I'm leanrning Banach Steinhaus Theorem and saw this on Wikipedia on one of the corollary of Banach Steinhaus Theorem:

Since $\{Tn\}$ is bounded in operator norm, and the limit operator $T$ is continuous, a standard "3-ε" estimate shows that $T_n$ converges to $T$ uniformly on compact sets.

I'm trying to prove this as an exercise: but here I don't understand how does being on a compact set and limit operator being continuous might help.

My thought is the followings:

If a sequence of bounded operators $A_n$ converges pointwise, that is, the limit of $\{A_n(x)\}$ exists for all $x$ in $X$, then these pointwise limits define a bounded operator $A$. This directly follow from the Banach Steinhaus Theorem.

Now let $\epsilon > 0$ be given. Since $T_n$ is bounded, so there exists $\delta > 0$ such that $\lVert A_n(x) - A_n(y) \rVert < \frac{\epsilon}{3}$ for $\lvert x - y \rvert < \delta$. Similarly, we can have $\lVert A(x) - A(y) \rVert < \frac{\epsilon}{3}$ and $\lVert A_n(x) - A(x) \rVert < \frac{\epsilon}{3}$ since the limit is continuous and $A_n$ converges point-wise. Therefore, we can have $\lVert A_n(x) - A(x) \rVert < \epsilon$. But now when taking the $\sup$ over a compact set, I'm not exactly sure why the result should follow.

Any help or comments are welcome!


Given compact set $K$ and $\epsilon>0$, since $T$ is continuous on $K$, for each $x\in K$, there is a $\delta_{x}>0$ such that $|T(y)-T(x)|<\epsilon/3$ for all $y\in B_{\delta_{x}}(x)$, where all such $\delta_{x}$ are chosen such that $\delta_{x}<(\epsilon/3)\cdot 1/(\sup_{n}|T_{n}|+|T|+1)$. Let $K\subseteq B_{\delta_{x_{1}}}(x_{1})\cup\cdots\cup B_{\delta_{x_{N}}}(x_{N})$. Now find an $n_{0}$ such that $|T_{n}(x_{i})-T(x_{i})|<\epsilon/3$ for all $n\geq n_{0}$, $i=1,...,N$.

For all $y\in K$ and $n\geq n_{0}$, then there is an $i$ such that $y\in B_{\delta_{x_{i}}}(x_{i})$, so \begin{align*} |T_{n}(y)-T(y)|&\leq|T_{n}(y)-T_{n}(x_{i})|+|T_{n}(x_{i})-T(x_{i})|+|T(x_{i})-T(y)|\\ &\leq(\sup_{n}|T_{n}|)|y-x_{i}|+\epsilon/3+|T||x_{i}-y|\\ &<\epsilon/3+\epsilon/3+\epsilon/3\\ &=\epsilon. \end{align*}

  • $\begingroup$ Wow this really makes sense! Thank you! $\endgroup$ – nekodesu Mar 25 '18 at 20:56

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