# Let $X,Y$ be banach spaces $T,T_n: X\to Y$ and let $T_n \to T$ pointwise, show $T_n \to T$ uniformly on all compact sets

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?

• Your proof is correct. Commented Jul 14, 2020 at 0:09

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.