Is there a (nonconstant) 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?

$\begingroup$ Typo: $\psi$ has to be be replaced by $f$. $\endgroup$– Dieter KadelkaJan 25, 2021 at 12:03

$\begingroup$ @DieterKadelka Correct. Thanks $\endgroup$– HiroJan 25, 2021 at 13:54

$\begingroup$ [deleted earlier comment since my "gluing argument" was insufficient] $\endgroup$– Yemon ChoiJan 25, 2021 at 14:27

$\begingroup$ [also deleted earlier comments which were rendered obsolete by the OP's improvement of their question] $\endgroup$– Yemon ChoiJan 25, 2021 at 18:15
1 Answer
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 CauchySchwarz inequality, for $x\in[b,1]$ we have $$f'''(x)\le\int_b^xf''''(t)\,dt \le\Big(\int_b^x f''''(t)^2\,dt\Big)^{1/2}\Big(\int_b^x dt\Big)^{1/2} \le c(xb)^{1/2},$$ whence $$f''(x)\le\int_b^xf'''(t)\,dt\le c\,\frac23\,(xb)^{3/2},$$ $$f'(x)\le\int_b^xf''(t)\,dt\le c\,\frac23\,\frac25\,(xb)^{5/2},$$ $$f(x)\le\int_b^xf'(t)\,dt\le c\,\frac23\,\frac25\,\frac27\,(xb)^{7/2},$$ $$\int_b^1f(t)\,dt\le c\,\frac23\,\frac25\,\frac27\,\frac29\,(1b)^{9/2} \le cc_1(1b),$$ where $$c_1:=\frac23\,\frac25\,\frac27\,\frac29<\frac1{\sqrt2}.$$
Similarly, $\int_0^af\le cc_1a$. So, $$\frac c{\sqrt2}=\f\_1=\int_0^af+\int_b^1f\le cc_1a+cc_1(1b)<cc_1<\frac c{\sqrt2},$$ a contradiction. $\Box$
(This answer is essentially the same as this previous one.)

$\begingroup$ 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: $$ f(x) \leq \frac{1}{(k  1/p)(k11/p)\cdots(21/p)}\f^{(k)}\_{L^p} xb^{k1/p}$$ and $$ f(x) \leq \frac{1}{(k+\alpha)(k1+\alpha)\cdots(1+\alpha)} \f^{(k)}\_{C^{0,\alpha}} xb^{k+\alpha} $$ $\endgroup$ Jan 25, 2021 at 15:15

$\begingroup$ @WillieWong : Good idea. I had something like that in my handwritten notes for the previous answer. :) $\endgroup$ 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? $\endgroup$– HiroJan 25, 2021 at 18:37