Show that $f$ is a constant function.

Question

Problem

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

My proof

Take $t=1$. Then for any fixed $r \in \mathbb{R}$, it's clear that $$f(r)=f \left(\frac{r}{e}\right)=f \left(\frac{r}{e^{2}}\right)=f \left(\frac{r}{e^{3}}\right)=\cdots=f \left(\frac{r}{e^{n}}\right).$$Take the limits of both sides as $n \to \infty$. Notice that $\dfrac{r}{e^n} \to 0$ as $n \to \infty$. By the continuity of $f(x)$, we have $$f(r)=\lim_{n \to \infty}f(r)=\lim_{n \to \infty}f \left(\frac{r}{e^{n}}\right)=f\left(\lim_{n \to \infty}\frac{r}{e^{n}}\right)=f(0).$$This shows that $f(x) \equiv f(0)$.

Please Correct me if I'm wrong.

Answer 1

Good job. Your working seems fine.

It might be not clear to someone as pointed out in the comment about letting $t=1$. In that case, just state what your $x$ is explicitly. Let $x=\frac{r}{e^{m}}, m \ge 0$ and hence we have $$f\left( \frac{r}{e^{m}}\right)=f\left(e\cdot \frac{r}{e^m} \right)$$ and $$f\left(e\cdot \frac{r}{e^m} \right)=f\left( \frac{r}{e^{m}}\right)$$

$$f\left(\frac{r}{e^{m-1}} \right)=f\left( \frac{r}{e^{m}}\right)$$

Remark:

Your working shows that the claim remains true for example if we restrict $t$ to be nonnegative integers.

Answer 2

Let me present alternative approach (assuming $f$ is not only continuous, but also differentiable).

Differentiate expression $f\left(x\right)=f\left(e^tx\right)$ with respect to $x$, get

$$ f\left(x\right)=f\left(e^tx\right) \implies f'\left(x\right)=f'\left(e^tx\right) e^t $$

Differentiate the same expression $f\left(x\right)=f\left(e^tx\right)$ with respect to $t$ we get

$$ f\left(x\right)=f\left(e^tx\right) \implies f\left(x\right)=f'\left(e^tx\right) xe^t $$

Then we conclude that either $f'\left(x\right) = 0\implies f\equiv \operatorname{const}$, or $xf'\left(x\right)=f\left(x\right)$.

Elaborating the latter case we get

$$ f'\left(x\right) = \frac{f\left(x\right)}{x} \implies \frac{d f}{dx}=\frac{f}{x}\implies \frac{df}{f} = \frac{dx}{x}\implies f\left(x\right) = Cx $$

where $C$ is a constant which we can compute by enforcing $f\left(x\right)=f\left(e^tx\right)$ which holds for any $x$:

$$ Cx = Cxe^t, \quad t>0\;\implies \; C=0\; \iff\; f\left(x\right)\equiv 0 $$

In either case, we get $ f\left(x\right)\equiv \operatorname{const}$.

Answer 3

Let $x$ be any real positive number. $f(x) = f(e^tx)$ $\forall t \ge 0$ can be rewritten as $f(x) = f(x_0)$, $ \forall x \ge x_0$, $ \forall x_0 \gt 0$. So $f(x)$ is constant over the positives.

Similarly, we have that for negative $x$, $f(x) = f(e^tx)$ for all $t \ge 0$ implies that $f(x) = f(x_0)$, $\forall x \le x_0$ $\forall x_0 \lt 0$. So $f(x)$ is constant over the negatives too.

So our function is constant over the positives and the negatives. Also, our function is continuous, so we can't have a discontinuity at $x = 0$. So f is continuous

Writing this because I didn't understand your method

Edit: maybe what I did isn't so trivial,but basically $e^t$ is a positive constant bigger than or equal to $ 1$. So any positive or negative real $x$ can be expressed as $e^tx_0$, for some non-negative $t$, for all $x_0$ such that $|x| \ge |x_0|$ and both have they same sign.

Another edit: Now I get your method. I was just seitching the variables around.

Answer 4

Note that by simply setting $y=e^tx$ you get $f(y)=f(e^tx)=f(x)=f(e^{-t}y)$

So the restriction on $t\ge 0$ is just artificial and can be extended to negative $t$ as well.

Now just make $e^t\to 0$ and get the result by continuity as well.