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Let $I$ be the open interval $(0, 1)$. Let $f, f_n \in H^1 (I)$. We consider

  • (S1) $f_n \to f$ in the weak topology of $H^1 (I)$.
  • (S2) $\|f_n -f\|_{L^2} \to 0$.

Is it true that (S1) $\implies$ (S2) or that (S2) $\implies$ (S1)?

Thank you so much for your elaboration!

Akira
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  • We do not have $(S2)\Rightarrow (S1)$ simply because we have no control over the derivatives. Let's consider $(-1,1)$ instead of $(0,1)$ (you can shift and rescale, but it would just make things look more complicated than they are). Pick $\varphi\in C_c^\infty(-1;1)\setminus {0}$ and consider $f_n(x)=n^{1/3}\varphi(nx)$. Then we have $$ \Vert f_n \Vert_{L^2(-1,1)}^2 = n^{-1/3} \int_{-1}^1 n \varphi(nx)^2 dx = n^{-1/3} \Vert \varphi \Vert_{L^2(-1,1)}^2 \rightarrow 0. $$ Thus, $f=0$ which is in $H^1(-1,1)$. Now pick $g(x)=x$ which is in $H^1(-1,1)$ and test for weak convergence. – Severin Schraven Nov 22 '23 at 22:55
  • [cont] We have $$ \langle f_n, g \rangle_{H^1(-1,1)} = \langle f_n, g \rangle_{L^2(-1,1)} + \langle f_n', g' \rangle_{L^2(-1,1)}. $$ We know the first term goes to zero by Cauchy-Schwarz. On the other hand, the second term blows up $$ \langle f_n', g'\rangle_{L^2(-1,1)} =\int_{-1}^1 f_n'(x) dx = \int_{-1}^1 n^{1+1/3} (\varphi')(nx)dx = n^{1/3}\int_{-1}^1 \varphi'(y) dy.$$ Thus, if we pick $\varphi$ in such a way that $\int_{-1}^1 \varphi'(x) dx \neq 0$ we will not have weak convergence. – Severin Schraven Nov 22 '23 at 23:01
  • @SeverinSchraven I wonder why you pick $\varphi\in C_c^\infty(-1;1)\setminus {0}$ but not simply $\varphi\in C_c^\infty(-1;1)$? – Akira Nov 22 '23 at 23:06
  • The implication $(S1) \Rightarrow (S2)$ is true. See here https://math.stackexchange.com/questions/4064577/weak-convergence-in-w1-2-implies-strong-convergence-in-l2 – Severin Schraven Nov 22 '23 at 23:07
  • I meant to write $\varphi\in C_c^\infty((-1,1)) \setminus {0}$. I just need that $\varphi$ is not identically zero (to get $\Vert \varphi \Vert_{L^2(-1,1)}\neq 0$). – Severin Schraven Nov 22 '23 at 23:08
  • @SeverinSchraven You solved my question thoroughly. Could you write your comment as an answer so that I can accept it? – Akira Nov 22 '23 at 23:10
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    I am a bit in a hurry, but maybe you could write it up? I also left out some details. – Severin Schraven Nov 22 '23 at 23:11
  • @SeverinSchraven It seems you have performed all the necessary computations... – Akira Nov 22 '23 at 23:12

1 Answers1

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We show first that $(S1) \Rightarrow (S2)$ using weak convergence in W1,2 implies strong convergence in L2 or Sobolev embedding: the injection of $H^1(I)$ into $L^2(I)$ is compact we get that $H^1(I)$ compactly embeds into $L^2(I)$. Thus, the weakly convergent sequence $(f_n)_n$ in $H^1(I)$ gets mapped to a convergent sequence in $L^2(I)$ (this is a general property of compact operators, see Compact operator maps weakly convergent sequences into strongly convergent sequences).

On the other hand, the converse implication is false. The underlying reason is that $L^2$ convergence is "unable to detect the behaviour of the derivatives". The easiest counterexample is to take a sequence which converges in $L^2$ to the zero function, but the $L^2$ norm of the derivatives explodes.

We start by choosing $\varphi\in C_c^\infty(I)$ and define $f_n(x)=n^{1/3} \varphi(nx)$. Then we compute \begin{align*} \Vert f_n \Vert_{L^2(I)}^2 &= \int_I n^{2/3} \vert \varphi(nx) \vert^2 dx = \int_\mathbb{R} n^{2/3} \vert \varphi(nx) \vert^2 dx \\ &=\int_\mathbb{R} n^{-1/3} \vert \varphi(y) \vert^2 dy = n^{-1/3} \Vert \varphi \Vert_{L^2(I)}^2. \end{align*} Thus, we get $\lim_{n\rightarrow \infty} \Vert f_n \Vert_{L^2(I)}=0$, i.e. $f_n$ converges to the zero function in $L^2(I)$.

Now we check that this sequence does not converge weakly. Recall that for weak convergence we need to check the convergence of $\ell(f_n)$ for all continuous linear functionals $\ell\in (H^1(I))^*$. However, our Sobolev space is a Hilbert space and hence, it is enough to check the convergence of $$ \langle f_n, g\rangle_{H^1(I)} = \langle f_n, g \rangle_{L^2(I)}+ \langle f_n', g' \rangle_{L^2(I)} \rightarrow \langle f, g\rangle_{H^1(I)} $$ for all $g\in H^1(I)$.

By Cauchy-Schwarz we readily see that $$\vert \langle f_n, g \rangle_{L^2(I)} \vert \leq \Vert f_n \Vert_{L^2(I)} \Vert g \Vert_{L^2(I)} \rightarrow 0.$$

For the gradient part we note that $f_n'(x)=n^{1+1/3} \varphi'(nx)$. Thus, the gradient is bad around the origin. Hence, in order to see this blow-up, we should choose a function $g$ which has nice, nonzero, derivative around the origin. This motivates the choice $g(x)=x$ which is in $H^1(I)$ and $g'(x)=1$. \begin{align*} \langle f_n', g'\rangle_{L^2(I)} &= \int_I n^{1+1/3} \varphi'(nx) dx \\ &= \int_\mathbb{R} n^{1+1/3} \varphi'(nx) dx \\ &= n^{1/3}\int_{\mathbb{R}} \varphi'(y) dy \\ &= n^{1/3} \int_I \varphi'(y)dy. \end{align*} Thus, if we choose $\varphi$ in such a way that $$ \int_I \varphi'(y) dy \neq 0, $$ then we get that $$ \vert \langle f_n', g'\rangle_{L^2(I)} \vert \rightarrow \infty. $$ This proves that $(f_n)_n$ does not converge weakly in $H^1(I)$.

Severin Schraven
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  • I really appreciate the detail of your answer. I don't see the advantage of working with the interval $(-1, 1)$ instead of $(0, 1)$... – Akira Nov 23 '23 at 07:59
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    You are right... there is none. I just need that I can rescale. Let me fix this. – Severin Schraven Nov 23 '23 at 08:01
  • You don't need to fix it. Your answer is flawless. – Akira Nov 23 '23 at 08:01
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    It is a bit confusing if I claim that it simplifies the argument when it actually doesn't. This was really out of habit as I typically want to make things symmetric. Here we would only have had an issue if the interval had not contained the origin. – Severin Schraven Nov 23 '23 at 08:06
  • I have a question on compact embedding of $W^{1, p} (I)$ here. If you don't mind, please have a look at it? – Akira Nov 24 '23 at 14:15