Take the tour ×  Let $X$ be a Banach space; $K\subset X$
nonempty, closed and convex; and $f:K\to \mathbb R$ lower
semicontinuous, convex functional. Let also $f$ be coercive,
i.e., $f(x)\to +\infty$ as $\|x\|\to +\infty$.  Now, it is well-
known that:  If $X$ is reflexive, then $f$ has a minimum.  The
proof goes essentially like this: One takes a sequence $(x_n)$
such that $(f(x_n))\to \inf_{x\in K} f(x)$, which is then
necessarily bounded and hence, due to reflexivity of $X$, also
weakly convergent, say to $x_0$. Then Hahn-Banach yields that
$x_0$ is actually in $K$, and then lower semicontinuity implies
that $f(x_0)\le \inf_{x\in K} f(x)$.  My question:  Does $f$
have a minimum if $X$ is merely a separable dual space?  I do
not see how the above proof could be modified: On one hand one
may still use Banach-Alaoglu to find a weak*-limit of the
sequence. But then I do not see how to conclude.  I am
interested at minimization of a certain functional in $\ell^1$.
Thanks, any help will be appreciated.  share|imp