \le N(u_n - \ell) \to 0\), so \(N_i(u_{i,n} - \ell_i) \to 0\) by squeezing: the \(i\)-th component converges to \(\ell_i\).
- Converse sense. Suppose each component converges, say \(u_{j,n} \to \ell_j\), i.e. \(N_j(u_{j,n} - \ell_j) \to 0\) for every \(j\). Then \(N(u_n - \ell) = \max_{1 \le j \le p} N_j(u_{j,n} - \ell_j)\) is a maximum of finitely many sequences each tending to \(0\), hence tends to \(0\). So \(u_n \to \ell\). The finiteness of \(p\) is essential to this direction: it is what lets the maximum inherit the limit \(0\) from its finitely many entries.
Example — A sequence studied componentwise
Determine, if it exists, the limit of the sequence \((u_n)\) of \(\mathbb{R}^3\) defined for \(n \ge 1\) by $$ u_n = \left( \frac{1}{n},\ \ \mathrm{e}^{-n},\ \ \frac{n}{n+1} \right), $$ where \(\mathbb{R}^3\) is equipped with the product norm \(\norme{\cdot}_\infty\).
The three component sequences converge: $$ \frac{1}{n} \xrightarrow[n \to +\infty]{} 0, \qquad \mathrm{e}^{-n} \xrightarrow[n \to +\infty]{} 0, \qquad \frac{n}{n+1} = 1 - \frac{1}{n+1} \xrightarrow[n \to +\infty]{} 1. $$ By the « convergence in a finite product » Proposition, \((u_n)\) converges, and $$ \lim_{n \to +\infty} u_n = (0,\ 0,\ 1). $$
Method — Show that a sequence converges
To prove that a sequence \((u_n)\) of a normed vector space converges: - guess the limit \(\ell\), then prove \(\norme{u_n - \ell} \to 0\) by bounding it above by a real sequence known to tend to \(0\);
- in a finite product equipped with the product norm, study each component sequence separately --- the sequence converges if and only if all components do, and the limit is read componentwise;
- to use the algebra (linear combinations), first establish convergence of the simpler building-block sequences, then combine.
Skills to practice
- Studying the convergence of a sequence
II.2
Subsequences and subsequential limits
Out of one sequence we can extract infinitely many terms, keeping their order, and form a new sequence --- a subsequence. The limits reachable that way are the subsequential limits of the original sequence. They are attached to a sequence; they should not be confused with the adherent points of a set, studied in Topology of a normed space.
Definition — Subsequence
Let \((u_n)\) be a sequence of \(E\). A subsequence (or extracted sequence) of \((u_n)\) is a sequence of the form \(\bigl(u_{\varphi(n)}\bigr)_n\), where \(\varphi \colon \mathbb{N} \to \mathbb{N}\) is strictly increasing; such a \(\varphi\) is called an extractor. Recall, from Real sequences, that a strictly increasing \(\varphi \colon \mathbb{N} \to \mathbb{N}\) satisfies \(\varphi(n) \ge n\) for every \(n\). Definition — Subsequential limit
Let \((u_n)\) be a sequence of \(E\). A subsequential limit of \((u_n)\) --- also called a cluster value of \((u_n)\) --- is the limit of some convergent subsequence of \((u_n)\). Proposition — Convergence and subsequences
A sequence \((u_n)\) of \(E\) converges to \(\ell\) if and only if every subsequence of \((u_n)\) converges to \(\ell\). In particular, the limit of a convergent sequence is its unique subsequential limit.
We prove the two implications, then read off the consequence.
- Direct sense. Suppose \(u_n \to \ell\), and let \(\bigl(u_{\varphi(n)}\bigr)\) be a subsequence. Fix \(\varepsilon > 0\); there is an \(N\) with \(\norme{u_n - \ell} < \varepsilon\) for \(n \ge N\). For \(n \ge N\), the extractor satisfies \(\varphi(n) \ge n \ge N\), hence \(\norme{u_{\varphi(n)} - \ell} < \varepsilon\). So \(\bigl(u_{\varphi(n)}\bigr)\) converges to \(\ell\).
- Converse sense. Suppose every subsequence converges to \(\ell\). The sequence \((u_n)\) is one of its own subsequences --- take the extractor \(\varphi = \operatorname{id}_{\mathbb{N}}\) --- so \((u_n)\) itself converges to \(\ell\).
Proposition — A sequence with two subsequential limits diverges
A sequence of \(E\) that has at least two distinct subsequential limits is divergent.
This is the contrapositive of the previous Proposition: a convergent sequence has its limit as unique subsequential limit, so it cannot have two distinct ones. A sequence with two distinct subsequential limits is therefore not convergent.
