Baire Category Theorem
Let \((X,d)\) be a complete metric space, and let \(O_1,O_2,\ldots\) be open dense subsets of \(X\). Then \(\displaystyle\bigcap_{n=1}^{\infty}O_n\) is dense in \(X\).

Recall that \(D\) is dense in \(X\) if the intersection of every nonempty open ball with \(D\) is non-empty. Equivalently, \(X\setminus D\) has empty interior. Taking complements gives a useful form of the Baire Category Theorem: if a nonempty complete metric space is a countable union of closed sets, then at least one of those sets contains an open ball.

Example: Graphs of Continuous Functions

Let \(P_n\) be the graph of a continuous function \(g_n\colon\mathbb{R}\to\mathbb{R}\). Each graph is closed in \(\mathbb{R}^2\) and has empty interior: moving vertically by any nonzero amount from \((x,g_n(x))\) takes us off the graph. Thus \(\mathbb{R}^2\setminus P_n\) is open and dense. Since \(\mathbb{R}^2\) is complete, the Baire Category Theorem gives

$$ \bigcap_{n=1}^{\infty}\bigl(\mathbb{R}^2\setminus P_n\bigr). $$

This says the intersection of all open dense sets is also a dense set. By De Morgan’s law, this is equivalent to

$$ \mathbb{R}^2\setminus\bigcup_{n=1}^{\infty}P_n. $$

While this one says the complement of the union of the graphs is dense. Note here the complement is all the points not lying on any graph.

Proof of the Uniform Boundedness Theorem

Uniform Boundedness Theorem
Let \(V_1\) be a Banach space, let \(V_2\) be a normed space, and let \(\{T_\alpha\}_{\alpha\in A}\) be a family of bounded linear operators from \(V_1\) to \(V_2\). If \[ \sup_{\alpha\in A}\|T_\alpha x\|_2\lt\infty \] for every \(x\in V_1\), then \[ \sup_{\alpha\in A}\|T_\alpha\|\lt\infty \]

That is if we have pointwise boundedness, we get a uniform operator norm if \(X\) is Banach.

Proof: For each positive integer \(n\), define

$$ E_n=\left\{x\in V_1: \sup_{\alpha\in A}\|T_\alpha x\|_2\le n\right\}. $$

An input \(x\) belongs to \(E_n\) precisely when every operator in the family sends it to a vector of norm at most \(n\). Note that by definition of the supremum, for any \(\alpha \in A\), we must have \(\| T_{\alpha}x \|\). Hence we can re-write this to be:

$$ E_n=\bigcap_{\alpha\in A} \left\{x\in V_1:\|T_\alpha x\|_2\le n\right\}, $$

Note that each set in this intersection is closed. Why? fix one \(\alpha\). Then let

$$ F_\alpha=\{x\in V_1:\|T_\alpha x\|_2\le n\} $$

First note that by assumption \(T_\alpha\) is a bounded linear operator and we have shown before that this implies \(T_{\alpha}\) is continous. Now, take a sequence \(x_k\in F_\alpha\) with \(x_k\to x\) in \(V_1\). We need to show \(x \in F_\alpha\) too.

$$ T_\alpha x_k\longrightarrow T_\alpha x. $$

By continuity of the norm,

$$ \|T_\alpha x_k\|_2\longrightarrow\|T_\alpha x\|_2. $$

Every number on the left is at most \(n\), so its limit is also at most \(n\). Thus \(x \in F_\alpha\), proving \(F_\alpha\) is closed. Finally, any intersection of closed sets is closed so \(E_n\) is closed. By hypothesis, every fixed \(x\) has some finite bound. Choosing an integer \(n\) at least as large as that bound gives \(x\in E_n\). Hence

$$ V_1=\bigcup_{n=1}^{\infty}E_n. $$

Next, since \(V_1\) is complete and the sets \(E_n\) are closed, the Baire Category Theorem says that at least one closed set \(E_N\) for some \(N\) has a non-empty interior. By definition, there exists a ball with center \(x_0\in V_1\) and a radius \(\delta>0\) such that

$$ B(x_0,\delta)\subseteq E_N. $$

But now recall that So if \(y\) is in that ball, then \(y\in E_N\). By the definition of \(E_N\), every operator sends this \(y\) to a vector of norm at most \(N\). That is

$$ \|T_\alpha y\|_2\le N \qquad \text{for all }y\in B(x_0,\delta) \text{ and all }\alpha. $$

Now we want to use that bound to control \(T_\alpha x\) for any nonzero \(x\in V_1\), even outside the ball. First, normalize \(x\) so let \(u=\frac{x}{\|x\|_1}\). Then \(\|u\|_1=1\), so \(u\) points in the direction of \(x\) but has length \(1\). Next, starting at the center \(x_0\), move a distance \(\delta/2\) in the direction of \(x\). The resulting point is

$$ z=x_0+\frac{\delta}{2}\frac{x}{\|x\|_1}. $$

Since \(\|u\|_1=1\),

$$ \|z-x_0\|_1 =\left\|\frac{\delta}{2}u\right\|_1 =\frac{\delta}{2}<\delta. $$

So \(z\) is inside our ball, and therefore \(\|T_\alpha y\|_2\le N\) for every \(\alpha\). Linearity and the triangle inequality give

$$ \begin{align*} \|T_\alpha(z-x_0)\|_2 =\|T_\alpha z-T_\alpha x_0\|_2 \le N+N=2N. \tag{1} \end{align*} $$

Now substitute

$$ \begin{aligned} z-x_0=\frac{\delta}{2\|x\|_1}x. \end{aligned} $$

By linearity,

$$ \begin{align*} \|T_\alpha(z-x_0)\|_2 &= \|T_\alpha \left(\frac{\delta}{2\|x\|_1}x\right) \|_2 \\ &= \frac{\delta}{2\|x\|_1}\|T_\alpha x\|_2 \\ &\le 2N. \quad \text{(by (1))} \end{align*} $$

Therefore,

$$ \begin{align*} \frac{\delta}{2\|x\|_1}\|T_\alpha x\|_2 \le 2N \end{align*} $$

Multiplying both sides by \(2/\delta\) gives

$$ \begin{aligned} \boxed{\frac{\|T_\alpha x\|_2}{\|x\|_1}\le\frac{4N}{\delta}.} \end{aligned} $$

Taking the supremum over nonzero inputs gives

$$ \|T_\alpha\| =\sup_{x\ne 0}\frac{\|T_\alpha x\|_2}{\|x\|_1} \le \frac{4N}{\delta}. $$

And \(N\) and \(\delta\) do not depend on \(\alpha\), so

$$ \|T_\alpha\| \sup_\alpha\|T_\alpha\|\le\frac{4N}{\delta}<\infty. \le \frac{4N}{\delta}. $$

as we wanted to show. \(\ \blacksquare\)

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi