This is a cross post in continuation to [this question](https://math.stackexchange.com/q/4472299/) 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!



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


