3
$\begingroup$

It is given that $|f''(x)|<=M$ for all $x$, and $f(0)=f'(0)=0$. Then prove that $|f(x)|\leq\frac{Mx^2}{2}$ for all $x$.

What I think about was by using the Mean Value Theorem twice, and then I will get $|f(x)|\leq Mx^2$ with no $\frac{1}{2}$. And I do not know if I use the MVT twice whethere there would be any mistake.

Is there any way other than the MVT to solve this problem? What I also considered were the Fundamental Theorem of Calculus and the Taylor's Theorem. But still, feel not too good about them.

Anyone offers a hint?

2 Answers 2

2

It is an immediate consequence of Taylor's theorem.

You have $f(x) = f(0) + f'(0)x + \int_0^x f''(t) (x-t)dt$. Using the facts that $f(0) = f'(0) = 0$ and $|f''(t)| \leq M$, you have $|f(x)| \leq M \int_0^x (x-t) dt = \frac{1}{2} M x^2$, as required.

1

Here's the hint that I can think of :

$\left|f(x)\right|=\left|\int_0^x\int_0^y f'(t)\; dt\,dy\right|\leqslant \int_0^x\int_0^y|f''(t)|dt\,dy\leqslant\int_0^x\int_0^yM\;dt\;dy$