Is the maximum of derivatives of a function in (s,2)-Sobolev space (an RKHS) bounded by their norms?

Question

Let $f(x) \in W^{s,2}(\Omega) \equiv H^s$, where $\Omega \subseteq \mathbb{R}^d$, $s > d/2$ and $W^{s,2}$ is a $(s,2)$-Sobolev space. Clearly, $W^{s,2}$ is an Reproducing Kernel Hilbert Space (RKHS) and therefore $|f(x)| \le M\left\|f\right\|_{H^s}$ for $M > 0$ holds.

The question is: does this also hold for derivatives of $f$ (they're in $L^2(\Omega)$, but does $|\nabla_x f(x)| \le M'\left\|\nabla_x f\right\|_{L^2}$ hold for some $M' > 0$?)?

Thank you

Update

Following @NateEldredge counter-example, if we tighten the requirements s.t. $s \ge 2$ then it'd seem that a bound on $|\nabla_x f(x)|$ in terms of norms does exist, albeit with a different norm.

Consider the case $d = 1$. As stated above, we assume that $f \in H^s(\Omega)$ and $s > d/2$. By requiring that $s \ge 2$, it follows that $g(x) \equiv f'(x) \in H^{s-1}(\Omega)$. Therefore, $g(x)$ is in an RKHS (albeit a different from $f(x)$'s one). Thus, $|g(x)| = |f'(x)| \le M_g \left\| f' \right\|_{H^{s-1}(\Omega)}, M_g>0$.

A $H^s$ norm verifies $\left\| f \right\|_{H^s}^2 \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha f \right\|_{L^2(\mathbb{R}^d)}^2$. Therefore, $$ |f'(x)|^2 \le M_g^2(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$

This process can be used with any $s \ge 2$.

Is this reasoning correct?

Thx

Answer 1

Consider $d=1$, $s=1$, $\Omega=(-1,1) \subset \mathbb{R}^1$. We can find a function $g \in L^2(-1,1)$ (or even continuous) for which $g(0)$ is arbitrarily large but $\|g\|_{L^2(-1,1)}$ is arbitrarily small. Set $f(x) = \int_{-1}^x g(t)\,dt$; then clearly $f \in H^1(-1,1)$ with $f'=g$, and we can violate any proposed bound of the form $|f'(0)| \le M' \|f'\|_{L^2(-1,1)}$.