Sobolev estimates $\|\nabla\phi\|_{\infty}\leq C\|\phi\|_{H^2}$

Question

This is a cross post in continuation to this question on Mathematics Stack Exchange. I wanted to know if this inequality holds true in two or three dimensions,

$\|\nabla\phi\|_{L^{\infty}(\Omega)}\leq C\|\phi\|_{H^2(\Omega)}.$

Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.

Thank you!

Answer 1

In $d \geq 3$ the answer is no from scaling argument.

WLOG we can assume $0\in \Omega$ (by translation) and that $B(0,r_0)\subset\Omega$. Take $\phi\in C^\infty_0(B(0,r_0))\subset C^\infty_0(\Omega)$. Define

$$ \phi_{\lambda}(x) = \lambda^{2 - d/2}\phi(x/\lambda) $$

Note that when $\lambda \in (0,1)$ the function thus defined is still in $C^\infty_0(\Omega)$, and that $\|\phi_\lambda\|_{H^2}$ is uniformly bounded.

However,

$$ \nabla \phi_\lambda(x) = \lambda^{1 - d/2} \phi'(x/\lambda) $$ and we see that this goes to $\infty$ as $\lambda \searrow 0$.

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When $d = 2$ scaling doesn't help. The answer is still negative. The construction is slightly more involved.

Fix $\psi_0$ a Schwartz function such that its Fourier transform satisfies:

• $\mathrm{supp}(\hat{\psi}_0) \subseteq B(0,2)\setminus B(0,1)$
• $\int \xi_1 \hat{\psi}_0(\xi_1, \xi_2) ~d\xi_1~d\xi_2 = \alpha > 0$. Note that this value, up to some complex constant, is equal to $\partial_{x_1}\psi_0(0,0)$.

Let $\beta = \|\nabla^2 \psi_0\|_{L^2}$.

Define $\psi_k$ by $\hat{\psi}_k(\xi) = \hat{\psi}_0(2^{-k}\xi)$. Observe that

• $\partial_{x_1} \psi_k(0,0) = 2^{3k} \alpha$
• $\|\nabla^2 \psi_k\|_{L^2} = 2^{3k} \beta$.

Now let $\gamma_k$ be a sequence of positive real numbers to be determined, and suppose

$$ f_k = \sum_{j = 0}^k \gamma_j \psi_j $$

Then we have

• $\partial_{x_1} f_k(0,0) = \alpha \sum_{j = 0}^k 2^{3j}\gamma_j$
• $\|\nabla^2 f_k(0,0)\|_{L^2} = \beta \left( \sum_{j = 0}^k 2^{6j}\gamma_j^2\right)^\frac12$

So you get a counterexample if you choose $\gamma_j$ to be a sequence such that $2^{3j}\gamma_j$ is square summable but not summable (so something like $\gamma_j = 2^{-3j}\frac{1}{j}$) (this gives a sequence of Schwartz functions on $\mathbb{R}^2$ whose $H^2$ norm is uniformly bounded and whose value $\partial_{x_1}f_j(0,0)$ is unbounded).

If you want to make them have compact support, just truncate all functions by the same smooth cut-off function.