Let $\Omega $ be a bounded open subset of $R^n,\; n\ge 1.$
I m asking about the existence of a subregion $\omega\subset \Omega$ such that the map $y\to y_\omega $ from $H^2(\Omega)$ into $L^\infty(\omega)$ is continuous?.
Let $\Omega $ be a bounded open subset of $R^n,\; n\ge 1.$ I m asking about the existence of a subregion $\omega\subset \Omega$ such that the map $y\to y_\omega $ from $H^2(\Omega)$ into $L^\infty(\omega)$ is continuous?. 


I think it would help to specify what you mean by subregion $\omega \subset \Omega$. In general, it is absolutely true for $\omega$ a lower dimensional object. For example, when $n=2$ and $u \in H^1(\Omega)$, you can expect that for most (Lebesgue almost every) slices the restriction $u_\omega$ is absolutely continuous (and therefore continuous and bounded). If you want a subregion of full measure, then the dimension and exponent are important. A nice example is a function like $x^a$, whose derivative is $ax^{a1}$ and second derivative is $a(a1)x^{a2}$. Computing the $L^2$ norm of this second derivative near zero, we find that we require $\int_0^1 r^{2(a2)}\;r^{n1}dr <\infty$ which is equivalent to saying $2(a2)+n1>1$, so that we should have $a>2\frac{n}{2}$. Therefore, if $n > 4$, the function $x^{\epsilon}$ is in $H^2(B)$ but not in $L^\infty(B)$. Then, you can define $u(x):= \sum_n \frac{1}{2^n} xx_n^{\epsilon}$, where $(x_n)_n$ is a dense set and obtain a function which is unbounded in every open set and still in $H^2$, by the above calculation. Also, YangMills comment is right, since we see when $n=4$ it should still be possible, but one should use logarithms. 

