corollary of Mazur's lemma I want to show  that if $f$ is convex and lower semi continuous (L.S.C),then it is weakly lower semi continuous (W.L.S.C), so :
By  defenition of l.s.c I have
If $$v_n\rightarrow u~~then~~f(u)\leq lim ~inf~f(v_n) $$and by Mazur's lemma I have ,if $\{v_n\}$converges weakly to u, and $u_n$ is a convex combination of $\{v_n\}$ then $\{u_n\}$converges strongly to u.
Since $f$ is convex
$$u_n=\sum_{i=n}^{N(n)}\lambda_i^{(n)}f(v_i)~~then~~f(u_n)\leq \sum_{i=n}^{N(n)}\lambda_i^{(n)}f(v_i)~~$$
so 
$$lim ~f(u_n)\leq ~lim~\sum_{i=n}^{N(n)}\lambda_i^{(n)}f(v_i)~~ as ~n~\rightarrow \infty $$
I need to conclude that
$$f(u)\leq lim~ inf~f(v_n) ~as ~n~\rightarrow \infty $$
Please help me.
Thanks
 A: The way it is proved for instance in »Direct Methods in the Calculus of Variations« by Giusti is the following:
Since $f$ is lower semi-continuous, its epigraph $\mathrm{Epi}(f) = \{(x,y) \in X \times \mathbb{R} \: | \: y \geq f(x)\}$ is closed. By convexity of $f$, it is also convex. By Mazur's Lemma, it is thus weakly closed in $X \times \mathbb{R}$. But then $f$ is lower semi-continuous with respect to the weak topology.
By the way, your application of Mazur's Lemma is wrong. If $(v_n)_n$ converges weakly to $u$, then there exists a sequence $(u_n)_n$ of convex combinations of the $v_n$ such that $u_n \to v$ strongly.
Edit: Ok, let me show what lower semi-continuity has to do with epigraphs for the case of weak convergence: If $f: X \rightarrow \mathbb{R}$ is lower semi-continuous with respect to weak convergence, i. e. if $x_n \rightharpoonup x$ then $\liminf_{n \to \infty} {f(x_n)} \geq f(x)$. If now $(x_n,y_n) \in \mathrm{Epi}(f)$, i. e. $y_n \geq f(x_n)$ for some $y_n \in \mathbb{R}$ and $x_n \in X$ with $y_n \to y$ and $x_n \rightharpoonup x$ (this is the meaning of weak convergence in $X \times \mathbb{R}$, since weak and strong convergence are equivalent in $\mathbb{R}$), then $$y = \lim_{n \to \infty} {y_n} \geq \liminf_{n \to \infty} {f(x_n)} \geq f(x),$$
so $(x,y) \in \mathrm{Epi}(f)$, which shows that $\mathrm{Epi}(f)$ is closed.
On the other hand, if this is the case and $x_n \rightharpoonup x$, and we assume by contradiction that $y := \liminf_{n \to \infty} {f(x_n)} < f(x)$, then there exists a subsequence $(x_{n_k})_k$ such that $y = \lim_{k \to \infty} {f(x_{n_k})} < f(x)$. But then $\mathrm{Epi}(f) \ni (x_{n_k},f(x_{n_k})) \rightharpoonup (x,y)$, so $(x,y) \in \mathrm{Epi}(f)$, too, which means $y \geq f(x)$, a contradiction.
