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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 $$ |f(1/2)| \quad \mbox{and} \quad \|f\|_2^2 = \int_0^1 f^2. $$ In particular, one way to address this question is to find the optimal exponent $\rho_t$ such that there exists a positive finite constant $C_t > 0$ such that $$ |f(1/2)| \leq C_t \|f\|_2^{\rho_t}, $$ holds uniformly for $f \in \mathcal{H}(t)$.

In this question, it was shown that $$ \rho_t \leq \begin{cases} \frac{2t}{2t + 1} & t \leq 1 \\ \frac{2}{3} & t \geq 1. \end{cases} $$ (Actually, in one answer to the question linked above, it is additionally shown that $\rho_t \leq \frac{2t - 1}{2t}$ for $t$ a positive integer.)

There are many natural questions, but the following are all essentially related:

(i) Can the estimate above be improved so that $\rho_t \leq \tfrac{2t}{2t + 1}$ for all $t > 0$?

(ii) Can we lower bound $\rho_t$, ideally by the same quantity?

Intuitively, we expect $\rho_t = 1 - o(1)$ as $t \to \infty$, simply because $\mathcal{H}(t)$ is getting smaller and becoming closer to the set of constant functions as $t \to \infty$.

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The following is a copy of my earlier answer which resolves this problem for integer $t$

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

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


The version with Holder endpoint is also true! (This result must be classical, but I am away from my office and library, so cannot easily check for a reference.) Below is a quick sketch of proof; the full argument is a bit messy to type in MarkDown so I put them in a PDF on my website.

The key first step is to observe the following interpolation bound:

If $h\in C^\infty_0(\mathbb{R})$, then $$ \|h\|_\infty^{p\beta + 1} \lesssim \|h\|_p^{p\beta} |h|_{0,\beta} \tag{A} $$ here $\|\cdot\|_p$ is the $L^p$ norm and $|\cdot|_{0,\beta}$ is the $C^\beta$ Holder seminorm.

To see this is true, let $x_0$ be a point where $h$ attains its maximum. Let $y$ be the nearest point to $x_0$ where $h(y) = \frac12 h(x_0)$. The fact that $h$ is Holder means that $$ \frac12 | h(x_0)| = |h(x_0) - h(y)| \leq |h|_{0,\beta} |x_0-y|^{\beta} \tag{B} $$ Observe now that we have the lower bound $$ |\int_{x_0}^y h^p | \geq \frac12 |h(x_0)|^p |x-y| $$ and the two combines to give the desired result.

Now suppose we know that $h = f'$ for some $f\in C^\infty_0(\mathbb{R})$. We also have a different way to estimate (B): writing it as $$ \|h\|_\infty^{1 + 1/\beta} \lesssim |h|_{0,\beta}^{1/\beta} |h(x_0)| |x_0-y| $$ using that $h$ is signed on the interval we have $$ |h(x_0)| |x_0 - y| \leq | \int_{x_0}^y h | = |f(y) - f(x_0)| \leq 2 \|f\|_\infty $$ The $\|f\|_{\infty}$ term we can estimate using (A) with $\beta = 1$ and that $|f|_{0,1} \leq \|f'\|_\infty$ to conclude, for any $p$, $$ \|f'\|_\infty \lesssim \|f\|_{p}^{\frac{\beta p}{\beta p + p + 1}} |f'|_{0,\beta}^{\frac{p+1}{\beta p + p + 1}} \tag{C}$$

After this, we just need to "climb the ladder" (so to speak). A standard induction argument (similar to what is used to prove Gagliardo-Nirenberg-Sobolev using elementary means) gives us the following improvement of (A) and (C): $$ \|f^{(k)}\|_\infty^{p(\beta + k)+1} \lesssim \|f\|_p^{\beta p} |f^{(k)}|_{0,\beta}^{pk+1} \tag{D} $$ which when combined with the GNS inequality listed at the beginning of this answer yields

$$ \|f\|_\infty^{p(\beta + k)+1} \lesssim \|f\|_{p}^{p(\beta + k)} |f^{(k)}|_{0,\beta} \tag{E} $$

Swapping (E) into the argument that is given above the cut gives your desired control.

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  • $\begingroup$ Your update, if I understood correctly establishes the exponent $\rho_t = 2t/(2t+1)$ using the case $p=2$ and your lower bound which doesn’t require $t$ to be an integer, right? Just making sure I understood correctly. $\endgroup$
    – Drew Brady
    Commented Dec 8, 2023 at 15:01
  • $\begingroup$ @DrewBrady: correct. This provides the sharp estimate. $\endgroup$ Commented Dec 9, 2023 at 4:49

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