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Let $X$ be a normed space and $(x_n)$ is a Cauchy sequence in the norm sense. Also assume the $x_n \rightarrow x_0 $ weakly. Then $x_n \rightarrow x_0 $ in norm.

What I did:Take $ \varepsilon >0 $ since $x_n$ is cauchy there a $n_0$ such that $ |\!| x_n-x_m |\!|< \epsilon$ $\forall n,m \geq n_0$ From Hahn Banach there are $x^{*} _n\, \in X^{*}$ such that $|\!| x_n-x_0 |\!| = | x^{*}_n (x_n-x_0)|$ and $|\!| x^{*}_n |\!|=1$. Hence \begin{align} |\!| x_n-x_0 |\!| &= | x^{*}_n (x_n-x_0)|\\ &=| x^{*}_n (x_n-x_m+x_m-x_0)| \\ &\leq | x^{*}_n (x_n-x_m)|+| x^{*}_n (x_m-x_0)|\\ &\leq \varepsilon+| x^{*}_n (x_m-x_0)|. \end{align} Since the last inequality hold $\forall m \geq n_0$ we can take the limit in respect of $m$ and then we get $|\!| x_n-x_0 |\!| \leq \epsilon $ (Since $x_n \rightarrow x_0 $ weakly). And then by definition we are done. Where I saw this exercise there was a hint.

Hint Observe that $x_n \in x_m +\varepsilon B_X$ and $x_m+\varepsilon B_X$ is weakly closed. How do we proceed from there?

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Your solution looks good, but let me make sure I understand: Your actual question is how to use the hint? –  Jonas Meyer Jun 27 '12 at 19:43
    
Yes, yes exactly –  clark Jun 27 '12 at 19:44
1  
Here's another way to think of it that probably doesn't answer your question: Let $Y\supseteq X$ be the (Banach space) completion of $X$, and let $y\in Y$ be the norm limit of $(x_n)$. Then $(x_n)\to y$ weakly in $Y$. Since $X$ is dense in $Y$, $(x_n)\to x_0$ weakly in $Y$. Since the weak topology is Hausdorff, $y=x_0$, so $(x_n)\to x_0$ in norm. –  Jonas Meyer Jun 27 '12 at 19:47

3 Answers 3

up vote 1 down vote accepted

Fix $\varepsilon>0$. Since $\{x_n:n\in\mathbb{N}\}$ is a Cauchy sequence we can find $N\in\mathbb{N}$ such that $n\geq m\geq N$ implies $x_n \in x_m+\varepsilon B_X$

Since $\{x_n:n\geq N\}\subset x_m+\varepsilon B_X$ and $x_m+\varepsilon B_X$ is weakly closed, then $$ x_0=w\lim\limits_{n\to\infty} x_n\in x_m+\varepsilon B_X. $$ Thus $x_0\in x_m+\varepsilon B_X$, which can be reformulated as $\Vert x_m-x_0\Vert\leq \varepsilon$.

Finally for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $m\geq N$ implies $\Vert x_m-x_0\Vert\leq \varepsilon$, hence $$x_0=\lim\limits_{m\to\infty} x_m$$

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For a fixed $\epsilon>0$, there is an $m$ so that for all $n\ge m$ we have $x_n\in x_m+\epsilon B_X$. Then we have $(x_j)_{j\ge m}\subset x_m+\epsilon B_X$; from this and the fact that $x_m+\epsilon B_X$ is weakly closed, we must have $x_0\in x_m+\epsilon B_X$. But then it follows that $\Vert x_n-x_0\Vert<\epsilon$ for all $n\ge m$.

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We have that $x_0\in x_m+\varepsilon B_X$. Indeed, if it's not the case, $x_0$ is in a weakly open set, and we can find a neighborhood $V$ of $x_0$ for the weak topology such that $x_0\in V\subset \complement (x_m+\varepsilon B_x)$. But $x_n$ should be in this neighborhood for $n$ large enough.

But in fact, both approach use Hahn-Banach (in the second it's used for the weak-closeness of the closed unit ball).

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