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Let $X$ be a subset of $L^2([0,1])$ which contains only Lipschitz function.
Also, the member of $X$ are uniformly bounded $$ |x(t)| < M, \text{ for all $x \in X$ and $t \in [0,1]$}. $$ Let $F: X \to [a,b]$ be a bounded Lipschitz functional $$ |F(x) - F(x')| \le \| x - x' \|, \text{ for all } x \in X. $$ The minimum of $F$ can be approximate by forming a finite $\epsilon$-net $D_m=\{x_i\}_{i=0}^m$ of $X$, and then taking $$ \min \left\{F(x_i) : x_i \in D_m \right\} , \tag{1} $$ as an approximation to $\min\left\{ F(x) : x \in X \right\}$.

What is a good way to construct the $\epsilon$-net $D_m$?

Since the function in $X$ are Lipschitz, I would use truncated Fourier series of order $n$ to build $D_m$.
Is it then true that for any $x \in X$, there is an $x_i \in D_m$ for which $$ \|x - x_i \| \le \frac1n? \tag{2} $$ Or is (2) delivered only by using truncated Chebyshev polynomial of order $n$?

Are there other good choices for building $D_m$ than trigonometric polynomial?

My greatest fear is that $X$ is not compact. So that no $\epsilon$-net can gives (1).
Is this fear justified?
From the different assumptions, I think it can be dispelled, but I don't see the proof.

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    $\begingroup$ A uniformly bounded set of Lipschitz functions is totally bounded even in $C[0,1]$ by Ascoli-Arzela. $\endgroup$
    – Bill Johnson
    Commented Oct 21, 2013 at 3:03
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    $\begingroup$ For your question about (2), see for example en.wikipedia.org/wiki/… $\endgroup$
    – Mark Meckes
    Commented Oct 21, 2013 at 8:03
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    $\begingroup$ @ Bill Johnson: the statement is that each $x\in X$ is Lipschitz, but unless I misunderstood something we have no control on the Lipschitz constants uniformly in $x$ so Ascoli doesn't apply. @Nicolas': your assumption $|x|\leq M$ is ambiguous: do you mean mean $|x(t)|\leq M$ uniformly in $t\in (0,1)$ and $x\in X$, or rather $|x|_{L^2}\leq M$ (in which case Ascoli fails)? From what I see I rather recommend Rellich-Kondrachov ($X\subset H^1\subset\subset L^2$) or Riesz-Frechet-Kolmogorov, but in any case you need a uniform bound on the Lipschitz constants! $\endgroup$
    – leo monsaingeon
    Commented Oct 21, 2013 at 9:07
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    $\begingroup$ Ok, then if you can get uniform Lipschitz bounds Ascoli works as suggested by @BillJohnson and you relative compactness of $X$ in the $L^{\infty}$ topology. But this still doesn't help to construct the best posisble $\epsilon$-net, sorry... $\endgroup$
    – leo monsaingeon
    Commented Oct 21, 2013 at 14:08
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    $\begingroup$ If you do not impose a bound on the Lipschitz constant, $X$ may fail to be relatively compact ; its $L^2$ - closure could be the whole closed ball of radius $M$ of $L^\infty$. Possible remedy: Find a smaller domain which is compact but where $F$ has the same infimum. Or use a weaker topology that makes $X$ compact and $F$ lower semicontinuous. $\endgroup$
    – Pietro Majer
    Commented Oct 21, 2013 at 21:33

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