Does there exist a smooth compactly supported function $$f \in C^{\infty}_c((0,1))$$ such that $$ \|D^k f\|_{L^{2}(0,1)} \leq \left\lfloor{\alpha\,k}\right \rfloor! \quad \forall\, k\in \mathbb N$$ for some $\alpha \in (0,1)$?
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1$\begingroup$ That implies that the Taylor series of $f$ is convergent, i.e., $f$ is analytic, which cannot also be compactly supported. $\endgroup$– Fan ZhengCommented Dec 23, 2019 at 16:27
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$\begingroup$ I don't see how. The estimate is not point-wise along the whole interval $(0,1)$. $\endgroup$– AliCommented Dec 23, 2019 at 16:31
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$\begingroup$ Hint: Sobolev embedding. $\endgroup$– Fan ZhengCommented Dec 23, 2019 at 16:48
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1 Answer
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For all natural $k$, we have $\|D^kf\|_1\le\|D^kf\|_2\le k!$, where $\|\cdot\|_p:=\|\cdot\|_{L^p(0,1)}$. So, for all $x\in(0,1)$ we have $$|(D^kf)(x)|\le\int_0^x |(D^{k+1}f)(u)|\,du\le\|D^{k+1}f\|_1\le(k+1)!. $$ So, for the Lagrange remainder $$R_n(a,x)=\int_a^x (D^{n+1}f)(u)\frac{(x-u)^n}{n!}\,du $$ for the $n$th-order Taylor approximation of $f$ at $a\in(0,1)$ we have $|R_n(a,x)|\le(n+2)|x-a|^{n+1}\to0$ as $n\to\infty$, for any $a$ and $x$ in $(0,1)$. So, $f$ is real analytic on $(0,1)$. Since $f$ is compactly supported, it follows that $f=0$.
answered Dec 23, 2019 at 18:51