Let $f_i \in L^1 ([0, 1])$ be a sequence of functions equibounded in $L^1$ norm - that is, there exists some $M > 0$ such that $\|f_i\|_{L^1} < M$.

Define the functional $F: L^1([0, 1]) \to \mathbb R$ by

$$F(h) = \limsup_{i \to \infty} \|f_i - h\|_{L^1}.$$

Question: Does this functional admit a minimiser? Is the minimiser unique whenever it exists?

Remarks:

What I have tried so far is to attempt to apply the direct method of the calculus of variations.

Since the $f_i$ are equibounded in $L^1$, it can be shown that $F$ is coercive, thus any minimising sequence is bounded in $L^1$ norm. In particular we have a weakly-* converging subsequence, say $h_n \overset{*}{\to} h$.

The result would follow if we had weak-* sequential lower semi continuity of $F$ at the minimiser - that is, that

$$\liminf_{n \to \infty} F(h_n) \geq F(h).$$

I could neither disprove this with a counterexample, nor prove it in generality.

Let $f_i \in L^1 ([0, 1])$ be a sequence of functions equibounded in $L^1$ norm - that is, there exists some $M > 0$ such that $\|f_i\|_{L^1} < M$.

Define the functional $F: L^1([0, 1]) \to \mathbb R$ by

$$F(h) = \limsup_{i \to \infty} \|f_i - h\|_{L^1}.$$

Question: Does this functional admit a minimiser? Is the minimiser unique whenever it exists?

Remarks:

What I have tried so far is to attempt to apply the direct method of the calculus of variations.

Since the $f_i$ are equibounded in $L^1$, it can be shown that $F$ is coercive, thus any minimising sequence is bounded in $L^1$ norm. In particular we have a weakly-* converging subsequence, say $h_n \overset{*}{\to} h$.

The result would follow if we had weak-* sequential lower semi continuity of $F$ at the minimiser - that is, that

$$\liminf_{n \to \infty} F(h_n) \geq F(h).$$

I could neither disprove this with a counterexample, nor prove it in generality.

Edit: As pointed out in the comments, weak-$*$ convergence to an $L^1$ function isn’t guaranteed, only convergence to a measure.