Functions such that $ \Vert\tfrac{d^4}{dx^4}f\Vert_{L^2(0,1)} < \sqrt{2} \Vert f \Vert_{L^1(0,1)}$ 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?
 A: 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.)
