existence of a minimum for a convex functional on a non-reflexive space

Question

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.

Answer 1

An equivalent formulation of your question is to prove that

$f:X^\prime \to \mathbb{R}$ be written as the supremum of affine functions, i.e.

$f(x^\prime)=\sup_{i \in \mathbb{N}} \langle x^\prime,x_i\rangle+c_i$.

This is equivalent to weak star lower semicontinuity (cf Ambrosio, Fusco, Pallara Functions of Bounded Variation and free discontinuity problems). Perhaps you have some representation that implies this? You don't need the full strength of weakly star lower semicontinuity then (though they are equivalent), since if $x_n^\prime \to x^\prime$ weak star, then we have \begin{align} \liminf_n f(x_n^\prime) &\geq \liminf_n \langle x_n^\prime,x_i\rangle+c_i \newline &=\langle x^\prime,x_i\rangle+c_i \end{align}

and taking the supremum on the right hand side we obtain sequential lower semicontinuity with respect to weak star convergence, which along with your subsequence implies existence of a minimizer. What does your functional look like?