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!
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 $H^1(I)$.