## § Uniform Boundedness Principle / Banach Steinhauss

• Consider a set of bounded linear operators $\mathcal F$. If $\mathcal F$ is pointwise bounded, that is, $sup_{T \in \mathcal F}\{ ||T(p)|| \}$ exists for all $p \in X$, then the family is norm-bounded: $\sup_{T \in \mathcal F} \{ ||T|| \}$ exists.

### § Proof 1: Based on an ingenious inequality

• Reference: A really simple elementary proof of the uniform boundedness theorem by Alan D Sokal

#### § Ingenious Inequality, Version 1

• Let $T: X \to Y$ be a bounded linear operator. Then for any $r \geq 0$ we have $\sup_{ ||x|| \leq r } ||Tx|| \geq ||T||r$.
• Proof: recall that $||T|| \equiv \sup_{||x|| = 1} ||Tx||$.
• Now see that $\sup_{ ||x|| \leq r } ||Tx|| \geq \sup_{ ||x|| = r } ||Tx||$.
• This can be rewritten as $r \sup_{ ||x|| = r } || T(x/r) ||$, but this $r \sup_{ ||\hat x|| = 1 } T(\hat x) = r ||T||$.

#### § Ingenious Inequality, Version 2

• Let $T: X \to Y$ be a bounded linear operator, let $p \in X$ be any basepoint. Then for any $r \geq 0$ we have $\sup_{ y' \in B(p, r) } ||Ty'|| \geq ||T||r$.
• We rewrite the optimization problem as $\sup_{ ||x|| \leq r } ||T(p + x)||$.
• First, consider: $\max{||T(p + x)||, ||T(p - x)||} \geq 1/2 [||T(p + x)|| + ||T(p - x)|| \geq ||T(x)||$.
• The last inequality follows from $||\alpha|| + ||\beta|| = ||\alpha|| + ||-\beta|| \leq ||\alpha + (-\beta)||$, that is, triangle inequality.
• Now we see that:
\begin{aligned} &\sup_{||x|| \leq r} ||T(p + x)|| = \sup_{||x|| \leq r} \max(||T(p + x)||, ||T(p - x)||) &\sup_{||x|| \leq r} \max(||T(p + x)||, ||T(p - x)||) \geq \sup_{||x|| \leq r} ||T(x)|| &\sup_{||x|| \leq r} ||T(x)|| = ||T||r \end{aligned}
• and thus we get the bound that $||\sup_{||x|| \leq r} ||T(p + x) \geq ||T||r$.

#### § Proof of theorem

• Suppose for contradiction that $\sup_{T \in \mathcal F} ||T|| = \infty$, which it is indeed pointwise bounded (for all $p$ $\sup_{T \in \mathcal F} ||Tp||$ is bounded).
• Then choose a sequence $T[n]$ such that $||T[n]|| \geq 4^n$. This is possible since the set is unbounded.
• Next, create a sequence of points, such that $x[0] = 0$, and $||x[n] - x[n - 1]|| \leq 3^{-n}$ (that is, $x[n]$ is a $3^{-n}$ radius ball around $x[n-1]$.
• See that this sequence is cauchy, and thus converges. In particular, let the limit be $L$. Then we can show that $||L - x[n]|| \leq 3^{-n}(1 - 1/3)) = 2/3 3^{-n}$.
• Also see that we have the bound $||T_n L || \geq 2/3 3^{-n} 4^n = 2/3 (4/3)^n$.
• Thus, $\lim_{n \to \infty} ||T_n L|| \to \infty$.
• But this contradicts the pointwise boundedness of $\mathcal F$ at the point $L$. Hence proved.

### § Proof 2 using Baire category

• Suppose that for every $x \in X$, $\sup_{T \in \mathcal F} ||T(x)|| < \infty$.
• We want to show that $\sup_{T \in \mathcal F} ||T|| < \infty$.
• For every integer $n \in \mathbb N$, we build the subset $X_n \equiv \{ x \in X : \sup_{T \in \mathcal F} ||T(x)|| \leq n \}$.
• Since for every $l \in X$, there is some $n_l$ such that $||T(l)|| < n_l$ (by assumption, $\mathcal F$ is pointwise bounded), we know that the sets $X_n$ cover $X$.
• Furthermore, each $X_n$ is closed: A cauchy sequence of points such that $||T x_n|| \leq k$ will converge to a limit $L$such that $||T L|| \leq k$.
• Thus, by the baire category theorem, there is a ball $B(p, r) \subseteq X_m$ for some $m \in \mathbb N$, $r > 0$.
• Now this means that the set $B(p, r)$ is norm bounded as $\leq m$.
• But this is a linear space, once we trap one ball we trap them all. By rescaling and translation, we can move the norm boundedness of $B(p, r)$ into the norm boundedness of $B(origin, 1)$ at which point we have proven that $||T(x)|| \leq infty||$.
• Now let $||u|| \leq 1$ and $T \in \mathcal F$. Calculate:
\begin{aligned} &||Tu|| \\ & = 1/r ||T (p + r u) - T(p)|| \\ & \text{(triangle inequality:)} \\ & \leq 1/r (||T(p + ru)|| + || T(p)|| & \text{(p + ru, p \in B(p, r))} \\ & \leq 1/r (m + m) \end{aligned}
• This bound of $2m/r$ does not in any way depend on $T$ or $u$, then $\sup_{T \in \mathcal F} ||T|| < \infty$, which establishes the bound.