A typical application of the direct method for calculus of variations is finding a minimizer of a coercive and weakly lower semi-continuous function $f$ over a reflexive Banachspace $X$.
The proof uses the result of Eberlein–Šmulian and now I'm wondering, if it is possible to give up the assumption that $X$ has to be reflexive and instead require that $X$ is the dual space of a seperable Banachspace.
Of course, one can use the direct method again, if $f$ is weakly-* lower semi-continuous.
I would be glad, if you could show me that not every weakly lower semi-continuous function is weakly-* lower semicontinuous. Thanks
Let us consider $X=l^1$. We denote an element in $l^1$ with $x=(x^1,x^2,...)=(x^i)$. Note that $l^1$ is not reflexive, but it is isomorphic to the dual of a separable Banach space, that is $c_0$, the subspace of $l^\infty$ of sequences tending to zero.
Let us consider $L:l^1\to\mathbb{R}$ given by $L(x)=\sum_{i=1}^\infty x^i$. This is actually the functional given by the element $(1,1,1,...)\in l^\infty$ on the space $l^1$, hence $L$ is linear and continuous, in particular it is weakly-lower semicontinuous on $l^1$. Now we show that $L$ is not weakly*-lower semicontinuous on $l^1$. In fact let us take the sequence $x_n=(0,...,0,-1,0,...)\in l^1$ with a $-1$ in the $n^{th}$ position. Then $L(x_n)=-1$ for all $n$, $L((0,0,0,...))=0$, and the sequence $x_n$ weakly*-converges to $(0,0,0,...)$, in fact for all $y=(y^i)\in c_0$we have:
$$\langle x_n,y \rangle_{l^1,c_0}=-y^n\to0 \qquad n\to\infty.$$
So $0>\liminf_n L(x_n)=-1$, and $L$ is not weakly*-lower semicontinuous on $l^1$.