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Assume $f\in C^{\infty}(0,1)\times(0,1)$. Is the following true?
$$\Bigg(\int_0^1 \Big|\frac{\partial^nf(t,y)}{\partial y^n}\Big|^2dy\Bigg)^{1/2}\leq c^{n+1}n!,\,\,\,\,n=1,2,\cdots,$$ for all $t\in(0,1),$ where $c>0$ is a constant.

BigM
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  • did you try proving it when $f(t,y) = \left(\sum_{m=0}^\infty b_{m} t^m\right)\left( \sum_{n=0}^\infty a_n y^n\right)$ then extending it to any $C^\infty$ function ? – reuns Feb 24 '16 at 23:47
  • another more direct way (by the Parseval equality) would be to go to the Fourier series : $f(t,y) = \sum_{m,n} c_{n,m} e^{ i \pi (n t+m y)}$, where (because of the smoothness of $f$) the $c_{n,m}$ are more than polynomially decreasing. when $k\to \infty$: $c_{ak,bk} = o(k^{-\delta})$ for every $\delta > 0,a,b$ – reuns Feb 24 '16 at 23:56
  • @user1952009 I tried that. becomes a little messy though. Intuitively, it seems to be true. – BigM Feb 25 '16 at 00:48
  • with the Fourier series, the question becomes : if a sequence $(a_n)$ is more than polynomially decreasing (i.e. $\sum_{n=1}^\infty n^{2k} |a_n|^2$ converges for every $k$) does there exists a constant $c$ such that for every $k$ : $$\sum_{n=1}^\infty n^{2k} |a_n|^{2} < c^{2k+2} (k!)^2$$ – reuns Feb 25 '16 at 01:11

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This won't work, not even for fixed $t$. As already suggested in the comments, I can show this by finding a sequence $a_k$ such that $$ S(n):=\sum k^{2n}|a_k|^2 < \infty $$ for all $n\ge 1$, but no matter what $C>0$ you try, it won't be true that $S(n)\le C^{2n+2}n!^2$ for all $n$.

Most of my $a_k$'s will be zero. I start out with $a_{N_1}=1$, where $N_1\gg 1$ is chosen such $N_1^2>1^{2+2}1!^2$, thus ruling out all $C\le 1$. Next, I will I choose $N_2\gg N_1$ and $a_{N_2}$ such that: (1) $C\le 2$ is similarly ruled out, by considering $n=2$ and the two terms of $S(2)$ I have so far; (2) $N_2^2 |a_{N_2}|^2\le (1/2) N_1^2$, say (this will help ensure that $S(1)<\infty$).

I continue in this way, always using the next (small) $a_{N_m}$ to make $S(m)$ huge, while not making the earlier $S(j)$'s, $j<m$, much larger than they already were. This can be done because the contribution to $S(m)$ has an extra factor $k^2$, which can be made arbitrarily large by moving out to infinity.

  • :perhaps that was too much to ask. seems to me that under real analyticity assumption of $f$, the inequality holds for all $t\in(0,t)$. – BigM Feb 25 '16 at 16:09