# Thread: Integral inequality from David van Dantzig

1. ## Integral inequality from David van Dantzig

$f(x) : [-1, \; 1]\to\mathbb{R}$ is differentiable on $(-1, \; 1)$ and continuous on $[-1, \; 1]$.
Furthermore, $f(x)$ satisfies following condition :
$f'(x)\ge-1\;\;\forall x\in (-1, 1)\\ f(-1)\ge f(1)\\ \int_{-1}^{1}f(x)\mathrm{d}x = 0.$
Then, prove $\int_{-1}^{1}(f(x))^{2}{d}x\le\frac{2}{3}$

2. ## Re: Integral inequality from David van Dantzig

Originally Posted by Cruella
$f(x) : [-1, \; 1]\to\mathbb{R}$ is differentiable on $(-1, \; 1)$ and continuous on $[-1, \; 1]$.
Furthermore, $f(x)$ satisfies following condition :
$f'(x)\ge-1\;\;\forall x\in (-1, 1)\\ f(-1)\ge f(1)\\ \int_{-1}^{1}f(x)\mathrm{d}x = 0.$
Then, prove $\int_{-1}^{1}(f(x))^{2}{d}x\le\frac{2}{3}$

The answer is practically given to you, you just have to analyze the known data.

1. $f(x) : [-1, \; 1]\to\mathbb{R}$ is differentiable on $(-1, \; 1)$ and continuous on $[-1, \; 1]$
All this is really saying is that the function is only defined on $[-1, \; 1]$ (i.e. that is its domain). Not much useful info.

2.
$f'(x)\ge-1\;\;\forall x\in (-1, 1)\\ f(-1)\ge f(1)\\ \int_{-1}^{1}f(x)\mathrm{d}x = 0.$

This is where the answer is hidden under. Firstly, it's saying that its derivative never goes under -1 (i.e. the lowest possible slope at any point x in [-1,1] is -1). However it says that the value of f(-1) is greater than or equal to f(1). This should indicate to you that f(x) is most likely decreasing on the interval [-1,1] since f(-1)>=f(1), and they've hinted that it's derivative can be negative.

Most importantly it says the integral of f(x) from [-1,1] is 0 which can only mean part of f(x) has to be positive and part of f(x) has to be negative (in order to achieve that integral). However, the slope never goes below -1 at any point.

In that case, let's consider the simplest scenario, in which f(x) = -x . You can verify that this function satisfies the above conditions. Try to figure out the rest of the problem from here.