Arzela-Ascoli for L_p-norm

Since I am from a different mathematical field and couldn't find it: Is there something which would be best called an Arzela-Ascoli version for the $$L_p$$-norm, namely:

Let $$X,Y$$ be two nice measurable/normed spaces (i.e. compact, locally compact, hausdorff and whatever properties you might want), e.g. $$X$$ a closed interval and $$Y= \mathbb R$$.

Let now $$f_n$$ be a family of piecewise differentiable maps from $$X$$ to $$Y$$ such that their $$L_p$$-norm converges, i.e. $$||f_n'||_p \to \lambda$$. Does there exist a converging subfamily $$f_k \to f$$ such that $$f$$ is piecewise differentiable and $$||f'||_p=\lambda$$?

The question arose from the following setting: Let $$X$$ and $$Y$$ be two finite metric graphs (you can think of them as closed intervals glued together at their endpoints with metric & measure from $$\mathbb R$$). Given a family of piecewise differentiable maps $$f_n: X \to Y$$, does there exist a piecewise differentiable map $$f: X \to Y$$, such that the $$L_p$$-norm of its derivative $$||f'||_p$$ attains the infimum of $$||f'_n||_p$$?

Here the family $$f_n$$ is the set of all piecewise differentiable maps homotopic to a given function. Hence I think we can even assume that the $$f_n$$ are continuous and piecewise linear, since averaging along edges decreases the $$L_p$$-norm of its derivative , i.e. do we have $$||f'_n||_p \geq |\frac{f_n(b)-f_n(a)}{b-a}|$$ for $$f_n:[a,b] \to \mathbb R$$?

For $$p=\infty$$ such an $$f$$ exists by Arzela-Ascoli, but does this also hold for other $$L_p$$-norms?

EDIT: In the first version of the question the $$f_n$$ and $$f$$ in the box were only continuous and $$||f_n||_p$$ converged to $$\lambda=||f||_p$$.

• No way, e.g. consider $f_n(x) = \sin(nx)$ on $[0,2\pi]$. But Arzela-Ascoli has an equicontinuity assumption, do you have anything like that? – Nik Weaver Nov 15 '19 at 0:57
• In my setting I actually want the derivatives to converge, so practically yes (my spaces are compact). I just recognized that I might have asked the wrong question in the box by trying to make it more understandable. In the box should be that the $f_n$ are piecewise differentiable and the derivatives of $f_n$ converge in $L_p$ norm. I assume it is now too late to change that? – ctst Nov 15 '19 at 1:20

For your interest in a minimal $$f$$, you might want to read a beginners textbook on Sobolev-Spaces and the calculus of variations, especially on the direct method, which is all about this. The beginner-texts won't deal with metric graphs, but I am pretty sure that you can see how to generalize to this from the solution of the problem on intervals (or rather general domains in $$\mathbb{R}^n$$) that you will find there. Let me give you a short overview for $$p>1$$ (the case $$p=1$$ is a bit more tricky in the analysis and it turns out that any monotone function is a minimizer):

The direct method in a simple example

Let $$I=[a,b]$$ be an interval and consider the Sobolev-space $$W^{1,p}(I;\mathbb{R})$$, which is the closure of $$C^1(I;\mathbb{R})$$ in the norm

$$\|f\|^p_{W^{1,p}} = \|f'\|_{L^p}^p + \|f\|_{L^p}^p.$$

In general for every $$f\in W^{1,p}(I,\mathbb{R})$$, which you can approximate in the norm using smooth $$(f_k)_k$$, the weak-derivative $$f' := \lim_{k\to\infty} f_k'$$ is a well defined $$L^p$$ function. It won't be the pointwise derivative, but in many cases you can almost treat it as such. Furthermore one can prove that $$W^{1,p}$$ is a reflexive space (here $$p>1$$ is needed).

Now let's say, we are interested in the class of functions $$A := \{f\in W^{1,p}: f(a) = 0, f(b)=1\}$$. (One needs to make sure that $$f(a)$$ and $$f(b)$$ are well defined here, this is in fact due to the trace-theorem, that you will also find in any relevant text.) And we want to minimize $$\|f'\|_p$$ in this class. (This method is a bit of overkill for such a simple problem, where we could guess the solution.)

