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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 is $\left\| f \right\|_{H^s} \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha \right\|_{L^2(\mathbb{R}^d)}$. Let $0 < M_f' \le M_f, 0 < M_g' \le M_g$ be constants that verify the RKHS property (squared) as equalities in the two RKHSs, respectively. Then, $$ |f'(x)|^2 = M_g'(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ |f(x)|^2 = M_f'(\left\|f\right\|_{L^2}^2 + \left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$ From which it follows that $$ |f'(x)|^2 = \frac{M'_g}{M'_f}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2 - (M'_f - 1)\left\|f\right\|_{L^2}^2) $$ Which can be made into $$ |f'(x)|^2 \le \tilde{M}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2)$$

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

Is this reasoning correct?

Thx

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

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 is $\left\| f \right\|_{H^s} \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha \right\|_{L^2(\mathbb{R}^d)}$. Let $0 < M_f' \le M_f, 0 < M_g' \le M_g$ be constants that verify the RKHS property (squared) as equalities in the two RKHSs, respectively. Then, $$ |f'(x)|^2 = M_g'(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ |f(x)|^2 = M_f'(\left\|f\right\|_{L^2}^2 + \left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$ From which it follows that $$ |f'(x)|^2 = \frac{M'_g}{M'_f}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2 - (M'_f - 1)\left\|f\right\|_{L^2}^2) $$ This process can be used with any $s \ge 2$.

Is this reasoning correct?

Thx

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 is $\left\| f \right\|_{H^s} \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha \right\|_{L^2(\mathbb{R}^d)}$. Let $0 < M_f' \le M_f, 0 < M_g' \le M_g$ be constants that verify the RKHS property (squared) as equalities in the two RKHSs, respectively. Then, $$ |f'(x)|^2 = M_g'(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ |f(x)|^2 = M_f'(\left\|f\right\|_{L^2}^2 + \left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$ From which it follows that $$ |f'(x)|^2 = \frac{M'_g}{M'_f}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2 - (M'_f - 1)\left\|f\right\|_{L^2}^2) $$ Which can be made into $$ |f'(x)|^2 \le \tilde{M}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2)$$

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

Is this reasoning correct?

Thx

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 is $\left\| f \right\|_{H^s} = \sum_{|\alpha|_1 \le s} \left\| D^\alpha \right\|_{L^2(\mathbb{R}^d)}$. Let $0 < M_f' \le M_f, 0 < M_g' \le M_g$ be constants that verify the RKHS property (squared) as equalities in the two RKHSs, respectively. Then, $$ |f'(x)|^2 = M_g'(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ |f(x)|^2 = M_f'(\left\|f\right\|_{L^2}^2 + \left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$ From which it follows that $$ |f'(x)|^2 = \frac{M'_g}{M'_f}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2 - (M'_f - 1)\left\|f\right\|_{L^2}^2) $$ This process can be used with any $s \ge 2$.

Is this reasoning correct?

Thx

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 is $\left\| f \right\|_{H^s} \doteq \sum_{|\alpha|_1 \le s} \left\| D^\alpha \right\|_{L^2(\mathbb{R}^d)}$. Let $0 < M_f' \le M_f, 0 < M_g' \le M_g$ be constants that verify the RKHS property (squared) as equalities in the two RKHSs, respectively. Then, $$ |f'(x)|^2 = M_g'(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ |f(x)|^2 = M_f'(\left\|f\right\|_{L^2}^2 + \left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2) \\ $$ From which it follows that $$ |f'(x)|^2 = \frac{M'_g}{M'_f}(\left\|f'\right\|_{L^2}^2 + \left\|f''\right\|_{L^2}^2 - (M'_f - 1)\left\|f\right\|_{L^2}^2) $$ This process can be used with any $s \ge 2$.

Is this reasoning correct?

Thx