# Discuss strong and weak convergence of a sequence in $W^{1,p}$ Sobolev space

Discuss the strong and weak convergence of the sequence of functions

$$u_n(x)=\frac{1}{n}\sin nx+2\sqrt{x}$$

in the $$W^{1,p}(0,1)$$ Sobolev space.

Pointwise limit is $$u(x)=2\sqrt{x}$$ and can be easily shown that the sequence converge strongly in $$L^p(0,1)\ \forall\ 1\le p \le +\infty$$.

The derivative of the sequence is $$u'_n(x)=\cos nx +\frac{1}{\sqrt{x}}$$ and the pointwise limit $$u'(x)=\frac{1}{\sqrt{x}}$$ (right?).

For $$p=1$$ I would say that

$$\int_0^1 \cos nx\ dx = \frac{\sin nx}{n}\to 0\text{ as }n\to+\infty$$

so the sequence would converge strongly in $$W^{1,1}$$, but actually

$$||u'_n(x)-u'(x)||_{L^1} = \int_0^1 |\cos nx|\ dx$$

and $$\cos nx$$ does not have a fixed sign for $$x\in[0,1]$$ since $$n$$ is going to infinity so how to solve it?

For $$p=+\infty$$

$$||u'_n(x)-u'(x)||_{L^{\infty}} = \max|\cos nx|=1\nrightarrow 0$$

so the sequence doesn't converge strongly in $$W^{1,+\infty}$$.

For $$1, since $$\max(\cos nx)=1$$

$$||u'_n(x)-u'(x)||_{L^p}^p \le 1$$

so we don't have strong convergence, but since $$L^p$$ is reflexive for $$1, by Banach-Alaoglu we can extract a subsequence of $$u'_n$$ converging weakly (but where?). In particular in $$L^2$$ $$\cos nx$$ is a subsequence of the trigonometric basis, which converges weakly to $$0$$ in $$L^2$$. But for others $$p$$ where does the subsequence extracted from $$u'_n$$ weakly converge?

First notice that $$u_n$$ only belong to some of the $$W^{1,p}$$ spaces, as we have $$u_n'=\cos (nx)+\frac{1}{\sqrt{x}}$$ and $$\cos(nx)+\frac{1}{\sqrt{x}}\in L^p(0,1)\iff \frac{1}{\sqrt{x}}\in L^p(0,1)\iff |x|^{-p/2}\in L^1(0,1)\iff p<2$$ Therefore it is only meaningful to discuss convergence in $$W^{1,p}(0,1)$$ for $$p\in [1,2)$$.

There is no pointwise limit for $$u_n'(x)=\cos (nx)+\frac{1}{\sqrt{x}}$$ because there is no pointwise limit for $$\cos(nx)$$ except for $$x=0$$. The sequence keeps on oscillating around $$\frac{1}{\sqrt{x}}$$.

Your Banach-Alaoglu + reflexivity argument works to prove the existence of a weakly converging subsequence if $$p\in (1,\infty)$$. The issue is that it says nothing about the weak limit aside from existence.

To understand the weak limit, there is a general result (see my answer here) - if $$g:\mathbb{R}\to \mathbb{R}$$ is a bounded periodic function such that its mean over a period is $$\alpha$$, then if $$v_n(x):=g(nx)$$ we have $$v_n\rightharpoonup \alpha$$ weakly-star in $$L^{\infty}(\mathbb{R})$$ for $$p<\infty$$. In this case, we have $$g(x)=\cos x$$ whose mean over a period is $$\alpha=\int_0^{2\pi}\cos x\, dx = 0$$. Therefore, $$\cos (nx)\rightharpoonup 0$$ weakly-star in $$L^{\infty}(\mathbb{R})$$, hence $$\cos (nx)\rightharpoonup 0$$ weakly-star in $$L^{\infty}(0,1)$$ and thus (since $$L^{\infty}(0,1)\hookrightarrow L^p(0,1)$$ for $$p\in [1,\infty]$$) we also have $$\cos (nx)\rightharpoonup 0$$ weakly-star (and hence weakly, since they are reflexive) in $$L^p(0,1)$$ for all $$p\in (1,\infty)$$. Finally, we can include $$p=1$$ as weak convergence in $$L^p(0,1)$$ for $$p>1$$ implies weak convergence in $$L^1(0,1)$$.

In conclusion, the above argument shows that the sequence $$u_n$$ is weakly convergent to $$2\sqrt{x}$$ in $$W^{1,p}(0,1)$$ for all $$p\in [1,2)$$.

To study strong convergence, notice that as we have proved, the weak limit of $$\cos (nx)$$ is $$0$$. Therefore, since the strong limit must agree with the weak limit when it exists, by contradiction if $$u_n'\to u$$ strongly in $$L^p(0,1)$$, then we would have $$\cos (nx)\to 0$$ strongly in $$L^p(0,1)$$ and in particular in $$L^1(0,1)$$, but if $$2(k+1)\pi\geq n\geq 2k\pi$$, then \begin{align*}\int_0^1|\cos (n x)|\,dx&= \frac{1}{n}\int_0^n|\cos (y)|\,dy\geq \frac{1}{2(k+1)\pi}\int_0^{2k\pi}|\cos y|\,dy\geq \\ &\geq \frac{k}{2(k+1)\pi}\int_0^{2\pi}|\cos y|\,dy=\frac{2}{(1+1/k)\pi}\not \to 0 \end{align*} a contradiction.

• How can we say that $\cos nx$ does not admit a.e. convergent subsequence? Is it obvious? Of course $\cos nx$ does not converge a.e. ... – Song Jan 18 at 12:00
• I agree that this is not entirely obvious. Because for instance, if we fix $x$ then the sequence $\left\{\cos (nx)\right\}$ is bounded in $\mathbb{R}$ and hence has a converging subsequence. The issue is in finding a converging subsequence which works for a.e. $x\in [a,b]$. With a diagonalization argument you can find a converging subsequence which works for countably many $x$ but a countable set will still have $0$ measure so that doesn't help. – Lorenzo Quarisa Jan 18 at 12:11
• It is true that there is no such subsequence (even on a set of positive measure). But isn't it what you should show in your answer? – Song Jan 18 at 12:14
• 1. The issue is that there is no pointwise limit for $\cos (nx)$, the $\frac{1}{\sqrt{x}}$ term is fine. 2. I think your mistake is that you obtain a multiple of $x^{-p/2+1}$ as a primitive and then you plug $0$ into it to obtain $0^{-p/2+1}=0$. Actually, if $p>2$ then $-p/2+1<0$ and so $\lim_{x\to 0^+}x^{-p/2+1}=+\infty$ which shows that the singularity is non-integrable. – Lorenzo Quarisa Jan 18 at 14:06
• 3. Yes that's a mistake, what is correct is that there is no limit for $\cos(nx)$ except for $x=0$, and so there is no a.e. limit for $u_n'(x)$. 4. This is nothing but the Banach-Alaoglu's theorem since you are deducing weak (sequential) compactness from the boundedness of the sequence $\left\{u_n\right\}$ in $W^{1,p}$ (the result you linked says more than that though). And, it still says nothing about what the limit is. – Lorenzo Quarisa Jan 20 at 20:10