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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.

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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.

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  • $\begingroup$ There is no continuous function with that property! @George Dewhirst $\endgroup$ – Kavi Rama Murthy Apr 19 at 23:18
  • $\begingroup$ 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 $\endgroup$ – George Dewhirst Apr 20 at 9:49
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Apr 20 at 11:27
  • $\begingroup$ Yeah true I will look at it in a bit, feel free to edit it $\endgroup$ – George Dewhirst Apr 20 at 11:31
  • $\begingroup$ 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 $f(\epsilon) = g(\epsilon)$ $\endgroup$ – George Dewhirst Apr 20 at 11:35

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