TITLE: Prove that $\left|30240\int_{0}^{1}x(1-x)f(x)f'(x)dx\right|\le1$.
QUESTION [28 upvotes]: Let $f\in C^{3}[0,1]$such that $f(0)=f'(0)=f(1)=0$ and $\big|f''' (x)\big|\le 1$.Prove that $$\left|30240\int_{0}^{1}x(1-x)f(x)f'(x)dx\right|\le1 .$$

I couldn't make much progress on this problem. I thought that maybe I should try using polynomial interpolation since I have a bound for $|f'''|$, but I can't determine the interpolation polynomial and I am quickly stuck (there is also the problem that I am dealing with both $f$ and $f'$ under the integral). Apart from this, I don't think that there is much we can do, the solution probably relies on this technique, but I can't make further progress.

REPLY [10 votes]: First, using integration by parts, and noting that $f(0)=f'(0)=0$ we get
$$
f(x)=\frac{x^2}{2}f''(0)+\frac{1}{2}\int_0^x(x-t)^2f^{(3)}(t)dt\tag{1}
$$
And from $f(1)=0$ we get also
$$
0=f(1)=\frac{1}{2}f''(0)+\frac{1}{2}\int_0^1(1-t)^2f^{(3)}(t)dt
$$
Hence
$$\eqalign{
0&=\frac{x^2}{2}f''(0)+
\frac{1}{2}\int_0^xx^2(1-t)^2f^{(3)}(t)dt\cr
&+
\frac{1}{2}\int_x^1x^2(1-t)^2f^{(3)}(t)dt
\tag{2}}
$$
Subtracting from $(1)$ from $(2)$ we get
$$f(x)=-\int_0^1k(x,t)f^{(3)}(t)dt\tag{3}$$ with
$$k(x,t)=\cases{\dfrac{x^2(1-t)^2-(x-t)^2}{2}& if $t\leq x$\cr
\dfrac{x^2(1-t)^2}{2}& if $ x\leq t$
}$$
(Note that $k(x,t)\ge 0$ for $0\leq x,t\leq 1$).
Now, we note that
$$\eqalign{I&=\int_0^1x(1-x)f(x)f'(x)dx=\int_0^1\left(x-\frac{1}{2}\right)(f(x))^2dx\\
&=\int_0^1\int_0^1\int_0^1\left(x-\frac{1}{2}\right)k(x,t)k(x,s)f^{(3)}(t)f^{(3)}(s)dtdsdx\\
&=\int_0^1\int_0^1H(t,s)f^{(3)}(t)f^{(3)}(s)dtds\tag{4}
}
$$
with
$$H(t,s)=\int_0^1\left(x-\frac{1}{2}\right)k(x,t)k(x,s)dx\tag{5}$$
Now, $H(s,t)=H(t,s)$ and for $0\leq t\leq s\leq 1$ we find
$$H(t,s)=\frac{1}{240} (s-1)^2 t \left(s^4 (t-2)+s^3 (t+2)+s^2 (1-4 t)+s t+(t-1) t^4\right)$$
It is not very easy but we can show (see the remark below,) that $H$ is nonnegative, and this is the crucial point. As it is depicted in the next figure

It follows from the condition $|f^{(3)}|\leq 1$ that
$$|{I}|\leq \int_0^1\int_0^1H(t,s)dtds=\frac{1}{30240}.$$
Remark. Let me add a proof of the positivity of $H(t,s)$. Indeed, It is a matter of verification that for $0<t\leq s\leq 1$ we have
$$\eqalign{\frac{240 }{(1-s)^2 t}H(t,s)&=
(1-t) \left(s^2-t^2\right)+2 s^2 (1-s) (s-t)\\&
+s (s+1) (1-s)^2 t+(1-t)^2 t^2 (t+1)}$$
which is  a sum of positive terms.