$v(x) := \sum_{i\in\Bbb N}\frac1{2^i}u(x − x_i) \in W^{1,p}(B(0, 1))$

Question

We want to prove that there exists a Sobolev function that is unbounded on each open set. Let $N \ge 2$, and $p \in [1, N )$. Then:

(i) For $α ∈ (0, ∞)$, consider the function $u : B(0, 1) → R$ defined as $u(x) := |x|^{−\alpha},$ for $x \neq 0 $ Find for what $p$'s and $\alpha$'s, it holds that $ u \in W^{1,p}(B(0, 1))$;

(ii) Let ${x_i}_{i\in\Bbb N} \subseteq B(0, 1)$ be a dense subset. Let $p, \alpha $ be such that the function u in the previous point is in $W^{1,p}(B(0, 1))$. Define the function

$v : B(0, 1) → R $ as

$v(x) := \sum_{i\in\Bbb N}\frac1{2^i}u(x − x_i)$.

Prove that $v \in W^{1,p}(B(0, 1))$ (Hint: Use (i) and series)

My try:

**I computed in the first part that $0<\alpha<N/p -1$ and p as given in the statement. I am having trouble solving (ii):

To prove that a function is in W^{1,p}(B(0, 1)), I have to do 3 things:

(1) prove that the $L^p$ norm is finite

(2) prove that the partial weak derivatives exists

(3) prove that the $L^p$ norm of the partial derivatives is finite

which are the same things that I did in (i) to derive $0<\alpha<N/p -1$

First of all I don't know if I can exchange integration and differentiation with the series. I know a sufficiente condition for exchanging differentiation and series is uniform convergence and for for exchanging differentiation and integration would be the hypothesis of Lebesgue's dominated convergence theorem but I don't know if that is the case or how to justify it here. Moreover I don't know how to used the density of the given set

Assuming I can exchange differentiation and integral:

$\|v\|^2_p=\sum_{i=1}^{\infty}\int_{|x|<1}\frac1{2^i}|u(x-x_i)|^pdx=\sum_{i=1}^{\infty}\int_{|y+x_i|<1}\frac1{2^i}|u(y)|^pdx\le \sum_{i=1}^{\infty}\int_{|y|\le 2}\frac1{2^i}|u(y)|^pdx$

where the last step is because the set $\{|y+x_i|<1\}\subseteq \{|y|<2\}$, since

by the reverse triangle inequality: $|y|-|x_i|\le |y+x_i|<1\implies |y|\le |x_i|+1< 2$

$=\sum_{i=1}^{\infty}(\int_{|y|<1}\frac1{2^i}|u(y)|^pdx+ \int_{1\le|y|<2}\frac1{2^i}|u(y)|^pdx)\le\sum_{i=1}^{\infty}\frac1{2^i}C\le \infty$

where C is the constant bounded the integral (the first integral is bounded because of the result in (i) and the second because it is an integral over a compact set of a continuous function). IF I assume I can interchange differentiation and series then I can prove something analogous for the partial derivatives, which would prove (1).

How can I prove that I can exchange differentiation and integration, use the density of the set and how do I show (2)?

Answer 1

I would say you don't need to do so if you use the completeness of $W^{1,p}(B(0,1))$. Define $v_n:= \sum_{i=1}^n\frac{u(x-x_i)}{2^i}$. Now, let us prove that $\{v_n\}$ is a Cauchy Sequence in $W^{1,p}(\Omega)$. This is easy to prove: $$\|v_n-v_m\|_{W^{1,p}(B(0,1))}=\|\sum_{i=n+1}^{m}\frac{u(x-x_i)}{2^i}\|_{W^{1,p}(B(0,1))}\leq \sum_{i=n+1}^{m}\|\frac{u(x-x_i)}{2^i}\|_{W^{1,p}(B(0,1))}\leq \frac{\|u\|_{W^{1,p}(B(0,1))}}{2^n}$$ Then, the sequence defined is a Cauchy sequence and, as $W^{1,p}(B(0,1))$ is a complete space, it must converge $v_n\to v\in W^{1,p}(B(0,1))$ and you don't have to worry about the existence of weak derivatives or those things because it is guaranteed.

However, in this case we have understood that the convergence of the series (which is nothing but the convergence of a limit) is understood in $W^{1,p}(B(0,1))$. One could argue that maybe the pointwise limit of the series ($w(x):=\sum_{i=1}^\infty \frac{u(x-x_i)}{2^i}$) may be a different function to $v$ (the limit in $W^{1,p}(B(0,1))$). However, this is not the case because the limit in $L^p(B(0,1))$ implies the limit almost everywhere of a subsequence (see Can the limit of a sequence of functions in $L_p$ and the limit in almost everywhere convergence be different?).