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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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    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 $Mx_n$ ?2012-12-25

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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.

  • 0
    That is true! thanks2012-12-26