Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

## C. Finite Dimensional Spaces

In this section we will characterize the normed linear spaces of finite dimension.

Proposition I.62

Any two norms on a finite dimensional normed linear space induce the same topology; in other words, any two norms are equivalent.

Proof. Let $$$X$$$ be a finite dimensional normed linear space. Let $$$\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc, {e}_{n}\,\right\}\mathclose{}$$$ be a basis for $$$X$$$. Then any $$$x\in X$$$ can be written $$$\sum_{i=1}^{n}{}{α}_{i}{e}_{i}$$$ with each $$${α}_{i}\in \mathbb{C}$$$. Define a Euclidean norm $$$ρ$$$ on $$$X$$$ by $$ρ\mathopen{}\left( x\right)\mathclose{}= ρ\mathopen{}\left( \sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right)\mathclose{}= {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \text{.}$$ We want to show that the norm on $$$X$$$ and $$$ρ$$$ determine the same topology, i.e. that any $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}$$$-open set contains a $$$ρ$$$-open set and vice versa. $$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right\rVert\mathclose{}\leq \sum_{i=1}^{n}{}\mathopen{}\left\lVert{}{α}_{i}{e}_{i}\right\rVert\mathclose{}= \sum_{i=1}^{n}{}\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}\leq {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \text{.}$$ Let $$$M= {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}}$$$ Thus, $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq Mρ\mathopen{}\left( x\right)\mathclose{}$$$ and so any $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}$$$-open $$$ε$$$-ball contains a $$$ρ$$$-open $$$\frac{ε}{M}$$$-ball.

Now we want to show that any $$$ρ$$$-open ball contains a $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}$$$-open ball. Define $$$f : {\mathbb{C}}^{n} \to \mathbb{R}$$$ by $$$f\mathopen{}\left( \mathopen{}\left({α}_{1}, {α}_{2}, \dotsc, {α}_{n}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\lVert{}\sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right\rVert\mathclose{}$$$. This function is continuous in the Euclidean topology (use the inequality above). The unit sphere in $$${\mathbb{C}}^{n}$$$, $$$\mathopen{}\left\{\, \mathopen{}\left({α}_{1}, {α}_{2}, \dotsc, {α}_{n}\right)\mathclose{}\,\middle\vert\, , \sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}= 1, \,\right\}\mathclose{}$$$, is compact, hence $$$f$$$ attains a minimum $$$m\gt 0$$$.

So for any $$$x\in X$$$ with $$$ρ\mathopen{}\left( x\right)\mathclose{}= 1$$$ we have $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\geq mρ\mathopen{}\left( x\right)\mathclose{}$$$. This extends to all of $$$X$$$ by properties of the norm. If $$$ρ\mathopen{}\left( x\right)\mathclose{}\gt 0$$$, then $$$ρ\mathopen{}\left( \frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right)\mathclose{}= 1$$$, and $$$\mathopen{}\left\lVert{}\frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right\rVert\mathclose{}\geq mρ\mathopen{}\left( \frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right)\mathclose{}$$$. Therefore, by multiplying both sides of the inequality by $$$ρ\mathopen{}\left( x\right)\mathclose{}$$$, we get the desired inequality $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\geq mρ\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x$$$ in $$$X$$$. Thus, any any $$$ρ$$$-open ball contains a $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}$$$-open ball.

Thus, we have shown that any norm on a finite dimensional normed linear space $$$X$$$ is equivalent to the Euclidean norm. In fact, $$$X$$$ is homeomorphic to $$${\mathbb{C}}^{n}$$$ for some $$$n$$$.

A direct consequence of this is the following proposition.

Proposition I.63

If $$$Y$$$ is a finite dimensional subspace of a normed linear space $$$X$$$, then $$$Y$$$ is closed.

Proof. Including the above proof, we know the following about $$$Y$$$: Any norm of $$$Y$$$ is equivalent to the Euclidean norm on $$${\mathbb{C}}^{n}$$$; Every Cauchy sequence in $$$Y$$$ is Cauchy in $$${\mathbb{C}}^{n}$$$; $$${\mathbb{C}}^{n}$$$ is a complete metric space, thus every Cauchy sequence converges; Convergence in $$${\mathbb{C}}^{n}$$$ implies convergence in $$$Y$$$; $$$Y$$$ is a complete subspace of the metric space $$$X$$$, thus $$$Y$$$ is closed.

It follows from this proposition that every finite dimensional normed linear space is complete. Next question to ask: Is every subspace of a normed linear space closed? And if we were to find a non-closed subspace of a normed linear space where would we have to look?

