A normed space $X$ is said to be weakly complete if every weakly Cauchy sequence weakly converges in $X$. Is there any example of a space which is weakly complete, but not complete?
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2Check that every norm Cauchy sequence is a weak Cauchy sequence, and that any weakly convergent sequence that is also norm Cauchy is norm convergent. – s.harp Aug 23 '20 at 16:45
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@harp Thanks! I realized that the completion $\bar{X}$ of $X$ contains both weak and strong limit of any norm Cauchy sequence. And they must coincide, therefore it's contained in $X$. – F.Flower Aug 24 '20 at 07:13
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But is there any way to prove it without considering the completion? – F.Flower Aug 24 '20 at 07:14
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To provide an answer without using the norm completion, consider a norm-Cauchy sequence $x_n$ that is weakly convergent to some $x\in X$. By considering $x_n-x$ we may assume that $x_n$ converges weakly to $0$.
Now suppose that $x_n$ is not norm-convergent to $0$, i.e. there is some $\epsilon>0$ with $\|x_n\|>\epsilon$ for infinitely many $n$. By rescaling the sequence with $1/\epsilon$ and by throwing away some terms we may assume $\|x_n\|>1$ for all $n$. Additionally we may pass to a subsequence to get: $$\|x_n-x_2\|≤\frac12.$$ Now let $f$ be some dual element so that $f(x_2)=\|x_2\|≥1$ and $\|f\|=1$. Then for any $n≥2$ you have $f(x_n) = f(x_2)+f(x_2-x_n) ≥\|x_2\|-\|x_2-x_n\| ≥ 1-2^{-1}=1/2$. As consequence $f(x_n)\not\to0$, contradicting that $x_n\to0$ weakly.
Daniel Fischer
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s.harp
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