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 $1<p<+\infty$, 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 $1<p<+\infty$, 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.

  • $\begingroup$ 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. ... $\endgroup$ – Song Jan 18 at 12:00
  • $\begingroup$ 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. $\endgroup$ – Lorenzo Quarisa Jan 18 at 12:11
  • $\begingroup$ 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? $\endgroup$ – Song Jan 18 at 12:14
  • 1
    $\begingroup$ 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. $\endgroup$ – Lorenzo Quarisa Jan 18 at 14:06
  • 1
    $\begingroup$ 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. $\endgroup$ – Lorenzo Quarisa Jan 20 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.