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Is there a (non-constant) function $f \in C^4((0,1))$ that is zero in an interval $(a,b) \subset (0,1)$ and such that the inequality $$\Vert\tfrac{d^4}{dx^4}f\Vert_{L^2(0,1)} < \sqrt{2}\Vert f \Vert_{L^1(0,1)}$$ holds?

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  • $\begingroup$ Typo: $\psi$ has to be be replaced by $f$. $\endgroup$ Commented Jan 25, 2021 at 12:03
  • $\begingroup$ @DieterKadelka Correct. Thanks $\endgroup$
    – Hiro
    Commented Jan 25, 2021 at 13:54
  • $\begingroup$ [deleted earlier comment since my "gluing argument" was insufficient] $\endgroup$
    – Yemon Choi
    Commented Jan 25, 2021 at 14:27
  • $\begingroup$ [also deleted earlier comments which were rendered obsolete by the OP's improvement of their question] $\endgroup$
    – Yemon Choi
    Commented Jan 25, 2021 at 18:15

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The answer is no.

Indeed, let $c:=\sqrt2\,\|f\|_1\in(0,\infty)$, where $\|f\|_p:=\|f\|_{L^p(0,1)}$. Suppose that $f\in C^4(0,1)$ and $f=0$ on $(a,b)$, where $0\le a<b\le1$. Suppose that the inequality in question holds: $$\|f''''\|_2<c.$$

Then, using the Cauchy--Schwarz inequality, for $x\in[b,1]$ we have $$|f'''(x)|\le\int_b^x|f''''(t)|\,dt \le\Big(\int_b^x f''''(t)^2\,dt\Big)^{1/2}\Big(\int_b^x dt\Big)^{1/2} \le c(x-b)^{1/2},$$ whence $$|f''(x)|\le\int_b^x|f'''(t)|\,dt\le c\,\frac23\,(x-b)^{3/2},$$ $$|f'(x)|\le\int_b^x|f''(t)|\,dt\le c\,\frac23\,\frac25\,(x-b)^{5/2},$$ $$|f(x)|\le\int_b^x|f'(t)|\,dt\le c\,\frac23\,\frac25\,\frac27\,(x-b)^{7/2},$$ $$\int_b^1|f(t)|\,dt\le c\,\frac23\,\frac25\,\frac27\,\frac29\,(1-b)^{9/2} \le cc_1(1-b),$$ where $$c_1:=\frac23\,\frac25\,\frac27\,\frac29<\frac1{\sqrt2}.$$

Similarly, $\int_0^a|f|\le cc_1a$. So, $$\frac c{\sqrt2}=\|f\|_1=\int_0^a|f|+\int_b^1|f|\le cc_1a+cc_1(1-b)<cc_1<\frac c{\sqrt2},$$ a contradiction. $\Box$

(This answer is essentially the same as this previous one.)

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  • $\begingroup$ Might as well record the general versions, no? Theorem Let $f\in C^k((0,1))$ be such that there exists $b\in (0,1)$ where $f(b) = f'(b) = \ldots = f^{(k)}(b) = 0$. Then: $$ |f(x)| \leq \frac{1}{(k - 1/p)(k-1-1/p)\cdots(2-1/p)}\|f^{(k)}\|_{L^p} |x-b|^{k-1/p}$$ and $$ |f(x)| \leq \frac{1}{(k+\alpha)(k-1+\alpha)\cdots(1+\alpha)} \|f^{(k)}\|_{C^{0,\alpha}} |x-b|^{k+\alpha} $$ $\endgroup$ Commented Jan 25, 2021 at 15:15
  • $\begingroup$ @WillieWong : Good idea. I had something like that in my handwritten notes for the previous answer. :-) $\endgroup$ Commented Jan 25, 2021 at 15:33
  • $\begingroup$ Thank you. I asked a new more particular version of this question at mathoverflow.net/questions/382160/… Maybe in this case there is a positive result? $\endgroup$
    – Hiro
    Commented Jan 25, 2021 at 18:37

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