Relating function value to $L^2$ norm in Holder space

Question

Suppose that $f \colon [0, 1] \to \mathbb{R}$ is $k$ times continuously differentiable and Holder in the sense that for some $t = k + \beta$, where $\beta \in (0, 1]$ and $k$ is a nonnegative integer and: $$ \max_{j \leq k} \sup_{x \in [0, 1]} |f^{(j)}(x)| \leq 1 \quad \mbox{and} \quad \sup_{x \neq y} \frac{|f^{(k)}(x) - f^{(k)}(y)|}{|x - y|^\beta} \leq 1. $$ We refer to above set of functions as the set $\mathcal{H}(t)$.

Question: I would like to know how we can relate $$ |f(1/2)| \quad \mbox{and} \quad \|f\|_2^2 = \int_0^1 f^2. $$ For instance, is it true that for every $t > 0$ that there exists some $C = C_t$ such that $$ |f(1/2)| \leq C_t \|f\|_2^{2t/(2t + 1)} \quad \mbox{for all}~f \in \mathcal{H}(t) \qquad\qquad (1) $$ holds?

Example for $t \in (0, 1]$: Here, we have $$ |f(1/2)| \leq |f(1/2) - f(x)| + |f(x)| \leq |1/2 - x|^t + |f(x)| $$ Integrating on the set $A(\delta) = \{x : |1/2 - x| \leq \delta\}$ we have (for $\delta \leq 1/2$) $$ |f(1/2)| \leq \delta^t + \sqrt{\frac{1}{2\delta} \int_{A(\delta)} |f(x)|^2} \leq \delta^t + \sqrt{\frac{1}{2\delta}\|f\|_2^2}. $$ If we take $\delta = \|f\|_2^{2/(2t+1)}/2$, then we see that (1) holds for all $t \in (0, 1]$ with $C_t \equiv 2$.

Does this generalize to other Hölder classes? I tried to naively apply a Taylor expansion around $x$, but integrating that seems to lead to a loose bound.

Answer 1

Take any real $t\ge1$, so that $|f|\le1$ and $|f'|\le1$. Let $h:=|f(1/2)|$ and $y:=\|f\|_2$. Then $y\in[0,1]$, $h\in[0,1]$, and $|f(x)|\ge g(x):=\max(0,h-|x-1/2|)$ for all $x\in[0,1]$. So, $$y^2\ge\int_0^1 g^2\asymp h^3.$$ So, for some real $c>0$, $$|f(1/2)|=h\le cy^{2/3}\le cy^{1/2}\le cy^{t/(2t+1)} =c\|f\|_2^{t/(2t+1)}.\quad\Box$$

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If now $t\in(0,1]$, then similarly $|f(x)|\ge g(x):=\max(0,h-|x-1/2|^t)$ for all $x\in[0,1]$ and, for some real $c_t>0$, $$|f(1/2)|\le c_t\|f\|_2^{2t/(2t+1)} \le c_t\|f\|_2^{t/(2t+1)}.\quad\Box$$

Answer 2

Edit

For $t > k \in \mathbb{N}$, we can get the conjectured rate. I leave my original non-sharp answer below the cut.

Start with the Gagliardo-Nirenberg-Sobolev interpolation inequality, which in particular states that for compactly supported $C^k$ functions $g$ on $\mathbb{R}$, there exists a constant $C_k$ such that

$$ \|g\|_{\infty} \leq C_k\|g^{(k)}\|_{\infty}^\theta \|g\|_2^{1-\theta} $$

where $\theta = \frac{1}{1+2k}$.

Now given $f$ in $\mathcal{H}(t)$, fix $\phi\in C^\infty_c((0,1))$, then we can apply GNS to $g = \phi f$.

Observe that $g^{(k)}$ is an expression involving up to $k$ derivatives of $\phi$ and up to $k$ derivatives of $f$. The derivatives of $\phi$ are bounded by some constant $M_k$ once we fixed $\phi$. The derivatives of $f$ are all bounded by $1$. And so GNS implies, $$ \|g\|_\infty \leq C'_k \|g\|_2^{\frac{2k}{2k+1}} $$

If we choose $\phi$ such that $\phi$ takes values in $[0,1]$ and $\phi(1/2) = 1$, we have

$$ |f(1/2)| = |g(1/2)| \leq \|g\|_{\infty} \leq C'_k \|g\|_2^{\frac{2k}{2k+1}} \leq C'_k \|f\|_2^{\frac{2k}{2k+1}} $$

as conjectured.

For non-integer orders, I would check standard references (Adams, or maybe Leoni) to see if there are GNS interpolation inequalities between $L^2$ and a Holder seminorm. (I don't have my copies near me right now.)

Sharpness

The rate $2k/(2k+1)$ is sharp.

Fix $\phi\in C^\infty_0((0,1))\cap \mathcal{H}(t)$, then for $\lambda > 1$, the function $\phi_\lambda(x) = \frac{1}{\lambda^{t}} \phi(\lambda (x-\frac12) + \frac12)$ is also in $C^\infty_0((-,1))\cap \mathcal{H}(t)$. And we have

$$ \phi_\lambda(1/2) = \lambda^{-t} \phi(1/2) $$

while

$$ \|\phi_\lambda\|_2 = \lambda^{-t-\frac12} \|\phi\|_2 $$

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The following rate is probably not sharp, but does converge to 1 as $t$ approaches $\infty$.

1

As a preliminary, observe that if $g$ is a smooth function compactly supported in $(0,1)$, we have

$$ \| g'\|_2^2 \leq \|g\|_2 \|g''\|_2 $$

via integration by parts.

By induction, one proves that

$$ \|g'\|_{2}^{k} \leq \|g\|_{2}^{k-1} \|g^{(k)}\|_2 $$

2

Again for a smooth function compactly supported in $(0,1)$, you have

$$ \|g^2\|_{\infty} \leq 2\|g\|_2 \|g'\|_2 $$

from the fundamental theorem of calculus. So combined with the first step you have

$$ \|g^2\|_{\infty} \leq 2\|g\|_2^{\frac{2k-1}{k}} \|g^{(k)}\|_2 $$

3

Now fix $\phi\in C^\infty_0((0,1))$, with $\phi$ positive, taking values in $[0,1]$, and $\phi(1/2) = 1$. Given $f\in \mathcal{H}(t)$, with $k < t$, consider the function $g = f\phi$. We have

$$ |f(1/2)| = |g(1/2)| \leq 2\|g\|_\infty \leq \|g\|_2^{\frac{2k-1}{2k}} \|g^{(k)} \|_2^{\frac12} $$

Since all of $f$'s derivatives are bounded by 1, and $\phi$ is a fixed function, we have that there exists some $C_k$ (which grows horrendously in $k$) such that $\|g^{(k)}\|_2^{\frac12} \leq C_k$ for all $f\in \mathcal{H}(t)$ with $t > k$.

As $|g| \leq |f|$ pointwise, we find finally

$$ |f(1/2)| \leq 2C_k \|f\|_2^{\frac{2k-1}{2k}} $$

which is

• a shade worse than your initial conjecture
• only applicable for integer $k$.