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Is the set of functions $f$ in $C^1([0,1])$ equipped with $\lVert f \lVert = \lVert f\lVert_\infty+\lVert f'\lVert_\infty$ (Case 1) or $\lVert f\lVert=\lVert f \lVert_\infty$ (Case 2) satisfying $\lvert f(0.5) \lvert \leq 1$ and $\int_0^1 \lvert f'(x)\lvert^2 dx \leq1$ relatively compact in $C^1([0,1])$?

I think this might be an application of the Arzela-Ascoli theorem. So I'd like to show that the set of functions is uniformly bounded and equicontinous. Boundedness is easy since

\begin{align}\lvert f(x)\lvert \leq 1+\int_{0.5}^x \lvert f'(t)\lvert dt\leq1+0.5+\int_0^1\lvert f'(t)\lvert^2 dt \quad \forall x. \end{align}

But equicontinuity seems harder since $\int_0^1 \lvert f'(x)\lvert^2 dx \leq1$ doesn't imply anything about the boundedness of $f'(x)$.

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@DavideGiraudo This is a nice counterexample –  no identity Jun 24 '12 at 22:31

1 Answer 1

up vote 3 down vote accepted

Let $$K:=\{f\in C^1[0,1]\mid |f(0.5)|\leq 1\mbox{ and }\int_0^1|f'(t)|^2\leq 1\}.$$ This set have a compact closure, since for $f\in K$, $x,y\in [0,1]$: $$|f(x)-f(y)|\leq \int_x^y1\cdot |f'(t)|dt\leq \sqrt{x-y},$$ by Cauchy-Scharz inequality. Uniform boundedness have already be shown by the OP, but can also be deduced from equi-continuity and boundedness at one point.

But $K$ hasn't a compact closure for the norm $\lVert f\rVert:=\sup_{0\leq x\leq 1}|f(x)|+\sup_{0\leq x\leq 1}|f'(x)|$. Indeed, consider $f_n(x):=\frac{\sin(nx)}n$. Then $f_n\in K$ for all $k$, and assume that $\{f_{n_k}\}$ is a converging subsequence for $\lVert\cdot\rVert$, to $f$ for example. In particular, $f_{n_k}'$ converges uniformly to $f'$. But $f'_{n_k}(x)=\cos(n_kx)$ and we should have, by uniform convergence, $$f'(0)=\lim_{k\to +\infty}f'_{n_k}\left(\frac{2\pi}{n_k}\right)=\lim_{k\to +\infty}f'_{n_k}\left(\frac{\pi}{2n_k}\right).$$ But the first sequence is always $1$ whereas the other one is always $0$.

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very nice answer! didn't think of applying Cauchy-Schwarz... –  Abhillascha Jun 24 '12 at 22:45

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