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We can prove that in Hilbert space, bounded space has a weakly convergent subsequence, as we can see here or there. Does the converse hold?

That is, in an inner product space $H$, if every bounded sequence has a weakly convergent subsequence, is $H$ complete?

My work:

I thought it is a good start to look at a bounded sequence in an incomplete inner product space. As we see here, the space of the sequence which has finitely non-zero element is incomplete. Then I tried to make a bounded (but not weakly convergent) sequence in this space, but I am failing to do so.

How can I proceed from here?

Kaira
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1 Answers1

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Let $K$ be the completion of $H$. Let $(x_n)$ be a Cauchy sequence in $H$. Then $x_n$ tends to some element $y$ of $K$. Let $x_{n_k} \to x$ weakly in $H$. Let $F$ be a continuous linear functional on $K$. Then $\lim F(x_n)=F(y)$ . By weak convergence of the subsequence we must have $\lim F(x_n)=F(x)$ (since the restriction of $F$ to $H$ is a continuous linear functional on $H$). Thus $F(y)=F(x)$ for al $F\in K^{*}$ which implies $y=x \in H$.

Kavi Rama Murthy
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