real analysis - Given $f$ is continuous and $f(x)=f(e^{t}x)$ for all $xinmathbb{R}$ and $tge0$, show that $f$ is constant function

This question was asked in ISI BStat / BMath 2018 entrance exam:

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that
for all $x\in\mathbb{R}$ and $t\ge 0$, $$f(x)=f(e^{t}x)$$ Show that
$f$ is a constant function.

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My attempt:

Suppose that $f$ is not a constant function. Then $f(0)\ne f(x_0)$ for some $x_0 \in \mathbb{R}$. We eliminate the possibilities that $x_0>0$ and $x_0<0$, thus proving that our assumption was wrong.

Case 1: ($x_0>0$). Let $k$ be any real number between $f(0)$ and $f(x_0)$ (not inclusive). Then by the intermediate value theorem, there exists $y_0 \in (0, x_0)$ such that $f(y_0)=k$. But $f(y_0)=f\left( e^{\ln \left( \frac{x_0}{y_0}\right) } y_0\right) = f(x_0)$ which contradicts our assumption that $f(y_0)$ was between $f(0)$ and $f(x_0)$.

Case 2: ($x_0<0$). Let $k$ be any real number between $f(0)$ and $f(x_0)$ (not inclusive). Then by the intermediate value theorem, there exists $y_0 \in (x_0, 0)$ such that $f(y_0)=k$. But $f(x_0)=f\left( e^{\ln \left( \frac{y_0}{x_0}\right) } x_0\right) = f(y_0)$, a contradiction again.

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Is this proof correct? I was probably looking for a direct proof if there's any. Alternative proofs are welcome.

Answer

An alternative proof: put $x=e^{-t}$ to get $f(e^{-t})=f(1)$ for all $t \geq 0$. This im plies $f(x)=f(1)$ for all $x \in (0,1]$. Next note that $f(x)=f(2x)$ for all $x$: just take $t=\ln (2)$). You now see easily that $f(x)=f(1)$ for all $x >0$. Since $f(-x)$ satisfies the same hypothesis it follows that $f$ is a constant on $(-\infty, 0)$ also. By continuity the constant values for $x <0$ and $x >0$ must be the same.