uniform boundedness theorem

[[concept]]

Uniform Boundedness Theorem

Let $V$ be a vector space, $B$ be a {==1||Banach space==}, and $\{T_n\}$ a sequence in $\mathcal{B}(B,V)$. Then if for all $b\in B$ we have {==2||${\sup }_n||T_{n}b||<\infty$} (ie the sequence is {2||pointwise bounded}) then we have {3||${\sup }_n||T_n||<\infty$} (ie the {3||operator norms are bounded==})

Proof

For all $k\in \mathbb{N}$, define $C_k=\{b\in B:||b||\le 1,{\sup }_n||T_{n}b||\le k\}$ Then each set is closed because if $\{b_n\}\subset C_k$ and $b_n\to b$ then we have $||b||={\lim }_{n\to \infty }||b_n||=1$. And for all $m\in \mathbb{N}$ we have $||T_{m}b||={\lim }_{n\to \infty }||T_m{b}_n||$ since each of the operators $T_m$ are continuous. And $||T_m{b}_n||\le k$ because $b_n\in C_k$, so $C_k$ must be closed.

Now, we also have $\{b\in B:||b||\le 1\}={\bigcup }_{k\le n}{C}_k$ because for any $b\in B$, there is no $k$ such that ${\sup }_m||T_{m}b||\le k$.

So LHS is complete because it is a closed subset of $M$, and so by Baire’s Theorem, one of the $C_k$ must contain an open ball $B(b_0,{\delta }_0)$.

Thus for any $b\in B(0,{\delta }_0)$, we have that $b_0+b\in B(b_0,{\delta }_0)\subset C_k$ so

${\sup }_n||T_n(b_0+b)||\le k$ Then since both $b_0,b_0+b\in B(b_0,{\delta }_0)$, we have

\sup_{n} \lvert \lvert T_{n}b \rvert \rvert &= \sup_{n} \lvert \lvert -T_{n}b_{0} + T_{n}(b_{0}+b)\rvert \rvert \\ &\leq \sup_{n} \lvert \lvert T_{n}b_{0} \rvert \rvert + \sup_{n} \lvert \lvert T_{n}(b_{0}+b) \rvert \rvert \\ &\leq k +k \end{align}$$ Thus, rescaling, we have for any $n \in \mathbb{N}$ and all $b \in B$ with $\lvert \lvert b \rvert \rvert = 1$, we have $$\left\lvert \left\lvert T_{n}\left( \frac{\delta_{0}}{2}b \right) \right\rvert \right\rvert \leq 2k \implies \lvert \lvert T_{n}b \rvert \rvert \leq 4k$$ ie the [[Concept Wiki/operator norm]] of $T_{n}$ is bounded for all $n$, and therefore its $\sup$ is bounded as well. $$\tag*{$\blacksquare$}$$

References

functional

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