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?



Functions such that $ \Vert\tfrac{d^4}{dx^4}f\Vert_{L^2(0,1)} < \sqrt{2} \Vert f \Vert_{L^1(0,1)}$
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](https://mathoverflow.net/a/382110/36721).)


