Is $\|u\|_\infty\leq C\|u\|_{L^2}$ true for $u \in H^2(\mathbb{R}^{n})$ for $n>3$?

Question

The Sobolev Lemma says if $u \in H^m(\mathbb{R}^n)$ $(m\in \mathbb{N})$, $k\in \mathbb{N}_0$ such that $k<m-n/2$ then $u\in C^k(\mathbb{R}^n)$ and for $|\alpha|\leq k$

$$\sup\limits_x |\partial^\alpha u(x)|\leq C\|u\|_{L^2}$$

For some constant $C$. Especially if $m=2$ and $n\leq 3$ then $0<2-n/2$. Hence

$$\|u\|_\infty\leq C\|u\|_{L^2}\quad (*)$$

Using $(*)$ one can show that

For $n\leq 3$ and $V \in L^2(\mathbb{R}^n)+L^\infty(\mathbb{R}^n)$ and arbitrary $a>0$ there is a constant $b$ such that

$$\|Vu\|_{L^2}\leq a \|\Delta u \|_{L^2}+b\|u\|_{L^2}$$

for all $u \in H^2(\mathbb{R}^n)$.

I want to know if this is also true if $n>3$.

So my question would be

is $(*)$ true for $n>3$ or is the second statement true for $n>3$?

Is it maybe possible to apply $(*)$ to $u(x_1,\ldots,x_{3n})$ when we regard it as $u((x_1,\ldots,x_{n}),(x_{n+1},\ldots,x_{2n}),(x_{2n+1},\ldots,x_{3n}))$?

Answer 1

If there is $c>0$ such that $\|u\|_{L^\infty} \le c \|u\|_{L^2}$ for all $u\in H^m(\mathbb R^n)$, then it is true for all $u \in C_c^\infty(\mathbb R^n)$, hence by density it is true for all $u\in L^2$. This is absurd, so the claim cannot be true.

Then also your initial claim is wrong.