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

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

• Many things are not clear: what is a RKHS? If $f$ is a fixed function, where a linear operator appears? Is $x$ fixed? Please, edit. Aug 13 '20 at 10:25
• RKHS - Reproducing Kernel Hilbert Space; $f$ is not fixed, but has an argument $x$; $x \in \Omega \subseteq \mathbb{R}^d$ Aug 13 '20 at 12:08
• $x\mapsto f(x)$ is not linear. I do not understand what you are talking about. Aug 13 '20 at 12:17
• But I think your question has an immediate counterexample: take $d=1$, $\Omega=(0,1)$, $s=1$. Then $f'$ can be any $L^2$ function, but clearly you cannot control the pointwise values of an $L^2$ function in terms of its $L^2$ norm (they are not even well defined). Aug 13 '20 at 16:15
• I think that you may want to study the standard Sobolev inequalities and embedding theorems, as they give clear answers as to exactly what results of this kind are true. For example, they indicate why $s>d/2$ is necessary to get a RKHS. They are essentially the first thing that anybody working with Sobolev spaces needs to learn. Aug 13 '20 at 17:25

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 $$|f'(0)| \le M' \|f'\|_{L^2(-1,1)}$$.
• Thx. A question: Does this change if $s > d/2$? Aug 13 '20 at 17:23
• @qwer1304: My example does have $s>d/2$. You can modify it to use $s$ as large as you wish, if you take $g$ to be a smooth function instead. Aug 13 '20 at 17:24
• OK, clearly this cannot work with $\left\|\cdot\right\|_{L^2}$ since $L^2$ is not an RKHS. Can some reasonable norm be defined s.t. the bound does work? Aug 13 '20 at 17:31
• I think that you will have an inequality like $|\partial^\alpha f(x)| \le M'(x, \alpha) \|f\|_{H^s}$ under an assumption like $s > \frac{p}{2} + |\alpha|$. Again, it would come from classical Sobolev inequalities. Aug 13 '20 at 17:34