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Given a sequence of positive numbers $\{a_n\}$ and $1 < p < \infty$, $p\neq 2$, is it possible to build a function $f\in C^\infty(\mathbb R)$ so that $\|f^{(n)}\|_{L^p(\mathbb R)} = a_n$?

For what I have in mind, I would be happy with $f$ so that $\frac 1C \leq \frac{\|f^{(n)}\|_{L^p(\mathbb R)}}{a_n} \leq C$ for some $C>0$.

Thanks! Andy

PS I should have added that I'm really interested in sequences that grow in a very specific way, namely, for some fixed $A>0$ and $\beta>1$, $a_n = A^n n^{n\beta}$. However, I think the question is interesting regardless -- trying to know what sequences $a_n$ are allowable.

PPS I exclude $p=2$ because the function $f$ with Fourier transform $\hat f(t) = e^{-a |t|^{1/\beta}}$ works. You can compute $\|t^\ell \hat f\|_{L^2}$.

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    $\begingroup$ why you exclude $p=2$? What is the answer in this case? $\endgroup$ Commented May 19, 2015 at 19:35
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    $\begingroup$ In the case $p=2$, integration by parts gives you $\|f'\|_2^2 = - \int f f'' \; dx \le \|f\|_2 \|f''\|_2$. $\endgroup$ Commented May 19, 2015 at 21:09

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In the case $p=4$, a similar argument to that in my comment above gives

$$ \|f'\|_4^4 = -3 \int f (f')^2 f'' \le 3 \|f\|_4 \|f''\|_4 \|f'\|_4^{2}$$ so that $$ \|f'\|_4^2 \le 3 \|f\|_4 \|f''\|_4 $$

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    $\begingroup$ These inequalities are just 1-dimensional versions of what are known as Gagliardo-Nirenberg inequalities, which are easy to prove using integration by parts and the Holder inequality. In particular, given any $1 < p < \infty$ and $0 \le i \le j \le k$, there is such an inequality. $\endgroup$
    – Deane Yang
    Commented May 20, 2015 at 18:52
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    $\begingroup$ Hi. While these inequalities limit what sequences can be achieved, do you have any suggestions on how to actually build a function with prescribed derivatives in $L^p$? $\endgroup$
    – Andy Raich
    Commented May 21, 2015 at 13:26
  • $\begingroup$ For a finite set of $a_n$, consider linear combinations of a finite set of basis functions and try to solve a nonlinear system of equations in the coefficients. $\endgroup$ Commented May 21, 2015 at 15:58

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