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?
Best Answer
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 $$x_0=\lim\limits_{m\to\infty} x_m$$