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Suppose $X$ and $Y$ are Banach spaces and $T:X \rightarrow Y$ is a BLO and $K$ is a compact subset of $X$. Prove that:

$$\lim_{n\rightarrow \infty} T_n(x) = T(x) \iff \lim_{n\rightarrow \infty} \sup_{x \in K}\|T_n(x) - T(x)\| = 0$$

the '$\Leftarrow$' implication is trivial, but how do I even start with the $\Rightarrow$ implication?

A few hints I got:

We know $\sup_{n\geq1}\|T_n\| < \infty$ and for any $K$: there is a collection of points $\{x_1, \dots, x_l\}$. Such that: $$\sup_{x \in K} \inf_{1\leq i \leq l} \|x-x_i\| < \epsilon$$

Kees Til
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    You're missing for every compact; otherwise you may trivially take $K=\varnothing$. – C-star-W-star Jun 26 '16 at 11:43
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    Are you sure that your question is about the $\Leftarrow$ implication? The hint is much more suited for proving $\Rightarrow$. – saz Jun 26 '16 at 12:04
  • I agree: it is the $\Leftarrow$ which is trivial, once the correction of "every compact set" is done. Just take $K$ to be the singleton ${x}$. And while we are at it: the left side should say "for every $x \in X$". – GEdgar Jun 26 '16 at 12:18

1 Answers1

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As @AlexanderFrei pointed out, it should read

$$\forall x: \lim_{n \to \infty} T_n(x) = T(x) \iff \forall K \subseteq X \, \text{compact}: \lim_{n \to \infty} \sup_{x \in K} \|T_n(x)-T(x)\| = 0.$$

The implication "$\Rightarrow$" is trivial, just choose $K= \{x\}$ for fixed $x \in X$. It remains to prove "$\Leftarrow$".

Suppose that $T_n(x) \to T(x)$ for each $x \in X$ and let $K$ be a compact set. Then for any $\epsilon>0$ the family of open sets

$$B(x,\epsilon):= \{x' \in X; |x-x'| <\epsilon\}, \qquad x \in K,$$

is a covering of $K$, and therefore there exist $x_1,\ldots,x_k \in K$ such that

$$K \subseteq \bigcup_{i=1}^k B(x_i,\epsilon). \tag{1}$$

Since we know that $T_n(x_i) \to T_n(x_i)$ for all $i=,1\ldots,k$, we can choose $N \in \mathbb{N}$ such that

$$\|T(x_i)-T_n(x_i)\| \leq \epsilon \quad \text{for all $n \geq N$ and $i \in \{1,\ldots,k\}$.} \tag{2}$$

Now let $x \in K$. By $(1)$, there exists $i \in \{1,\ldots,k\}$ such that $x \in B(x_i,\epsilon)$. By the triangle inequality,

$$\|T_n(x)-T(x)\| \leq \|T_n(x)-T_n(x_i)\|+\|T_n(x_i)-T(x_i)\|+\|T(x_i)-T(x)\|.$$

Now use $(2)$, $\|x-x_i\| < \epsilon$ and the boundedness of the operator to conclude

$$\|T_n(x)-T(x)\| \leq \left( \sup_n \|T_n\|+\|T\| \right) \epsilon + \epsilon$$

for all $n \geq N$. Since $N$ does not depend on $x$, this proves $$\sup_{x \in K} \|T_n(x)-T(x)\| \xrightarrow[]{n \to \infty} 0.$$

saz
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  • I think that the problem is to proof that $T_n$ converge strongly to $T$ – Hamza Jun 26 '16 at 11:57
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    @Hamza I have edited my answer. I hope this clarifies any confusion. – saz Jun 26 '16 at 12:08
  • I don't really understand the part after: $||T_n(x) - T(x)|| \leq \dots$. I see that the inequality is true but why do the other two terms on the right hand side turn into $\left(sup||T_n|| + ||T||\right) \epsilon$? – Kees Til Jun 26 '16 at 14:47
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    @KeesTil Well... $|T_n(y) | \leq |T_n| \cdot |y|$ for any $y$, right? Now use this inequality for $y:= x-x_i$ and the linearity of $T_n$. This gives you an upper bound for the first term. The estimate of the third one is similar and the second one is less or equal tan $\epsilon$ because of $(2)$. – saz Jun 26 '16 at 17:01