Let f be a real valued continuous function such that for all real x and for all t≥0, f(x)=f(x•e^t). Show that f is a constant function.

Question

My approach to solving the problem stated in the title is as follows:

f(x)=f(x•e^t)=y(say)

Differentiating both sides w.r.t x, I got

f'(x)=e^t•f'(x•e^t)

This implies

e^t=1 [since f(x)=f(x•e^t) implies their derivatives are equal as well]

The above equation holds true for the case when t=0, which would trivially mean f(x)=f(x). To explain the above equality for all t≥0,

f'(x)=f'(x•e^t)=0, which means f is a constant function.

However, I'm concerned with the rationality of my last step, which makes me doubt my approach altogether. Can anyone explain if my approach is correct?

My background: High school level calculus. Once again, apologies for not using MathJax :)

Answer 1

For the function $e^x$ maps $[0,\infty]$ onto $[1,\infty]$. So given any $0<x<y$ we can choose $t$ such that $y=xe^t$ and then we have $f(x)=f(y)$. In other words we have $f(x)$ equals some constant $c_1$ on the positive reals (either $x=y,x<y$ or $y<x$).

Similarly, $f(x)$ is constant $c_2$ on the negative reals. But we are told it is continuous, so we must have $c_1=f(0)=c_2$ and $f$ is constant on the reals.