Let $f\in C^2(\mathbb{R})$ such that $\displaystyle f(1)=\int_0^1f(x)dx=0$.

Prove that

$$\left|\int_0^x f(t)dt\right|\le \frac{2}{81}\max_{0\le x\le1}|f^{''}(x)|\,\,\,\,\,\, \forall x\in [0,1].$$

Thanks in advance!

  • $\begingroup$ This reminds me of a Putnam problem. $\endgroup$ – IAmNoOne Mar 22 '14 at 5:27

The key idea is to use: $$\int f''(t)P(t)\,dt = P(t)f'(t)-P'(t)f(t)+ \int P''(t)f(t)\,dt \tag{1}$$

for a suitable polynomial $P$.

Fix an $x \in (0,1)$ and choose two degree $2$ polynomials $P_1$ and $P_2$ to write:

$$\int_0^x f''(t)P_1(t)\,dt = [P_1(t)f'(t)-P_1'(t)f(t)]_0^x+ \int_0^x P_1''(t)f(t)\,dt \tag{2}$$

$$\int_x^1 f''(t)P_2(t)\,dt = [P_2(t)f'(t)-P_2'(t)f(t)]_x^1 + \int_x^1 P_2''(t)f(t)\,dt\tag{3}$$

Subtracting the equations $(2)$ and $(3)$, we get:

\begin{align}&\int_0^x f''(t)P_1(t)\,dt - \int_x^1 f''(t)P_2(t)\,dt \\=\,\,&\left(P_1(x)+P_2(x)\right)f'(x)- P_1(0)f'(0) -P_2(1)f'(1) \\&-(P'_1(x)+P'_2(x))f(x)+ P'_1(0)f(0) +P'_2(1)f(1) \\&+ (P''_1(t)+P''_2(t))\int_0^x f(t)\,dt\tag{4}\end{align}

Since, $P_1,P_2$ are degree $2$, the double derivatives are constant.

Also, $$\displaystyle \int_0^x f(t)\,dt + \int_x^1 f(t)\,dt = 0$$

Now, we choose the polynomials such that,

\begin{align}P_1(x)+P_2(x) = 0\tag{i}\\P'_1(x)+P'_2(x) = 0\tag{ii}\\P_1(0)=0,P'_1(0)=0 \text { and }P_2(1)=0\tag{iii}\end{align}

Now, from (iii) we can infer $P_1(t)=ct^2$, (where, $c$ is some non-zero constant in $(0,1)$)

(i) and (ii) gives $P_1(t)+P_2(t) = (t-x)^2$.

Combining with the fact that $P_2(1)=0$ gives, $\displaystyle P_2(t)=(1-c)(t-1)\left(t-\frac{x}{2-x}\right)$.

Then we get $c=1-(1-x)^2$, that is $(1-c)=x(2-x)$

The fact that $f(1)=0$ then makes the expression $(4)$:

$$\int_0^x f''(t)P_1(t)\,dt - \int_x^1 f''(t)P_2(t)\,dt = (P''_1(t)+P''_2(t))\int_0^x f(t)\,dt$$

Observe that,

$P_1(t) > 0 \text{ for } t \in [0,x]$ and $\displaystyle 0 < \frac{x}{2-x} < x < 1$, so that $P_2(t) < 0 \text{ for } t \in[x,1]$.

Therefore, $$\left| 2\int_0^x f(t)\,dt \right| \le \sup\limits_{x \in [0,1]} |f''(x)|.\left| \int_0^x P_1(t)\,dt - \int_x^1 P_2(t)\,dt \right|$$

Now, $$\int_0^x P_1(t)\,dt - \int_x^1 P_2(t)\,dt = \frac{x^3}{3} + \frac{2}{3}(1-c) - \frac{1-c}{2-x}=\frac{x(1-x)^2}{3}$$

For $x \in (0,1)$, we have $$\frac{x(1-x)^2}{3}=\frac{2x(1-x)^2}{6} \le \frac{1}{6}\bigg(\frac{2x+(1-x)+(1-x)}{3}\bigg)^3 \le \frac{4}{81}$$

Thus, $$\left| \int_0^x f(t)\,dt \right| \le \frac{2}{81} \sup\limits_{x \in [0,1]} |f''(x)|$$

  • 1
    $\begingroup$ Nice solution. Many thanks! $\endgroup$ – math123 Mar 25 '14 at 13:32
  • $\begingroup$ @math123 welcome :) .. $\endgroup$ – r9m Mar 28 '14 at 1:48

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