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.