Find f if $\int_0^1 f(x)^2dx=\int_0^1 f(x)^3dx =\int_0^1 f(x)^4dx$

Question

$f:[0,1]\to \mathbb{R}$ is a continuous function such that $$\int_0^1 f(x)^2dx=\int_0^1 f(x)^3dx =\int_0^1 f(x)^4dx$$ Find all such $f$

My approach: Clearly $f=0,1$ are solutions (and only possible constant functions). Now let $f$ to be a non constant function. By CS inequality we have, $$\int_0^1 f(x)^2dx \times \int_0^1 f(x)^4dx \ge ( \int_0^1 f(x)^3dx )^2$$ Equality holds here.So $\int_0^t f(x)dx =k \int_0^t f(x)^2dx$ for all $t\in (0,1]$ for some $k$ . I don't know what to do from here Please help.

From CS it follows already that $f^2(x) = cf(x)$. – trancelocation Apr 09 '21 at 19:06 — trancelocation, Apr 09 '21 at 19:06

score 6 · Answer 1 · answered Apr 09 '21 at 19:33

Note that $f(x)^4 - 2f(x)^3 + f(x)^2 = f(x)^2(f(x)-1)^2$ so that $$\int_0^1 f(x)^2 (f(x) - 1)^2 \, dx = 0.$$ Since $f$ is continuous we conclude $f(x)^2 (f(x) -1)^2 = 0$ for all $x \in [0,1]$, and in particular that either $f(x) = 0$ or $f(x) = 1$ for all $x \in [0,1]$.

Finally it is not possible for a continuous function to take exactly two values. Thus either $f(x) = 0$ for all $x \in [0,1]$ or $f(x) = 1$ for all $x \in [0,1]$.

score 2 · Accepted Answer · answered Apr 09 '21 at 18:30

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Hint: Differentiate your last equality and use the fundamental theorem of calculus; you'll get: $$f(t)=k\cdot f^2(t).$$ What can you conclude about $f$?

answered Apr 09 '21 at 18:30

csch2

5,095

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There is nothing to be differentiated because from CS it follows already that $f^2(x) = cf(x)$. – trancelocation Apr 09 '21 at 19:07

Find f if $\int_0^1 f(x)^2dx=\int_0^1 f(x)^3dx =\int_0^1 f(x)^4dx$

2 Answers2