Banach-Steinhaus theorem

Let $X,Y$ be normed spaces, $X$ a Banach space. We denote the set of all linear bounded operators from $X$ to $Y$ as:

$$B(X,Y):=\{T:X\to Y\mid T\text{ linear and bounded}\}$$

We equip $B(X,Y)$ with the usual operator norm $\Vert \cdot {\Vert }_{X\to Y}$.

Note

Let $X,Y$ be normed spaces, $T:X\to Y$. A subset $M\subseteq B(X,Y)$ is bounded pointwise on $X$ if:

$$\forall x\in X\exists C_x\ge 0\text{ s.t. }\Vert T_{n}x{\Vert }_Y\le C_x\forall n$$

$M$ is uniformly bounded if:

$$\exists C>0\text{ s.t. }\Vert T_n{\Vert }_{X\to Y}\le C\forall n$$

As usual the difference is the order of the quantifiers.

Note

For every subset $M\subseteq B(X,Y)$, $M$ is bounded pointwise on $X$ iff $M$ is uniformly bounded.

A cool application:

Note

Let $X,Y$ be normed spaces, $X$ a Banach space. Let $T_n\in B(X,Y)$ with ${\lim }_{n\to \infty }{T}_{n}x$ exists for all $x\in X$. Then $T:X\to Y$ defined by $Tx:={\lim }_{n\to \infty }{T}_{n}x$ is linear and bounded

Note M := \{T_n\} is bounded pointwise on X. Using Banach-steinhaus we know that there is a C > 0 such that \Vert T_n \Vert \leq C for all n. Let's compute the operator norm of T:

$$\Vert T\Vert =\sup \{\Vert Tx{\Vert }_Y\mid \Vert x{\Vert }_X=1\}$$

using the definition of $T$

$$\Vert Tx{\Vert }_Y=\Vert \underset{n\to \infty }{\lim }{T}_{n}x{\Vert }_Y=\underset{n\to \infty }{\lim }\Vert T_{n}x{\Vert }_Y\le C$$

we can take out the limit since the norm is a continuous function. Hence $T$ is bounded (linearity is obvious.) $\square$