Let $\Omega$ be some open, bounded, smooth subset of $\mathbb{R}^n$. I'm wondering whether it is possible for a sequence of functions $f_n:\Omega \rightarrow \mathbb{R} $ to be strongly convergent to zero in $L^2$ but have non vanishing (e.g. constant) $H^1$ norm $||f_n||_{H^1}^2:=||f_n||_{L^2}^2+||Df_n||_{L^2}^2$.
I can't come up with an example, but I don't see how to prove that the $H^1$ norm should vanish either. Any help would be welcome. Hints are even better than full solutions.
On
Another example. Let $\Omega=(0,1)\subset\mathbb{R}$ and $f_n(x)=\sin(n\,\pi\,x)/n$. Then $$ \|f_n\|_2\le\frac1n\to0\text{ as }n\to\infty. $$ On the other hand $$ \|f'_n\|_2^2=\pi^2\int_0^1\cos^2(n\,\pi\,x)\,dx=\frac{\pi^2}{2}. $$
On
You can take any orthonormal sequence $\{f_n\}$ in $H^1$. Then you have $f_n \rightharpoonup 0$, but $\|f_n\|_{H^1} = 1$. Using the compact embedding from $H^1$ to $L^2$, you have $f_n \to 0$ in $L^2$.
More generally, you can construct a counterexample by using any weakly but not strongly convergent sequence in $H^1$.
I think @PhoemueX's answer will lead you to a general situation. It works, of course. But here let me provide you a quick and insight example on $R^1$.
Take $I=(0,1)$ and define $u_n$ in following way:
For each fixed $n$, we partition $I$ into $n$'s small subinterval with length $1/n$. Let's name those interval by $I^n_i:=(i/n,i+1/n)$ for $i=0,1,2,...,n-1$. That is, $I^n_1=(0,1/n)$, $I^n_2=(1/n,2/n)$ etc...
Now we define $u_n$. Here I am not going to write done the math formula for $u_n$ but will rather just describe it for you. I think it is better this way.
Anyhow, for fixed $n$, we take $I^n_1$ and we build a triangle on $I^n_1$ with length $1/n$ and hight $1/n$, do the same thing for every $I^n_i$. Hence, $u_n$ is just $n$ small triangles with length $1/n$ and hight $1/n$.
Clearly, $u_n\to 0$ in $L^2(0,1)$. But what is $\|u_n'\|_{L^2(0,1)}$? In each $I^n_i$ we have $$ \int_{I^n_i}u_n'^2dx=\frac{4}{n} $$ and hence $$ \int_{I}u_n'^2dx= 4 $$ and we have our result.