Prove that $f$ is constant.

Question

If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.

(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$

(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$

(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.

Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.

Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.

Answer 1

For each of these, you have something of the form $f(x) = f(g(x))$ where $g^n(x) \to 0$ (where $g^n$ is the repeated application of $g$) or $g^n(x) \to 1$ (e.g. $g$ is a contraction to some point, let's call it $y$ in general), you can use this and continuity to show that we have $$ f(x) = f(g^n(x)) \to f(y), \forall x, n\to \infty. $$

In this case, consider the following:

(1) Assume $|m| > 1$ then $g(x) = x/m$, now we know $f(x) = f(x/m) = f(x/m^2) = \dots \to f(0), \forall x$ by continuity of $f$. A similar case follows for $|m| < 1$, using $g(x) = mx$.

(2) Assume $g(x) = (x-1)/2$, then note that $g^n(x) \to 0, n\to \infty$ for every $x$, so a similar proof $f(x) = f(g^n(x)) \to f(0)$ holds.

(3) Assume that $g(x) = \sqrt{x}$ if $x>0$ else $-\sqrt{x}$ will work. Note that $g^n(x) \to 1, \forall x$ so here we are done as well by noting that $f(x) = f(g^n(x)) \to f(1)$ again by continuity of $f$.

Answer 2

1. Suppose $|m|>1$ $f(x)=f(mx)$ implies that $f(x/m)=f(m(m/x))=f(x)$. We deduce that for every integer $n$, $f({x\over m^n})=f(x)$, since $lim_n{x\over m^n}=0$ and $f$ is continue, we deduce that $f(x)=lim_{n\rightarrow +\infty}f({x\over m^n})=f(0)$.

The case $m<1$ consider the sequence $f(m^nx)$.

Answer 3

2. Let $y=x+1$, so that $\forall y \in \mathbb R, f(y-1)=f(2y-1)$. Then define $g(y)=f(y-1)$, so that $\forall y, g(y)=g(2y)$.

Using the result of 1., $g$ is constant and equal to $g(0)=f(-1)$, so $f$ is also constant and equal to $f(-1)$.

3. Note that $f(-x)=f((-x)^2)=f(x^2)=f(x)$ so $f$ is even.

Consider $x>0$. Prove by induction on $n$ that $f(x^{1/n})=f(x)$ and let $n\to \infty$. Since $x^{1/n} \to 1$, then $f(x^{1/n})\to f(1)$, hence $f(x)=f(1)$.

Since $f$ is even, $f(x)=f(1)$ everywhere except $0$, and since $f$ is continuous, $f(0)=f(1)$.

Answer 4

For 1: Suppose $|m|=0$, then$$f(x)=f(0) \ \forall \ x \in \mathbb{R}$$ So $f$ is constant.

Otherwise, suppose $|m| > 1$.

Take any $x, y \in \mathbb{R}$

Since $f$ is continuous on $\mathbb{R}$, $$\forall p \in \mathbb{R}, \forall \epsilon>0, \exists \delta>0:\forall q \in \mathbb{R} \ \mbox{where} \ 0<|q-p|\leqslant \delta,\\ |f(q)-f(p)|<\epsilon $$ Hence for any arbitrary $\epsilon$, we can chose $p=y \cdot m^{-n}$ for some appropriate $n$, $q=x \cdot m^{-k}$ for some appropriate $k$ (take the negative out for $|m|<1$) and $\delta>0$ so that $0<|q-p|\leqslant \delta$, and hence $|f(y \cdot m^{-n})-f(x \cdot m^{-k})|=|f(y)-f(x)|<\epsilon$.

But since $\epsilon$ is arbitrary, $f(x)=f(y)$

Answer 5

Here the answer to your first question, where use is made of one of the main corollaries for continuous functions (see $(4)$ below).

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ and $m\in\mathbb{R}$, $|m|\ne1$ $$f \text{ is continuous}\tag{1}$$ $$\forall x\in\mathbb{R},f(x)=f(mx)\tag{2}$$ Theorem. $f\text{ is constant}$

Proof. Let $A_x=\{\xi\in\mathbb{R}|\exists n\in\mathbb{Z},\xi=m^nx\}$ and $\mathcal{A}=\{A_x, x\in\mathbb{R}\}$

Being $(2)$, then:

$$\forall x\in\mathbb{R},f(x)=f(mx)\iff f(A_x)=\{f(x)\}$$

Then there must follow that:

$$\forall x\in\mathbb{R}, \exists \mathcal{B}_x\in\mathcal{A}, f^{-1}(x)=\cup\mathcal{B}_x\tag{3}$$

Being $(1)$, then:

$$f \text{ continuous }\implies\forall x\in\mathbb{R},f^{-1}(x)\text{ is closed}\tag{4}$$

Notice that every subfamily of $\mathcal{A}$ not including $A_0=\{0\}$ has a union that is not closed, while every one including it has a closed union, that is all the subfamilies with closed union are not disjoint.

$$\forall \mathcal{F}\subset\mathcal{A}, \cup\mathcal{F}\text{ is closed}\iff A_0\in\mathcal{F}\tag{5}$$

So from $(3)$, $(4)$, and $(5)$ it is: $$\forall x, f^{-1}(x)\supset A_0$$ But it is a fact that: $f^{-1}(x)\cap f^{-1}(y)=\emptyset\iff f(x)\ne f(y)$ so: $$\forall x, f(x)=f(0)$$