# Instructive proofs in functional analysis

I am beginning to learn functional analysis (from Folland and Royden), but I am from a non-mathematical background, so I often encounter techniques in proofs that I am not familiar with (for example the proof of the open mapping theorem).

An idle curiosity masquerading as a pedagogical question, what are proofs of some theorems in functional analysis that make use of techniques/constructions/results that one should be familiar with? I am happy to look up redirections to specific theorems in texts.

This question might as well be called - what is your favorite instructive proof in functional analysis?

• By the way, I don't mind the proof being for an obscure theorem or even the solution of an exercise. – tchakravarty Apr 18 '12 at 23:27
• I think the most instructive is a spectral theorem for normal operators on Hilbert spaces. But I don't know whether the OP appreciate it, since it is very difficult for beginners. – Norbert Apr 23 '12 at 17:30
• Well, if you've got the time to break it down for me, I will certainly try to appreciate it. :) – tchakravarty Apr 24 '12 at 22:57
• I haven't got so much time :) It will take me about two days to write a brief proof. Here you can find a small discussion math.stackexchange.com/questions/32815/… – Norbert Apr 24 '12 at 23:18
• Well, I awarded them the bounty, but was semantically disinclined to award correct answer to anyone. Obviously not acceptable practice. – tchakravarty Apr 28 '12 at 23:37

## 3 Answers

Here is a proof of Alaoglu's theorem which I find nice (it might actually be the standard proof):

If $X$ is a normed vector space, the closed unit ball $B^*=\{f\in X^* : \|f\|\leq 1\}$ in $X^*$ is compact in the weak* topology (this is taken from Folland's Real Analysis, Theorem 5.18).

Proof Given a point $x\in X$, consider the set $$D_x=\{z\in\mathbb{C}:|z|\leq\|x\|\}.$$ Note that $D_x$ is compact for all $x$. Let $D=\prod_{x\in X}D_x$. By Tychonoff's theorem, $D$ is also compact. Now we take a new point of view. Namely, we note that $D$ coincides with the set of all complex-valued functions $\phi$ on $X$ (not necessarily continuous) such that $|\phi(x)|\leq\|x\|$ for all $x\in X$, and $B^*$ is now a subset of $D$ consisting of those functions which are linear. By definition, the weak* topology on $B^*$ is the topology of pointwise convergence, which is exactly the topology $B^*$ inherits from $D$. Therefore, to conclude that $B^*$ is compact (in the weak* topology), it suffices to show that it is a closed subset of $D$.

Let $\{f_\alpha\}$ be a net in $B^*$ that converges to $f\in D$. Then, for all $x,y\in X$ and $a,b\in\mathbb{C}$, we have $$f(ax+by)=\lim f_\alpha(ax+by)=\lim (af_\alpha(x)+bf_\alpha(y))=af(x)+bf(y),$$ that is, a limit of linear maps is linear. Therefore, we conclude that $f\in B^*$, and so $B^*$ is closed in $D$.

The reason I like this proof is that you have to consider an object in two different ways, and it is the interplay between the different viewpoints that allows you to prove the result. When I first learned functional analysis (and I am still learning), the different topologies kept confusing me, and proofs like this one allowed me to understand these concepts.

• This is a very nice answer and, by coincidence, it came in very handy. Thanks! – t.b. Apr 18 '12 at 14:36
• That was exactly the kind of thing I was looking for. My head hurts, but I think I get it. :) Thanks. – tchakravarty Apr 19 '12 at 0:09

These are proofs of existence of completion for different objects of functional analysis without explicit construction. I think they are instructive because they are very easy and show usage of different standard results in functional analysis.

Defintion 1. A normed space $Y$ is called a completion of normed space $X$ if $Y$ is complete and there exists isometric operator $j:X\to Y$ with dense image

Theorem 1. Every normed $X$ space has a completion.

Proof. Consider second dual $X^{**}$ and standard embedding $i:X\to X^{**}$. Define $Y=\overline{\operatorname{Im}\;i}$. Since $Y$ is a closed subspace of complete space $X^{**}$ then $Y$ is complete. By corollary of Hahn-Banach theorem $i$ is isometric hence its corestriction $$j:X\to Y:x\mapsto i(x)$$ is isometric too. By construction $\operatorname{Im}\;j$ is dense in $Y$. Thus $Y$ is a completion of $X$.

Also there is a proof of existence of completion for metric spaces.

Defintion 2. A metric space $Y$ is called a completion of metric space $X$ if $Y$ is complete and there exist isometric map $j:X\to Y$ with dense image.

Theorem 2. Every metric $X$ space have a completion.

Proof. Consider space $C_b(X)$ of continuous functions on metric space $X$. Fix $x_0\in X$ and define map $$i:X\to C_b(X):x\mapsto(p\mapsto d_X(x,p)-d_X(x_0,p))$$ Since for all $p\in X$ we have $|i(x)(p)|\leq d_X(x,x_0)$, so $i(x)\in C_b(X)$ and $i$ is well defined. One can easily show that $\Vert i(x'')-i(x')\Vert_{C_b(X)}=d_X(x'',x')$, so $i$ is isometric. Define $Y=\overline{\operatorname{Im}\;i}$, then $Y$ is complete as closed subspace of complete metric space $C_b(X)$. It is remains to consider the corestriction $$j:X\to Y:x\mapsto i(x)$$ By construction $\operatorname{Im}\;j$ is dense in $Y$. Thus $Y$ is a completion of $X$.

• It might be fair to mention for someone new to functional analysis that you're using the Hahn-Banach theorem when claiming that $i$ is isometric. – t.b. Apr 21 '12 at 17:26

I have found the proof of $\text{Closed - Graph -Theorem}$ to be very instructive.

Proof of the Theorem from Planet Math.

Let $T\colon X\to Y$ be a linear mapping. Denote its graph by $G(T)$, and let $p_1\colon X\times Y\to X$ and $p_2\colon X\times Y\to Y$ be the projections onto $X$ and $Y$, respectively. We remark that these projections are continuous, by definition of the product of Banach spaces.

If $T$ is bounded, then given a sequence $\{(x_i, Tx_i)\}$ in $G(T)$ which converges to $(x,y)\in X\times Y$, we have that $$x_i = p_1(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_1(x,y) = x$$ and $$Tx_i = p_2(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_2(x,y) = y,$$ by continuity of the projections. But then, since $T$ is continuous, $$Tx = \lim_{i\to\infty} Tx_i = y.$$ Thus $(x,y) = (x,Tx)\in G(T)$, proving that $G(T)$ is closed.

Now suppose $G(T)$ is closed. We remark that $G(T)$ is a vector subspace of $X\times Y$, and being closed, it is a Banach space. Consider the operator $\tilde T:X\to G(T)$ defined by $\tilde Tx = (x,Tx)$. It is clear that $\tilde T$ is a bijection, its inverse being $p_1|_{G(T)}$, the restriction of $p_1$ to $G(T)$. Since $p_1$ is continuous on $X\times Y$, the restriction is continuous as well; and since it is also surjective, the open mapping theorem implies that $p_1|_{G(T)}$ is an open mapping, so its inverse must be continuous. That is, $\tilde T$ is continuous, and consequently $T = p_2\circ\tilde T$ is continuous.

• We have $G(T) = f^{-1}(0) = \ker{f}$, where $f: X \times Y \to Y$ is given by $f(x,y) = y - T(x)$, hence $G(T)$ is closed. – t.b. Apr 23 '12 at 11:37