Then first we note that there is an obvious bound from below and so we can pick a sequence $$(f_k)_k$$ such that $$\|f_k'\|_{L^p} \to \inf_{g\in A} \|g'\|_{L^p}$$. In particular then $$\|f_k'\|_{L^p}$$ is bounded. Now we need the next tool which is Poincare's inequality, which due to $$f_k(a) = 0$$ (which prevents shifting by constants) also implies that $$\|f_k\|_{W^{1,p}}$$ is bounded.

But then there is the Banach-Alaoglu theorem, which is kind of the convergence theorem you were looking for. This tells us that $$f_k$$ has a weak* (=weak, since $$W^{1,p}$$ is reflexive) converging subsequence with some limit $$f \in W^{1,p}(I;\mathbb{R})$$.

Now finally using a bit more of the toolbox: The trace-theorem also shows that the boundary conditions are stable under weak convergence, so $$f \in A$$. And weak convergence of $$f_k$$ in $$W^{1,p}$$ means weak convergence of $$f_k$$ and $$f_k'$$ in $$L^p$$. Finally the norm is lower semi-continuous under weak convergence. There is no norm convergence as you requested in your question, but the point is that lower semicontinuity is enough, as we get

$$\inf_{g\in A} \|g'\|_{L^p} \leq \|f'\|_{L^p} \leq \liminf_{k\to\infty} \|f_k'\|_{L^p} = \inf_{g\in A} \|g'\|_{L^p}$$

So $$f$$ is indeed a minimizer in the class $$A$$. Now the final step would be to use the Euler-Lagrange equation to show that $$f$$ is in fact affine (which would translate to piecewise affine in you case).

However...

All this being said there might be another solution to your problem requiring far less hard analysis. In your metric graphs you only ever have to consider piecewise affine functions. Let $$f:[a,b] \to Y$$ be a map from an interval to a metric graph. Now for fixed $$f(a),f(b)$$ and a fixed path from $$f(a)$$ through the vertices $$f(x_1) = p_1,..., f(x_n) = p_n$$ to $$f(b)$$, the minimal $$f$$ is the piecewise affine one. Then since the set of $$x_i$$ with $$a \leq x_1 \leq ... \leq x_n \leq b$$ is compact and $$\|f'\|_p$$ continuous with respect to that choice, there is a minimal $$x_1 < ... < x_n$$ in there. Now there are finitely many loop free paths to choose in a finite graph, so there is a finite choice and thus a minimal $$p_1,...,p_n$$ as well. But then for any $$f: X \to Y$$, with $$f(v_i)$$ fixed on the vertices $$v_i$$, you'll get a minimal $$f$$ by applying the previous reasoning. And now since $$Y$$ is compact, and $$\|f'\|_p$$ should be continuous under the choice of $$f(v_i)$$, you'll get your minimizer. I'm no expert on metric graphs, but I am quite sure something like this has already been written down though.

The characterization of norm compactness in $$L^p$$: you want the Frechét-Kolmogorov theorem, usually stated for $$X=\mathbb{R}^n$$ and $$Y:=\mathbb{R}$$. The proof follows quite easily from Ascoli-Arzelà by mollification. You can find it on Haïm Brezis' Functional Analysis.

• (I forgot to add, from the characterization it also follows that the assumptions in the rectangle are not sufficient for the conclusion) – Pietro Majer Nov 15 '19 at 1:01
• Yes, but for my case it seems to work out, since I actually want to consider the derivatives. In particular if they really can be considered as piecewise linear the assumptions seem to be satisfied (of course, my case is a little bit more complicated as $\mathbb R^n$). The theorem you stated looks like a good step to start with, but some work is left to do (i.e. it gives the derivative of the function I actually wanted). – ctst Nov 15 '19 at 1:08
• Yes, in this new version you do have a convergent subsequence, but you also need to ask convergence of $f_k(x_0)$ for at least one point (think of a divergent sequence of constants). Finally, the limit will be $W^{1,p}$ but not piecewise differentiable, and you only have $\|f\|_p\le p$. Even if the $f_k$ where polynomials on an interval $[a,b]$. – Pietro Majer Nov 15 '19 at 7:01
• Yes, in the general case you are right. The conditions in the box are not enough (so we might really need $Y$ to be compact or your more general condition). In my application it is, hence we have $f_k(x_0)$ converges for some subsequence, so this is fine. Also that the limit is "just" in $W^{1,p}$ seems good to me (I only need continuous and the limit). I don't really get what you meant with $||f||_p \leq p$ though. Did you mean $||f||_p \leq \lambda$? – ctst Nov 15 '19 at 17:32