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 $2$).


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.

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

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.