Boundedness of a sequence of functions

Question

Let $\Omega$ be a bounded domain and $f_n\in L^2(\Omega)$ be a sequence such that $$\int_\Omega f_nq\operatorname{dx}\leq C<\infty\qquad \text{for all}\quad q\in H^1(\Omega),\ \|q\|_{H^1(\Omega)}\leq1,\ n\in\mathbb{N}.\quad (1) $$ Is it then possible to conclude that $$ \sup_{n\in\mathbb{N}}\|f_n\|_{L^2(\Omega)}\leq C. $$

Here, $H^1(\Omega)$ denotes the Sobolev-Hilbert-Space $H^1(\Omega)$.

Obviously, this statement would be true if we were to replace (1) with $$\int_\Omega f_nq\operatorname{dx}\leq C<\infty\qquad \text{for all}\quad q\in L^2(\Omega),\ \|q\|_{L^2(\Omega)}\leq1,\ n\in\mathbb{N}. $$

and maybe the dense and compact embedding $H^1(\Omega)\hookrightarrow L^2(\Omega)$ is of help but I'm not sure of it.

Edit: By now I'm pretty sure, that this statement doesn't hold. We only have a bound in the dual of $H^1(\Omega)$. But until now I'm failing to compile a conclusive argument!

Answer 1

Take $\Omega:=(0,1)$ and $f_n(x):=(x+1/n)^{-1/2}$. Then for all $q\in H^1(0,1)$ and all integer $n$, using continuity of the trace operator $\Gamma$, $$\int_0^1 f_n(x)q(x)dx=-2\int_0^1\sqrt{x+1/n}q'(x)dx+2\Gamma(q)\leqslant C\lVert q\rVert_{H^1}.$$ But $\lVert f_n\rVert_2^2=\int_0^1\frac 1{x+1/n}dx=\log(n+1)-\log n+\log n=\log(n+1)$.

Answer 2

Davide's counterexample is absolutely valid.

I wish to add that, although the condition given in the question for this sequence $\{f_n\}$ does not guarantee $L^2$-boundedness, it DOES guarantee boundedness with respect to a weaker norm, namely the $H^{-1}$-norm, which is defined as $$ \|f\|_{H^{-1}(\Omega)}=\sup_{\substack{u\in H^1(\Omega), \\ \|u\|_{H^1}=1}}\int_{\Omega} f\,u\,dx. $$