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Let $X,Y$ be banach spaces $T,T_n: X\to Y$ and let $T_n \to T$ pointwise (weak*), show $T_n \to T$ uniformly on all compact sets.

I reason like this:

I claim that $T_n$ are equicontinuous. That is true as by uniform boundedness principle $\|T_n\|\leq M$ for all $n$. Thus $T_n$ are all lipshitz of constant less than $M$, which means they are equicontinuous. Now Pointwise+equicontinuity imply uniform on a compact sets, and so the result follows. Is this correct? IS there another solution to this problem?

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    $\begingroup$ Your proof is correct. $\endgroup$ Commented Jul 14, 2020 at 0:09

1 Answer 1

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Let us assume that $T_n$ does not converge uniformly on every compact set. Then there exists a compact set $K$, an $ε_0 >0$ and a subsequence $(T_n)_{n \in M \subset \mathbb N}$ of the original sequence, as well as a sequence $(x_n)_{n \in M} \subset K$ such that $|| T_n (x_n) -L(x_n)|| \geq ε_0$, for all $n \in M$. Since $K$ is compact, we may assume without loss of generality that there exists a $x \in K$ s.t $x_n \to x$. Notice that

\begin{align} ε_0& \leq ||(T_n-T) (x_n) || \leq || (T_n-T) (x) || + || (T_n-T)(x_n-x) || \\ &\leq || (T_n-T)(x)|| + ||T_n-T||_{op} || x_n-x||. ~~~~~~~~~(1) \end{align}

But, $||x_n -x ||\to 0$ and $|| (T_n-T)(x) || \to 0$. Furthermore, since the sequence of operators $(T_n)$ converges pointwise, it must be pointwise bounded. Applying Banach-Steinhaus thm, we get that $\sup_n ||T_n||_{op}< \infty$. Taking the $ \limsup_{n \to \infty} $ in (1) we end up with $ε_0 \leq 0$ which is absurd.

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