$\int^1_0 f (f(x)t) \,\mathrm{d}t =\frac{1}{2}f(x)$ for every $x$

Question

$$\int^1_0 f (f(x)t) \,\mathrm{d}t =\frac{1}{2}f(x)$$ for every $x$.

I have to find all linear functions that looks like: $f(x)=Ax+B$

I thought maybe to differentiate it but I get nothing ...

Thanks for your help !

Answer 1

Let us assume that $f(x) \neq 0$, using the change of variables $u=f(x) t$ you have: $$\int_0^1 f(f(x)t)dt=\frac{1}{f(x)} \int_0^{f(x)} f(u) du$$ so the equality rewrites as: $$\int_0^{f(x)} f(u) du=\frac{1}{2} f(x)^2$$ This equality is also true when $f(x)=0$ as it becomes $0=0$.

Differentiating you obtain:

$$f'(x) f(f(x))=f'(x) f(x)$$ so if $f'(x) \neq 0$ you obtain: $$f(f(x))=f(x)$$

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So if you search linear function verifying the equality then you have either constants $f(x)=B$ (as $f'(x)=0$) or $f(x)=Ax+b$ with: $$A(Ax+B)+B=Ax+B$$ so: $$A^2=A$$ $$AB+B=B$$ i.e, $A=\pm1,B=0$.

Answer 2

Although the answer has already been accepted one case has been neglected. If $f'(x)=0$ then $f=B$ and $$ \int_0^1 f(tf(x))dt = \int_0^1 B dt =B t |_0^1=B=\frac{B}{2}$$

So the zero function is also a solution