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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?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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