Let $X$ be a uniformly convex space with norm $\norm{\cdot}$. Suppose that $x_n$ converges weakly to $x$ and $\norm{x_n}\rightarrow \norm{x}$. Then $x_n$ converges strongly to $x$. //Proof// First, the [[http://en.wikipedia.org/wiki/Milman%E2%80%93Pettis_theorem|Milmanâ€“Pettis theorem]] states that every uniformly convex Banach space is reflexive (while the converse is not true). Without loss of generality, we assume that $\norm{x}=1$ (the case when $\norm{x}=0$ is trivial). We prove it by contradiction. There exists $\ep_0>0$ such that $\exists \{n_k\}$ satisfying $$\norm{x_{n_k} - x} \geq \ep_0.$$ We also know that $\norm{x_n}\rightarrow 1$. Hence by the definition of uniformly convexity of the space $X$, we have that $$(*)\hspace{20pt}\limsup_{k\rightarrow \infty} \norm{\f{x_{n_k} + x}{2}} \leq 1 - \delta,$$ for some $\delta >0$. We now use a fact that there exists a hyperplane which strictly separates $x$ and the ball $B_{1-\delta}(0)$ centered at $0$ with radius $1-\delta$. Thus, $\exists \ell\in X'$ such that $$\ell(x) < \ell(y),\quad\forall y\in B_{1-\delta}(0),$$ which contradicts with $(*)$ (may need some minor modification). === Remark The classical $L^p$ space are uniformly convex space for $1<p<+\infty$. === References [[http://en.wikipedia.org/wiki/Uniformly_convex_space|Uniformly convex space]] [[http://en.wikipedia.org/wiki/Clarkson's_inequalities|Clarkson's inequalities]]