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 $2$).



Optimal constant for a Sobolev-type inequality
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 


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




---


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. 


