3
$\begingroup$

Let $f:[0,1]\to\mathbb{R}$ be a continuous function then the maximum value of $\int_0^1 f(x)x^2 \,\text{d}x -\int_0^1x(f(x))^2\,\text{d}x$ for all such function(s) is? I first thought differentiating it but the limits are not variable so no use then I spent a lot of time thinking over it but literally have no idea how to begin. Please help.

$\endgroup$

2 Answers 2

6
$\begingroup$

Hint:

$$f(x)x^2 -xf(x)^2 = xf(x) ( x-f(x)) $$

Think of $f(x-f)$ as a quadratic in $f$. Complete the square to get $$\frac{x^2}{4} - \left(f-\frac x2\right)^2 $$

So, your integral is $$\int_0^1 \frac{x^3}{4} - x\left(f(x) -\frac x2\right)^2 dx $$

What function $f$ maximizes this?

$\endgroup$
2
  • $\begingroup$ @ Tavish thanks a lot for your support.:-) $\endgroup$ Mar 4, 2021 at 14:10
  • $\begingroup$ @sameedhussain Of course. $\endgroup$
    – Vishu
    Mar 4, 2021 at 14:23
2
$\begingroup$

Seems like functioncal calculus to me:

We want to optmize:

$$ g[f] = \int_0^1 \left[ f(x)x^2 - x f^2 (x) \right]dx$$

Add a peturbation to $g$: $$g[f+ \epsilon] = \int_0^1 \left[ (f + \epsilon)x^2 - x (f+\epsilon)^2 \right] dx$$

It is easy to find that terms of order $\epsilon$ are:

$$ \epsilon \int_0^1 \left[x^2 - 2 xf \right] dx$$

So, for the minimum condition, this term in bracket must be zero, which makes our function as:

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

Now you can integrate easily

All of this is based on this article that I read


Meta commentary:

This is completely analogous to how if we are a maxima point in single variable calculus , then first order variation is zero. I.e:

$$ f(x+h) = f(x) + h f'(x) + \text{junk}$$

Now, if you are at maxima then $f'(x)=0$ and hence if you increase $h$ by some teeny amount, the function doesn't change much. Similar notion here though more abstract.

Another notion is that, similar to how we neglect term of $dx^2$, we neglect terms of $\epsilon^2$ saying they are tinnier than tiny petrubations.

$\endgroup$
1
  • $\begingroup$ Thanks a lot for another approach to solve the problem. $\endgroup$ Mar 4, 2021 at 14:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .