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Consider the Banach space $X$ of Lipschitz functions $g:[0,1] \to \mathbb{R}$ such that $g(0)=0$, with the norm of $g$ given by its Lipschitz constant, i.e. $||g||_L = \sup_{x \neq y } \frac{g(x)-g(y)}{|x-y|}$.

Suppose that for some constant $K$ we have a bounded sequence of functions $\{f_n\} \subseteq X$ such that $||f_n||_L \leq K$ for all $n$. By the Arzelà–Ascoli theorem, there is a subsequence which converges (in the $C^0$ norm) to a continuous function $f:[0,1] \to \mathbb{R}$. However, this convergence is not necessarily in the $||\cdot ||_L$ norm. Nevertheless, the function $f$ is Lipschitz and $||f||_L \leq K$.

When could one expect the subsequence of $f_n$ to converge weakly to the function $f$?

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Do you want the whole sequence to converge? Why should it? –  Alex Degtyarev Mar 1 at 18:42
Your comment basically shows that your space is a dual space since its unit ball is compact for a suitable topology. The subsequence will then converge in the corresponding weak $\ast$ topology. –  alpha Mar 1 at 19:01
To clarify: do you mean weak convergence, or as @alpha suggests, weak-star convergence? –  Yemon Choi Mar 1 at 20:14
Implicit in my comment is that you can't expext weak convergence---otherwise the space would be reflexive which it certainly isn't. One should note that such spaces of Lipschitz functions have been extensively studied (see, e.g., the book by N. Weaver). –  alpha Mar 1 at 21:26

3 Answers 3

Since my comments above didn't evince any response, I have decided to elaborate them to an answer, especially as it is part of a longer and interesting story which goes back to the distinguished mathematicians Arens and Eells in (1955) (Pac. J. Math, vol. 78). They showed that a metric space can be embedded into a Banach space in such a way that a suitable universal property is satisfied (the special case of the unit interval is not particularly relevant here---the natural setting is that of a general metric space). In the language of category theory (which they did not use), this means that you can reflect a suitable category of metric spaces into one of Banach spaces. In order to simplify the notation, I shall assume that the metric space has a base point and its radius is at most $1$---then every Lipschitz function which vanishes at the base point is automatically bounded. It turns out that this space is precisely that predual of the space of Lipschitz functions in question which I mentioned in my comment (more or less a direct consequence of the universal property). Then the remarks in my comments apply. The sequence in your question will converge weak $\ast$ but not weakly, in general.

Your question of when it will converge weakly is a bit vague for me to answer but the following considerations suggest that it won't be easy to give a simple necessary and sufficient condition. One can embed your space in a natural way into the bounded, continuous functions on the square of the metric space minus the diagonal set. So your question can be reinterpreted as one concerning weak convergence in a $C^b(S)$. But the latter is the same as $C(\beta S)$ (the Stone-Čech compactification) and so this convergence is equivalent to uniform boundedness (no problem here) and pointwise convergence on $\beta S$. Interpreting this in your original setting might be a problem.

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Let me first recall the Fourier characterization of the Lipschitz functions: Take $S$ be a compactly supported function in $\mathbb R^n$, equal to 1 near 0. We note $S_\nu$ the Fourier multiplier $S(2^{-\nu} D_x)$ : a function $f$ is Lipschitz iff $$ \sup_{\nu\in \mathbb N}\Vert S_\nu \nabla f\Vert_{L^\infty}<+\infty. $$ A slightly larger space is the Besov space $B^1_{\infty,\infty}$, also called Zygmund class, whose Fourier characterization is $$ \sup_{\nu\in \mathbb N}2^{\nu}\Vert \phi_\nu(D) f\Vert_{L^\infty}<+\infty,\quad \sum_\nu \phi_ \nu(\xi)=1,\quad \phi_ \nu(\xi)=\phi (\xi 2^{-\nu})\quad \text{for $\nu\ge 1$},$$ $\phi$ supported in a ring $1/2\le \vert \eta\vert\le 2$. Now $B^1_{\infty,\infty}$ is a Banach space, dual space of $B^{-1}_{1,1}$. The latter is separable, so bounded subsets of $B^1_{\infty,\infty}$ are sequentially weakly-$\ast$ compact.

N.B. A direct characterization of $B^1_{\infty,\infty}$ is $ \vert f(x+h)+f(x-h)-2f(x)\vert\le C\vert h\vert. $

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Too long for a comment.

Via the distributional derivative, the space $X$ of Lipschitz functions null in zero is linearly isometric to $L_\infty$. The unique predual of this abelian von Neumann algebra is the separable $L_1$, so (as already observed) one has a compact metrizable unit ball of $L_\infty$ for the weak topology of this duality. But to have some kind of compactness in the weak topology of the natural duality with spaces larger than $L_1$ (all $\sigma$-additive Borel measures, or the ones finitely additive of bounded variation) one has to change $L_\infty$ with the abelian von Neumann algebra whose predual is the chosen extension of $L_1$ (hence: your sequence has the wanted property when the sequence of distributional derivatives is in the required von Neumann algebra).

An example of norm-bounded sequence in $L_\infty$ with no accumulation point for the weak topology of the duality with Borel measures (even Dirac's deltas): $f'_n$ is the function with value $1$ in $[0,1/2n)$, then $-1$ in $[1/2n,2/2n)$, then $1$ in $[2/2n,3/2n)$, then $-1$ in $[3/2n,4/2n)$, ...

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