# Approximating with a sequence convergent in multiple Sobolev Norms.

I'm stuck proving a Gagliardo-Nirenberg Interpolation-type inequality.

Typically authors prove the inequality for functions of their favorite regularity and try a density argument. This often requires them to approximate a Sobolev function with a sequence that converges to it in several different Sobolev norms,and I don't know how to guarantee the existence of such a sequence.

In particular, I'm looking for a reason why a function $f$ in $H^2(\Omega)\cap L^\infty(\Omega)$ can be approximated by a single sequence $u_n$ converging in $H^2$ and $L^\infty$. Here $\Omega$ is a bounded region in $\mathbb{R}^2.$

• Essentially, the approximation is performed by convolution. The convoultion maintains the $L^\infty$ bound of the limiting function. So if $f\in H^2\cap L^\infty$ then $\|f* \Phi_\varepsilon\|_\infty\leq \|f\|_\infty$ where $\Phi_\varepsilon$ is a sequence of mollifieres. Commented May 8, 2014 at 17:45
• It won't converge in $L^\infty$ since $f$ would be necessarily continuous which faild if the space dimension is large enough. Commented May 8, 2014 at 17:48
• Is your second comment recanting your first? (I'm a bit confused.) Commented May 8, 2014 at 17:51
• No it isn't. I just wanted to emphasize, that, though you may not find convergence in $L^\infty$, you can still expect, that the approximating sequence and the limit is still bounded in $L^\infty$. Nevertheless, I read over the condition $\Omega\subset \mathbb{R}^2$ and there you can apply the Sobolev embedding theorem as pointed out by gerw Commented May 8, 2014 at 20:28

If $\Omega \subset \mathbb R^2$, then $H^2(\Omega)$ is continuously embedded in $L^\infty(\Omega)$. Hence, the convergence in $H^2(\Omega)$ implies the convergence in $L^\infty(\Omega)$.
As pointed out by Quickbeam2k1 in the comments, one can not approximate a function $f$ from $H^2(\Omega) \cap L^\infty(\Omega)$ in the $L^\infty(\Omega)$-norm by continuous functions if the space dimension is large enough ($n \ge 4$). The reason is that $f$ might be discontinous, but every $L^\infty(\Omega)$-limit of continuous functions is continuous.