Optimal constant for a Sobolev-type inequality

Question

Let $\overset{\circ}{H^s}(\mathbb T)$, where $s\ge 0$, be the space zero average of $2\pi$-periodic functions $u(x)=\sum_{k\in\mathbb Z}\hat u_k\,\mathrm{e}^{ikx},$ such that $$ \lvert u\rvert_s = \left(\sum_{k\in\mathbb Z} \lvert k\rvert^{2s}\lvert \hat u_k\rvert^2\right)^{\!1/2}<\infty $$ Then, for $\,u\in \overset{\circ\,\,\,\,\,}{H^{n+1}}(\mathbb T)$, $n$ positive integer, we have that $$ \left|\int_{-\pi}^\pi u^{(n)}(x)\,\big(u(x)u'(x)\big)^{(n)}\,dx\right|\le c\lvert u\rvert_{L^\infty}\lvert u\rvert_{n+1/2}^2, \tag{1} $$ for some $c$ depending on $n$ and not on $u$. I managed to obtain that $c\le 2^{n+1}$, which is extreme crude!

Question. Is it possible to show that $c=\mathcal O(n)$?

My crude estimate is based on the fact that $\lvert u^2\rvert_s\le 2^{s+1}\lvert u\rvert_{L^\infty}\lvert u\rvert_s $, which can not improve significantly (perhaps by at most a factor of $2$).

Answer 1

Taking the Fourier transform and using $L^2$ orthogonality you are equivalently trying to estimate

$$ \sum_{i + j +k = 0} \hat{u}_i \hat{u}_j \hat{u}_{k} |i+j|^{n+1}|k|^{n} $$

Now from the equality $i + j +k = 0$ we have that either

1. $|k| >\max(|i|,|j|)$ in which case $i$ and $j$ have the same sign,
2. $|i| > \max(|j|,|k|)$ in which case $j$ and $k$ have the same sign,
3. same as 2 but with $i$ and $j$ swapped.

In cases 2 and 3 you have

$$ \sum_{\text{case 2}} \hat{u}_i \hat{u}_j \hat{u}_{k} |i+j|^{n+1}|k|^{n} \leq \sum_{\text{case 2}} |\hat{u}_i| |i|^{n+1/2} |\hat{u}_{-i-k}| |\hat{u}_{k}| |k|^{n+1/2}$$

which we control by

$$ \max_j |\hat{u}_j| (\sum_k |\hat{u}_k|^2 |k|^{2n+1}) \leq |u|_{L^1} |u|_{n+1/2} $$

Case 1 is the one in which we may have loss. Your $2^{n+1}$ factor corresponds to the estimate

$$ |i| + |j| \leq 2 \max(|i|,|j|) $$

Informed by this, we can construct a counter example.

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Now let $\kappa$ be real. Let $$u = \sum_{k \geq 0} 2^{2k\kappa} \exp(i 2^{2k} x) + 2^{(2k+1)\kappa}\exp(- i 2^{2k+1} x).$$ We have that $$ \hat{u}_k = \begin{cases} k^\kappa & k = 2^{2m} \\ k^\kappa & k = - 2^{2m+1} \\ 0 & \text{otherwise}\end{cases} $$ So that whenever case 1 of $i+j+k$ happens, we have $i = j$ and $k = -2j$ to saturate the problematic inequality.

It is clear that $|u(x)| \leq u(0) = \sum 2^{k\kappa} = (1 - 2^{\kappa})^{-1}$. Also $$ |u|_{n+1/2}^2 = \sum_{k \geq 0} 2^{2k(n+1/2 +\kappa)} = (1 - 2^{2n + 1 + 2\kappa})^{-1} $$

What we want to estimate can be simplified to

$$ \sum_{k\geq 0} \left(\underbrace{2^{(k+1)(2n+1)}}_{\text{Case 1}} + \underbrace{2\cdot 2^{(2n+1)k + 1}}_{\text{Cases 2 and 3}}\right) 2^{2k\kappa} 2^{(k+1)\kappa} = \left(2^{2n} + 1\right)2^{1+\kappa} \left(1 - 2^{2n+1 + 3\kappa}\right)^{-1}$$

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You asked for $u \in H^{n+1}$, which bounds $\kappa < -n - 1$. We consider the "limiting case" where $\kappa = -n-1$. You are then asking whether the quotient

$$ \frac{ 2^n + 2^{-n} }{1 - 2^{-n-2}} \cdot \frac12 \cdot (1 - 2^{-n-1}) \geq 2^{n-1} \frac{1 - 2^{-n-1}}{1 - 2^{-n-2}}$$

is bounded/linear as $n \to \infty$. The answer is no. In fact, your bound is basically sharp.