Example I.64

Let $$ denote the subspace of $$$\mathrm{l}^{0}$$$ consisting of those sequences which have only finitely many non-zero terms. For any $$$k\in \mathbb{N}$$$ let $${x}_{k}= \mathopen{}\left(1, \frac{1}{2}, \frac{1}{3}, \dotsc, \frac{1}{k}, 0, 0, \dotsc\right)\mathclose{} \text{.}$$ The sequence of these sequences, $$$\mathopen{}\left({x}_{k}\right)\mathclose{}_{k=1}^{\infty}$$$, is a subset of $$ that converges to $$x= \mathopen{}\left(1, \frac{1}{2}, \frac{1}{3}, \dotsc\right)\mathclose{}= \mathopen{}\left(\frac{1}{n}\right)\mathclose{}_{n=1}^{\infty} \text{.}$$ Note that $$$x$$$ is in $$$\mathrm{l}^{0}$$$ but not in $$; therefore $$ is not closed. Indeed, $$\lim_{k\to\infty}{}\mathopen{}\left\lVert{}{x}_{k}-x\right\rVert\mathclose{}= \lim_{k\to\infty}{}\mathopen{}\left\lVert{}\mathopen{}\left(0, 0, \dotsc, 0, \frac{1}{k+1}, \frac{1}{k+2}, \dotsc\right)\mathclose{}\right\rVert\mathclose{}= 0 \text{.}$$

The last property we need to show is that finite dimensional gives us closed and bounded equals compact and vice versa. We first need a lemma.

Lemma I.65

Let $$$Y$$$ be a closed, proper subspace of a normed linear space $$$X$$$. Then for any $$$r\gt 0$$$, there exists a $$$x\in X$$$ with $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1$$$ and $$$\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}\gt 1-r$$$ for all $$$y\in Y$$$.

Proof. Let $$${x}_{0}\in X\setminus Y$$$. The distance from $$${x}_{0}$$$ to $$$Y$$$, $$$\operatorname{d}\mathopen{}\left( {x}_{0}, Y\right)\mathclose{}$$$, is positive (else you can find a sequence of elements in $$$Y$$$ converging to $$${x}_{0}$$$, which would contradict that $$$Y$$$ is a closed subspace). Let $$$ε$$$ be a positive real number. There exists $$${y}_{0}\in Y$$$ such that $$$0\lt \mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}\leq \operatorname{d}\mathopen{}\left( {x}_{0}, Y\right)\mathclose{}+ε$$$. Let $$$x= \frac{{x}_{0}-{y}_{0}}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}}$$$. Then $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1$$$. We want to show that with the right choice of $$$ε$$$ we get the inequality we want. Because, for any $$$y\in Y$$$, $$\mathopen{}\left\lVert{}x, y\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\frac{{x}_{0}-{y}_{0}}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}}-y\right\rVert\mathclose{}= \frac{1}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}}\mathopen{}\left\lVert{}\mathopen{}\left({x}_{0}-{y}_{0}\right)\mathclose{}-\mathopen{}\left(\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}y\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε}= 1-\frac{ε}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε} \text{,}$$ we just need to choose $$$ε$$$ so that $$$\frac{ε}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε}\lt r$$$.

Theorem I.66

A normed linear space is finite dimensional if and only if every closed and bounded subset is compact.

Proof. Let $$$X$$$ be a finite dimensional normed linear space. Let $$$S\subseteq X$$$ be a closed and bounded subset. By Proposition I.62, closed and bounded in the norm implies closed and bounded in the Euclidean norm. So, the Heine-Borel Theorem applies and $$$S$$$ is compact.

Suppose every closed, bounded subset of $$$X$$$ is compact. By way of contradiction, suppose that $$$X$$$ is infinite dimensional. We'll construct a bounded sequence with no convergent subsequence to get the contradiction.

Let $$${x}_{1}\in X$$$ such that $$$\mathopen{}\left\lVert{}{x}_{1}\right\rVert\mathclose{}= 1$$$. Let $$${Y}_{1}= \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {x}_{1}\,\right\}\mathclose{}\right)\mathclose{}$$$. By Proposition I.63, $$${Y}_{1}$$$ is closed and by Lemma I.65, there exists an $$${x}_{2}\in X\setminus {Y}_{1}$$$ such that $$$\mathopen{}\left\lVert{}{x}_{2}\right\rVert\mathclose{}= 1$$$ and $$$\mathopen{}\left\lVert{}{x}_{1}-{x}_{2}\right\rVert\mathclose{}\gt \frac{1}{2}$$$. Let $$${Y}_{2}= \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {x}_{1}, {x}_{2}\,\right\}\mathclose{}\right)\mathclose{}$$$. As before, $$${Y}_{2}$$$ is closed and there exists an $$${x}_{3}\in X\setminus {Y}_{2}$$$ such that $$$\mathopen{}\left\lVert{}{x}_{3}\right\rVert\mathclose{}= 1$$$, $$$\mathopen{}\left\lVert{}{x}_{1}-{x}_{3}\right\rVert\mathclose{}\gt \frac{1}{2}$$$, and $$$\mathopen{}\left\lVert{}{x}_{2}-{x}_{3}\right\rVert\mathclose{}\gt \frac{1}{2}$$$. Continue forever, which you can do because $$$X$$$ is infinite dimensional. In this way, we get a bounded infinite sequence $$$\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ where $$$\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1$$$ for all $$$n$$$, and $$$\mathopen{}\left\lVert{}{x}_{i}-{x}_{j}\right\rVert\mathclose{}\gt \frac{1}{2}$$$ when $$$i\neq j$$$. Therefore, $$$\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ has no convergent subsequence, which contradicts the assumption that every closed, bounded subset is compact.

Corollary I.67

The closed unit ball in an infinite-dimensional, normed linear space is not compact.