# Maximum of $\int_0^1(x^p|f(x)|^q-x^q|f(x)|^p)dx$

Let $$p>q>0$$ and $$C=\{f:[0,1] \to \mathbb{R} \mid f \text{ is continuous} \}$$. Determine $$\max_{f \in C}\int_0^1(x^p|f(x)|^q-x^q|f(x)|^p)dx$$ and the functions for which this maximum occurs.

If $$f(x)=x^n$$, then the integral is $$a(n)=\frac{(n-1)(p-q)}{(p+nq+1)(np+q+1)}$$. The particular case $$p=2, q=1$$ appeared on Putnam 2006, but I don't know if it is of any help. However, the answer in that case is $$\frac{1}{16} \neq a(n)$$ and $$f(x)=\frac{x}{2}$$, hence we should search other types of functions.

I tried to write the integral as $$\int_0^1 x^q|f(x)|^q(x^{p-q}-|f(x)|^{p-q})dx$$ but it doesn't seem to lead to anything. Since there are so many options to choose $$f$$ from, I have no idea how to proceed.

We can start by asking what is the maximum of $$x^p |f(x)|^q - x^q |f(x)|^p$$ for some $$x \in [0,1]$$, and $$y=f(x)$$ a given number.

This is given by $$qx^p y^{q-1} - x^q p y^{p-1} = 0$$, i.e. $$y^{p-q} = \frac{qx^{p-q}}{p}$$. Check it is a maximum.

So then $$|f(x)| = (\frac{qx^{p-q}}{p})^{q-p}$$.

So thus any function $$f$$ with the above modulus (i.e. $$\pm f$$) will give us our maximum.

• There is no continuous function with that property! @George Dewhirst – Kavi Rama Murthy Apr 19 at 23:18
• It's only discontinuous at zero. Given that it is maximal over the rest of the range, we can smooth it to a fixed value at zero and it will be fine – George Dewhirst Apr 20 at 9:49
• I know. I just wanted to point out that your proof looks incomplete because the basic space is $c[0,1]$ and continuity is important. – Kavi Rama Murthy Apr 20 at 11:27
• Yeah true I will look at it in a bit, feel free to edit it – George Dewhirst Apr 20 at 11:31
• What we should do is extend this function over the range $[\epsilon, 1]$, and then find the best pair $(\epsilon, g)$ such that $g:[0,\epsilon]\to \mathbb{R}$ has $f(\epsilon) = g(\epsilon)$ – George Dewhirst Apr 20 at 11:35