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I have trouble answering the second part of the following exercise. Any help would be appreciated!

Let $(X, \| \cdot \|)$ be a reflexive Banach space. Let $\{ T_n \}_{n = 1}^\infty$ be a sequence of bounded linear operators from $X$ into $X$ such that $\lim_{n \to \infty} f(T_n x)$ exists for all $f \in X'$ and all $x \in X$.

(a) Show that $\sup_{n} \|T_n' \| < + \infty$

(b) Show that the map $S$ defined by $(Sf)(x) := \lim_{n \to \infty} (T_n'f)(x)$ is a bounded linear operator from $X'$ into $X'$.

What I've done so far:

(a): I have done this using the Uniform Boundedness Principle twice after first showing that $\sup_n \| T_n' (f)(x) \| < + \infty$

(b): I think that $\displaystyle \sup_n\|T_n' \|$ would make a good bound since we have just seen that it is finite, but I have not succeeded in proving that so far...

Thank you very much for any hints you can offer me!

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  • $\begingroup$ You mean $\sup\limits_n \lVert T_n'\rVert$ in (b), presumably. Well, for every $x$ and $f$, you have $\lvert S(f)(x)\rvert = \lim \lvert T_n'(f)(x)\rvert \leqslant \limsup \lVert T_n'\rVert\cdot \lVert f\rVert \cdot \lVert x\rVert$. $\endgroup$ – Daniel Fischer Jan 21 '14 at 15:55
  • $\begingroup$ What is $T_n'$? $\endgroup$ – copper.hat Jan 21 '14 at 16:18
  • $\begingroup$ @copper.hat $T_n':X' \to X'$ is the dual operator of $T_n$, given by $T_n'(f)(x) = f(T_nx)$. $\endgroup$ – Albert Jan 21 '14 at 21:19
  • $\begingroup$ @DanielFischer I'm not entirely sure what you mean.. Does your statement imply that $\| S(f)(x) \| \leq \limsup \|T_n'\| \cdot \|f \| \cdot \|x\|$, since $\| S(f)(x) \| = \sup\{ |S(f)(x)|: \|x \| \leq 1 \}$ ? $\endgroup$ – Albert Jan 21 '14 at 21:27
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    $\begingroup$ Let $N = \limsup \lVert T_n'\rVert$ (you could also take $\sup \lVert T_n'\rVert$, that's not important). You have shown $N < \infty$. Also, for all $x\in X, f \in X'$, you have $\lvert S(f)(x)\rvert \leqslant N\lVert f\rVert \cdot \lVert x\rVert$. That immediately implies $\lVert S(f)\rVert \leqslant N\lVert f\rVert$ for all $f\in X'$, and that alone implies that $S$ is continuous with $\lVert S\rVert \leqslant N$. You don't need any further application of Banach-Steinhaus. $\endgroup$ – Daniel Fischer Jan 21 '14 at 22:01
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After you have shown that $A := \sup\limits_n \lVert T_n'\rVert < \infty$, you have a known bound on $S(f)(x)$ for every $x\in X$ and $f\in X'$, namely

$$\lvert S(f)(x)\rvert = \lim_{n\to\infty} \lvert T_n'(f)(x)\rvert \leqslant \limsup_{n\to\infty} \lVert T_n'(f)\rVert\cdot \lVert x\rVert \leqslant \limsup_{n\to\infty} \lVert T_n'\rVert\cdot \lVert f\rVert\cdot \lVert x\rVert.$$

From $\lvert S(f)(x)\rvert \leqslant N\lVert f\rVert\cdot \lVert x\rVert$, we deduce by the definition of the norm on $X'$ that $\lVert S(f)\rVert \leqslant N\cdot \lVert f\rVert$, and this in turn shows that $S$ is a continuous (bounded) operator on $X'$ with $\lVert S\rVert \leqslant N$.

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