Infimum of $\frac{||u'||^p_{L^p}}{||u||^p_{L^p}}$ for $u \in W^{1,p}_0((0,1))$

Question

Good afternoon everyone!

It is very easy to show that the infimum mentioned in the title is strictly positive, but it seems much more difficult to show that it is attained within the Sobolev space of traceless functions.

I do not know what to do to show this, because the best we can do is to extract a minimizing weakly convergent sequence (for $p > 1$) $x_{k_n}$. This does not seem to help since the norm is not weakly continuous.

The only theorem I know for proving such results for weakly sequentially lower semicontinuous functions needs the set to be closed and convex, while the circumference is not.

Can anyone help?

Answer 1

Well, I think I found something. We want to show that the infimum of the norm of the derivative is attained on the intersection of our space with the unit circumference of $L^p$.

All we need is the Relich-Kondrachov theorem, which allows us to extract a weakly convergent minimizing sequence, which is strongly convergent in $L_p$. Therefore if $u$ is the limit, then $||u||_{L^p} = 1$ and $||u'||_{L^p} \leq m$, where $m$ is our infimum.

Moreover, trace is a continuous function with the weak topology of the Sobolev space (easy), therefore $u$ has also a zero trace. Now $||u'||_{L^p} \geq m$, therefore the two inequalities give $||u'||_{L^p} = m$ with $||u||_{L^p} = 1$ and we are done.

Answer 2

Your idea is right. But let me tell you what you are actually doing here. This is actually an minimizing problem in calculous of variation. (of course I believe you are already very well realized about this)

Let me re-write you problem in a more standard way in calculus of variation:

Finding the minimizer of functional $$ E[u]:= \|u'\|_{L^p(0,1)}$$ among the admissible set $$ \mathcal A:=\{u\in W_0^{1,p}(0,1),\,\,\|u\|_{L^p(0,1)}=1\}$$

The first thing you need to prove is $\mathcal A$ is weakly closed w.r.t $W_0^{1,p}$. Yes, it is true, and this conclusion comes from the fact that $W_0^{1,p}(0,1)$ is weakly closed together Relich compactness theorem.

But here I have two questions for you.

a): why $W_0^{1,p}$ is weakly closed? This is not trivial, you need something from functional analysis, actually the Hanh-Banach theorem, to prove this. You said "trace is a continuous function with the weak topology of the Sobolev space (easy)" is "wrong", because from trace estimation, you only have trace is continuous w.r.t "strong" convergence in $W^{1,p}$ norm.

b): Why could you apply Relich? You are actually using a property called "coercivity" of $E$ defined above. Please check it and make sure you understand.

Of course, your idea is right, extract subsequence which is converge to a function $u$, but why this $u$ is the minimizer? That is, why $\|u'\|_{L^p}\leq m$? This is the fact that $L^p$ norm is "weak lower semi-continuous" w.r.t weak convergence in $L^p$. Please make sure you know what I am talking about.

In the end, let me point out this whole progress is called direct method in calculus of variation.

Lastly, the minimum value you found, is actually the best constant $C$ of Poincare inequality which states that there exists $C>0$ such that $$ \|u\|_{L^p(0,1)}\leq C\|u'\|_{L^p(0,1)} $$ for all $u\in W^{1,p}_0(0,1)$