Example — Two subsequential limits
The real sequence \(\bigl((-1)^n\bigr)\) has the constant subsequence \((-1)^{2n} = 1\), of limit \(1\), and the constant subsequence \((-1)^{2n+1} = -1\), of limit \(-1\). It has the two distinct subsequential limits \(1\) and \(-1\), hence diverges. Example — No subsequential limit
The real sequence \(\bigl(n(-1)^n\bigr)\) has no subsequential limit at all: any subsequence \(\bigl(\varphi(n)(-1)^{\varphi(n)}\bigr)\) has terms of modulus \(\varphi(n) \ge n \to +\infty\), so it is unbounded and cannot converge. A sequence may thus fail to converge for a reason other than oscillation between two values --- here, by escaping to infinity. Method — Show that a sequence diverges
To prove that a sequence \((u_n)\) diverges, the most efficient route is via subsequential limits: - exhibit two subsequences of \((u_n)\) converging to two different limits --- by the Proposition above, \((u_n)\) then diverges;
- typical extractors are the even indices \(\varphi(n) = 2n\) and the odd indices \(\varphi(n) = 2n+1\), when the sequence alternates;
- alternatively, exhibit one subsequence \((u_{\varphi(n)})\) that is unbounded --- for instance with \(\norme{u_{\varphi(n)}} \to +\infty\) --- since a convergent sequence is bounded, the sequence then has no chance of converging.
Skills to practice
- Using subsequences and subsequential limits
III
Comparison of norms
III.1
Equivalent norms
The example \(f_n(t) = t^n\) of \S2.1 showed that whether a sequence converges to a prescribed limit can depend on the norm. This section names a sufficient condition --- equivalence --- under which two norms on \(E\) guarantee the same bounded sets and the same convergent sequences. It is stated as an implication « equivalent \(\Rightarrow\) same analysis »; the practical converse --- detecting non-equivalence by a sequence --- is the subject of \S3.2.
Definition — Equivalent norms
Two norms \(N\) and \(N'\) on the same vector space \(E\) are equivalent if there exist two reals \(\alpha > 0\) and \(\beta > 0\) such that $$ \forall x \in E,\quad \alpha\, N(x) \le N'(x) \le \beta\, N(x). $$ Proposition — Equivalence of norms is an equivalence relation
On the set of all norms of a vector space \(E\), the relation « \(N\) is equivalent to \(N'\) » is an equivalence relation: it is reflexive, symmetric and transitive.
Let \(N\), \(N'\), \(N''\) be norms on \(E\).
- Reflexivity. \(N\) is equivalent to itself: \(1 \cdot N(x) \le N(x) \le 1 \cdot N(x)\), so \(\alpha = \beta = 1\) works.
- Symmetry. If \(\alpha N \le N' \le \beta N\), then dividing by the positive constants gives \(\tfrac{1}{\beta} N' \le N \le \tfrac{1}{\alpha} N'\), so \(N'\) is equivalent to \(N\).
- Transitivity. If \(\alpha N \le N' \le \beta N\) and \(\alpha' N' \le N'' \le \beta' N'\), then for all \(x\), \(N''(x) \le \beta' N'(x) \le \beta' \beta\, N(x)\) and \(N''(x) \ge \alpha' N'(x) \ge \alpha' \alpha\, N(x)\). So \(\alpha' \alpha\, N \le N'' \le \beta' \beta\, N\), and \(N\) is equivalent to \(N''\).
Proposition — Equivalent norms preserve boundedness and convergence
Let \(N\) and \(N'\) be two equivalent norms on \(E\). - A subset \(A \subset E\) is bounded for \(N\) if and only if it is bounded for \(N'\).
- A sequence \((u_n)\) converges to \(\ell\) for \(N\) if and only if it converges to \(\ell\) for \(N'\).
Let \(\alpha, \beta > 0\) with \(\alpha N \le N' \le \beta N\). By symmetry of the relation it is enough to prove each « if » direction.
- Boundedness. If \(A\) is bounded for \(N\), there is an \(M\) with \(N(x) \le M\) for all \(x \in A\); then \(N'(x) \le \beta N(x) \le \beta M\), so \(A\) is bounded for \(N'\). The converse uses \(\alpha N \le N'\) in the same way.
- Convergence. \((u_n)\) converges to \(\ell\) for \(N\) means \(N(u_n - \ell) \to 0\); then \(0 \le N'(u_n - \ell) \le \beta\, N(u_n - \ell) \to 0\), so \(N'(u_n - \ell) \to 0\) by squeezing, i.e. \((u_n)\) converges to the same \(\ell\) for \(N'\). The converse uses \(\alpha N \le N'\).
Example — The three usual norms on \(\mathbb{K}^n\) are equivalent
Show that the norms \(\norme{\cdot}_1\), \(\norme{\cdot}_2\) and \(\norme{\cdot}_\infty\) are pairwise equivalent on \(\mathbb{K}^n\).
Let \(x = (x_1, \dots, x_n) \in \mathbb{K}^n\). We first compare each of \(\norme{x}_1\) and \(\norme{x}_2\) with \(\norme{x}_\infty\).
- Each \(|x_i| \le \norme{x}_\infty\), and one index reaches the maximum, so \(\norme{x}_\infty \le \norme{x}_1 = \sum_{i=1}^n |x_i| \le \sum_{i=1}^n \norme{x}_\infty = n\, \norme{x}_\infty\). Hence \(\norme{x}_\infty \le \norme{x}_1 \le n\, \norme{x}_\infty\).
