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Let $X$ be a Banach space and let $M: X \rightarrow X$ be a linear map. Prov that M is bounded iff there exists a set $S \subset X'$, dense in X', such that for each $\ell \in S$ the functional $m_l$ defined by $$m_\ell(x) = \ell M (x)$$ is continuous on X.

My try: If $M$ is bounded then $\ell M$ is bounded for all $\ell$, hence all $m_\ell$ are continuous, for all subsets $S$. So we need to find a dense one?

On the other hand: suppose all $m_\ell$ is continuous, since the weak limit is unique, $Mx_n \rightarrow y$ and $x_n \rightarrow x$ $\Longrightarrow$ $Mx = y$ and by the closed graph M is bounded/continuous.

It feels like I'm missing something with the denseness of $S$. Should I look at $\ell \in S^c$ also? and do some $\epsilon/2$ argument?

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Your argument "On the other hand: ..." has a gap: how do you argue that $Mx = y$? More precisely: What is that "weak limit" you are talking about? You only know about convergence for $\ell \in S$, not about all $\ell \in X'$. – Martin Dec 25 '12 at 10:10
take $s \not \in S$. $|m_s(x_n) - m_s(x)| = |m_s(x_n) - m_s(x) + \ell x_n - \ell x_n + \ell x - \ell x | \leq 3*\frac{1}{\epsilon}$ for some $\ell \in S$. Will this help me? Now we have weak convergens for $Mx_n$ ? – Johan Dec 25 '12 at 19:30
up vote 2 down vote accepted

Assume that $\lim\limits_{n\to\infty} x_n=x$ and $\lim\limits_{n\to\infty} M(x_n)=y$. Take arbitrary $\ell\in S$, then $$ \ell(y-M(x))= \ell(y)-\ell(M(x))= \ell(\lim\limits_{n\to\infty}M(x_n))-m_\ell(x)= \lim\limits_{n\to\infty}\ell(M(x_n))-m_\ell(x)= \lim\limits_{n\to\infty}m_\ell(x_n)-m_\ell(x)= m_\ell(\lim\limits_{n\to\infty}x_n)-m_\ell(x)= m_\ell(x)-m_\ell(x)=0 $$ Since $S$ is dense in $X^*$ then for all $\ell\in X^*$ wee have $$ \ell(M(x)-y)=0 $$ By corollary of Hahn-Banach theorem it follows that $M(x)-y=0$. Now from closed graph theorem we get that $M$ is continuous.

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great, can you please state that corollary, I cannot find it! – Johan Dec 26 '12 at 12:38
Corollary of Hahn Banach theorem: Let $X$ be a normed space, and $x\in X$. If for all $\ell\in X^*$ we have $\ell(x)=0$, then $x=0$. – Norbert Dec 26 '12 at 12:50
thanks, I was thinking about that, why do we need hahn banach for that? Is it not true without axiom of choice? I was thinking it would be true for all projections... – Johan Dec 26 '12 at 12:54
this is another question – Norbert Dec 26 '12 at 12:55
That is true! thanks – Johan Dec 26 '12 at 13:03

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