Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a function $f$, let

$$ a = \int_{0}^{1} x^2f(x) \mathrm{d}x\\ b = \int_{0}^{1} xf^2(x) \mathrm{d}x, $$

where $f$ is a continuous function from $[0,1]$ to $\mathbb{R}$. Then find $\text{max}\{a-b\}$ for all such $f$.

I am getting $\dfrac {1}{16}$. Note that it can be written as $\int_{0}^{1} \left({\dfrac{x^3}{4}-x(f(x)-\dfrac{x}{2})^2}\right)dx$. I guess that's less than $\int_{0}^{1} \dfrac{x^3}{4} dx$ which is $\dfrac {1}{16}$.

share|improve this question
I get the idea that some information is missing. I would start by noting that since $[0, 1]$ is compact, $f(x)$ can be bounded from above by some constant $M$, and then you can estimate the integral by replacing $f(x) \to M$, leading to something like $1/3M - 1/2M^2$ as a first upper bound. However, I don't quite see how to get rid of $M$ without more information on $f$. –  CompuChip Nov 18 '13 at 11:44
Your result is right, the maximum is $\frac{1}{16}$. You wrote the difference as the difference of a fixed function, and a square involving $f$ (times $x$), and the difference is maximised when the square - $(f(x) - \frac{x}{2})^2$ - is $0$. –  Daniel Fischer Nov 18 '13 at 12:55

1 Answer 1

up vote 4 down vote accepted

Here is another approach in the case we can't express the integral in the way you did (see the comment of @DanielFischer). Let $X=C([0,1])$ and $I:X\to\mathbb{R}$ be defined by $$I(f)=\int_0^1 x^2f(x)-\int_0^1xf(x)^2$$

Note that $I$ is a continuously differentiable function and $$\langle I'(f),g\rangle =\int_0^1 (x^2 g(x)-2xf(x)g(x)),\ \forall\ f,g\in X$$

We want to find $f\in X$ such that $$\langle I'(f),g\rangle=0,\ \forall\ g\in X$$

Therefore $$f(x)=\frac{x}{2}$$

To verify that $f$ is a maximum, note that $$I(f+h)=\int_0^1\left(\frac{x^3}{4}-xh(x)^2\right)$$

share|improve this answer
Can you please specify what g(x) is? –  CobraCobra Nov 18 '13 at 13:43
I have specified: $g\in X=C([0,1])$. –  Tomás Nov 18 '13 at 13:56
Ok.. Didn't note that.. Thanks ! –  CobraCobra Nov 18 '13 at 13:59
Ok, you are welcome @Apurv. Note that the equation $$\langle I'(f),g\rangle =0,\ \forall\ g\in X$$ comes from the fact that we want to find a critical point of $I$, i.e. a point $f$ satisfying $$I'(f)=0$$ –  Tomás Nov 18 '13 at 14:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.