- Similarly, \(\norme{x}_\infty^2 = \max_i |x_i|^2 \le \sum_{i=1}^n |x_i|^2 = \norme{x}_2^2 \le \sum_{i=1}^n \norme{x}_\infty^2 = n\, \norme{x}_\infty^2\); taking square roots, \(\norme{x}_\infty \le \norme{x}_2 \le \sqrt{n}\, \norme{x}_\infty\).
Method — Show that two norms are equivalent
To prove that two norms \(N\) and \(N'\) on \(E\) are equivalent, exhibit the two constants: - find \(\beta > 0\) with \(N'(x) \le \beta\, N(x)\) for all \(x\) --- bound \(N'\) above by \(N\);
- find \(\alpha > 0\) with \(\alpha\, N(x) \le N'(x)\) for all \(x\) --- equivalently, bound \(N\) above by \(N'\);
- it is often easier to compare both \(N\) and \(N'\) with a third, simpler norm, and then conclude by transitivity.
Skills to practice
- Proving that two norms are equivalent
III.2
Showing that two norms are not equivalent
Equivalence \(\alpha N \le N' \le \beta N\) is the conjunction of a lower inequality \(\alpha N \le N'\) and an upper inequality \(N' \le \beta N\). To disprove equivalence it is enough to break one of the two, and one well-chosen sequence does it.
Proposition — A criterion for non-equivalence
Let \(N\) and \(N'\) be two norms on \(E\). If there exists a sequence \((u_n)\) of \(E \setminus \{0_E\}\) such that $$ \frac{N'(u_n)}{N(u_n)} \xrightarrow[n \to +\infty]{} +\infty, $$ then \(N\) and \(N'\) are not equivalent. Since the relation of being equivalent norms is symmetric (\S3.1), the lower inequality \(\alpha N \le N'\) is broken by the same criterion with the roles of \(N\) and \(N'\) exchanged.
Argue by contradiction. Suppose \(N\) and \(N'\) were equivalent: there would be \(\beta > 0\) with \(N'(x) \le \beta\, N(x)\) for every \(x \in E\). Since each \(u_n \ne 0_E\), separation gives \(N(u_n) > 0\), so we may divide: $$ \frac{N'(u_n)}{N(u_n)} \le \beta \qquad \text{for every } n. $$ The ratio would then be bounded above by the constant \(\beta\) --- contradicting \(\dfrac{N'(u_n)}{N(u_n)} \to +\infty\). Hence \(N\) and \(N'\) are not equivalent. Only the upper inequality of equivalence was used.
Example — Two non-equivalent norms on a function space
On \(\mathcal{C}([0,1], \mathbb{R})\), the norms \(\norme{\cdot}_1\) and \(\norme{\cdot}_\infty\) are not equivalent. Take the family of \S2.1: \(f_n(t) = t^n\), which is non-zero. Then $$ \norme{f_n}_1 = \int_0^1 t^n\, \mathrm{d}t = \frac{1}{n+1}, \qquad \norme{f_n}_\infty = \sup_{[0,1]} t^n = 1, \qquad \text{so} \qquad \frac{\norme{f_n}_\infty}{\norme{f_n}_1} = n + 1 \xrightarrow[n \to +\infty]{} +\infty. $$ By the criterion --- with \(N = \norme{\cdot}_1\) and \(N' = \norme{\cdot}_\infty\) --- the norms \(\norme{\cdot}_1\) and \(\norme{\cdot}_\infty\) are not equivalent on \(\mathcal{C}([0,1], \mathbb{R})\). Method — Show that two norms are not equivalent
To prove that two norms \(N\) and \(N'\) on \(E\) are not equivalent: - decide which inequality to break --- the upper one \(N' \le \beta N\), or, after exchanging \(N\) and \(N'\), the lower one;
- construct a sequence \((u_n)\) of non-zero vectors for which the ratio \(N'(u_n) / N(u_n)\) tends to \(+\infty\);
- conclude by the criterion above. A single such sequence is enough; there is no need to handle every pair of constants.
One last remark closes the comparison. On the function space \(\mathcal{C}([0,1], \mathbb{R})\) --- which is infinite-dimensional --- two norms could fail to be equivalent. In a finite-dimensional space the situation is completely different: there, all norms turn out to be equivalent. That result, and the notion of compactness behind it, are established later, in Compactness, connectedness, finite dimension.
Going further
This chapter built the metric vocabulary of a normed vector space: norm, distance, ball, bounded set, convergent sequence, equivalent norms. Three chapters extend it directly. Topology of a normed space uses balls to define open and closed sets, interior, closure and frontier. Limits and continuity in a normed space defines limits and continuity of maps between normed vector spaces. Compactness, connectedness, finite dimension introduces compactness and proves the headline result announced just above: on a finite-dimensional space, all norms are equivalent.
Skills to practice
- Proving that two norms are not equivalent
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