[Math] How to understand the proof of the uniform boundedness principle

functional-analysisgeneral-topologymetric-spacesproof-explanationreal-analysis

The uniform boundedness principle is a quite important result in functional analysis. It can be stated as:

Let $X$ be a Banach space and $Y$ be a normed vector space. Suppose $F$ is a collection of continuous linear operators from $X$ to $Y$. If for all $x\in X$ one has $\sup_{T\in F}\|T(x)\|_Y<\infty$ then $\sup_{T\in F}\|T\|<\infty$.

This result can be proved as follows, using Baire category theorem, according to Wikipedia:

Suppose that for every $x$ in the Banach space $X$, one has:

$$\sup_{T\in F} \|T(x)\|_Y<\infty.$$

For every integer $n\in \mathbb{N}$, let

$$X_n=\{x\in X:\sup_{T\in F}\|T(x)\|_Y\leq n\}.$$

The set $X_n$ is a closed set and by the assumption,

$$\bigcup_{n\in \mathbb{N}}X_n=X\neq \emptyset.$$

By the Baire category theorem for the non-empty complete metric space $X$, there exists $m$ such that $X_m$ has non empty interior, i.e., there exists $x_0\in X_m$ and $\epsilon > 0$ such that

$$\overline{B_{\epsilon}(x_0)}=\{x\in X : \|x-x_0\|\leq \epsilon\}\subset X_m.$$

Let $u\in X$ with $\|u\|\leq 1$ and $T\in F$. One has that:

$$\|T(u)\|_Y=\epsilon^{-1}\|T(x_0+\epsilon u)-T(x_0)\|_Y$$

$$\leq \epsilon^{-1}(\|T(x_0+\epsilon u)\|_Y+\|T(x_0)\|_Y)$$

$$\leq \epsilon^{-1}(m+m), \quad \text{since $x_0+\epsilon u$, $x_0\in X_m$}$$

Taking the supremum over $u$ in the unit ball of $X$ it follows that

$$\sup_{T\in F}\|T\|\leq 2\epsilon^{-1}m < \infty.$$

Now I want to understand this proof but it is being a little bit hard to do it. There are some points to it:

  1. First, I can't get any intuition about that proof. How can I get some intuition on why to follow the steps from this proof?

  2. Second, we defined a countable collection of closed sets $X_n$ whose union equals $X$. Baire Category theorem says that if we have a countable collection of open dense sets has dense intersection. How does Baire Category theorem applies to the collection $X_n$ to show that there is $m$ such that $X_m$ has non-empty interior?

  3. Why would finding this $X_m$ be able to furnish the proof?

I mean, in general I can't understand what is going on here with this proof. I want to understand it correctly, and mainly, how Baire Category theorem is being used. How can one understand this proof?

Best Answer

  1. $\newcommand{\norm}[1]{\lVert#1\rVert}$ Intuitively, since $F$ is bounded "near" some $x_0$ (Represented by $X_m$, and the bound is $m$), and linear maps commute with scalings and translations, we wish to show the boundedness of $F$ in the unit ball by transforming the unit ball (Represented by splitting $u = \epsilon^{-1}(x_0 + \epsilon u) - \epsilon^{-1}x_0$) to some ball ($B_\epsilon(x_0)$) centred at $x_0$ for which $F$ is bounded on. The boundedness of $F$ on the entire space follows from this observation.

  2. Every complete metric space is a Baire space, and every Baire space is not a countable union of nowhere dense sets. Hence one of the $X_m$ is not nowhere dense, and contains an open ball. Since $X_m$ is closed, the open ball's closure (A closed ball) is also contained.

  3. The existence of $X_m$ is necessary in this step:

$$ \epsilon^{-1}\norm{T(x_0 +\epsilon u)} \leq \epsilon^{-1} m $$ because $B_\epsilon(x_0)^- \subseteq X_m$ implies $x_0 + \epsilon u \in X_m$ for all $\norm{u} \leq 1$.

Related Question