1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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*}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.