Prove that $\left|30240\int_{0}^{1}x(1-x)f(x)f'(x)dx\right|\le1$.

Question

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.

Answer 1

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.

Answer 2

Just an idea put $f(x)=\frac{g(x^2(1-x)^2)}{x(1-x)}$ we have :

$$f'(x)=-\frac{(2x-1)(2(x-1)^2x^2g'((x-1)^2x^2)-g((x-1)^2x^2))}{x^2(1-x)^2}$$

So we have :

$$f'(x)f(x)x(1-x)=-\frac{(2x-1)(2(x-1)^2x^2g'((x-1)^2x^2)-g((x-1)^2x^2))}{x^2(1-x)^2}x(1-x)\frac{g(x^2(1-x)^2)}{x(1-x)}$$

Or :

$$f'(x)f(x)x(1-x)=-(2x-1)\Big(2g'((1-x)^2x^2)g((1-x)^2x^2)-\frac{g^2(x^2(1-x)^2)}{x^2(1-x)^2}\Big)$$

Now we can integrate by parts but I can't go further .

Update :

As I'm stuck with the substitution above I propose another one :

Put :

$$f(x)=g(\ln(x)-\ln(1-x))$$

Then :

$$f'(x)f(x)x(1-x)=g(\ln(x)-\ln(1-x))g'(\ln(x)-\ln(1-x))$$

Now we can make the substitution $y=\ln(x)-\ln(1-x)$ or $x=\frac{e^y}{e^y+1}$ in the integral . Another substitution and we see that there is a link with the Laplace transform .

Update 2:

It's really my last idea on this problem .

We have the condition :

$$|f'''(x)|\leq 1\quad (1)$$

but :

$$1=(x+(1-x))^n$$

So we can use binomials theorem to expand the terms .

Now we put :

$$f(x)=\operatorname{B}_{i,n}(x)$$

Where we speak about the Bernstein polynomials .

See here (3.1) for the derivative of a Bernstein polynomial

Remains to compare the two quantities .

If it's really useless I shall delete .

Hope it inspire someone .

Answer 3

At first,

$$I=\int_0^1x(1-x)f(x)f'(x)\text{d}x \,\overset{IBP}{=\!=\!=}\,\frac12x(1-x)f^2(x)\bigg|_0^1+\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x,$$ $$I=\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x.\tag1$$

The given conditions mean that $f(x)$ has a root of multiplicity $2$ at $x=0$ and a simple root at $x=1.$

Taking in account the Besou theorem and assuming the condition $|f'''(x)| = 1,$ one can get $$|f_{opt}(x)| = a(x-0)^2(1-x) = a(x^2-x^3),\quad a = \frac16,\tag2$$

$$|I|\le\dfrac1{72}\int_0^1(2x-1)(x^2-x^3)^2\text{ d}x = \dfrac1{30240}.$$

$\color{green}{\textbf{EDIT of 12.08.20.}}$

Let us integrate the given inequality $$-1\le f'''(x) \le 1\tag{1n}$$ under the given conditions on the interval $(0,x),$ then $$ \begin{cases} -x\le f''(x) - f''(0)\le x\\[4pt] -\frac12x^2 \le f'(x) - xf''\left(0\right) \le \frac12x^2\\[4pt] -\frac16x^3 \le f(x) - \frac12x^2f''\left(0\right) \le \frac16x^3._{\Large\mathstrut} \end{cases}\tag{2n}$$

From $(2n.3)$ should $$|6f(x)-3f''(0)x^2| \le x^3,\quad -1\le -3f''(0)\le1,$$ $$6|f(x)|\le |x^3+ax^2|,\quad |a|\le 1,\quad f(1)=0,$$ $$6|f(x)|\le x^2-x^3 = h(x).$$

This confirms formulas $(2).$

Similarly, from $(2n.2)$ $$|f'(x)| \le \frac12 x^2 -\frac13x = h'(x),$$ wherein $h'(x)$ is synchronized with $h(x).$

Therefore, $$|I| \le \left|\int_0^1 x(1-x) h(x) h'(x) \text{ d}x\right| = \frac1{30240}.$$

Answer 4

$\color{green}{\textbf{Version of 15.08.20.}}$

At first,

$$I=\int_0^1x(1-x)f(x)f'(x)\text{d}x \,\overset{IBP}{=\!=\!=}\,\frac12x(1-x)f^2(x)\bigg|_0^1+\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x,$$ $$I=\frac12\int_0^1 (2x-1)f^2(x)\text{ d}x.\tag1$$

Let $$x = \frac{y+1}2,\quad \text{ d}x = \frac12\text{ d}y,\quad y = 2x-1,\quad g(y) = f\left(\frac{y+1}2\right), \tag2$$

then $$f(x) = g(2x-1) = g(y),\quad f'''(x) = 8g'''(2x-1) = g'''(y),\tag3$$ $$I = \frac14\int\limits_{-1}^{1} y g^2(y)\text{ d}y,\tag4$$ under the conditions $$g(-1) = g'(-1) = g(1) = 0,\quad |g'''(y)| \le \frac18.\tag5$$

Decomposition to the even and the odd parts $$g(y)=g^\,_+(y)+g^\,_-(y),\quad g^\,_\pm(y) = \frac12(g(y)\pm g(-y)),\quad g^\,_\pm(-y) = \pm g^\,_\pm(y),\tag6$$ gives $$I = \int\limits_{0}^{1} y g^\,_+(y)\,g^\,_-(y)\text{ d}y.\tag7$$

In accordance with the Shwartz inequality, $$I^2 \le \int\limits_{0}^{1} \big(y g^\,_+(y)\big)^2\text{ d}y\cdot \int\limits_{0}^{1} g^2_-(y)\text{ d}y,\tag8$$

wherein $(8)$ becames the equality if $$|g^\,_-(y)| = y\,g^\,_+(y).$$ Then from $(5)$ should $$g^\,_+(y) = (1-y^2)h(y),\quad \big|g^\,_-(y) \big| =(y-y^3)h(y)\tag{9}$$ Therefore, the function $$g(y)=(1+y)(1-y^2) h(y)$$ maximizes $|I|$ under the conditions $(5)$ near $y=\pm1.$

Taking in account the rest of the conditions, one can get $h(y) = \text{constant} =\frac1{48},$ $$g(y) = \frac1{48}(1+y)(1-y^2),\tag{10}$$

$$48^2I_{opt} = \int\limits_0^1 (y^3-y)^2\text{ d}y = \int\limits_0^1 (y^6-2y^4+y^2)\text{ d}y = \frac17-\frac25+\frac13 = \frac8{105},\tag{11}$$ $$\color{brown}{\mathbf{|I| \le \frac1{30240}}},$$ $$f_{opt}(x) = \pm g_{opt}(2x-1) = \pm \frac16 (x^2-x^3).$$