Suppose that $f(x)$ is continuous on $(0, \infty)$ such that for all $x > 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function.

Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x > 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?

4 Answers

Hint:

\begin{align}f(x^2)=f(x)=f(\sqrt{x})= \cdots=& f(x^{\frac{1}{2^n}}) \\ f((x+1)^2)=f(x+1)=f(\sqrt{x+1})= \cdots=& f((x+1)^{\frac{1}{2^n}}) \\ \vdots\hspace{ 1 cm }=\cdots=&\hspace{ 1 cm }\vdots \\ f((x+k)^2)=f(x)=f(\sqrt{x+n})= \cdots=& f((x+k)^{\frac{1}{2^n}}) \end{align}

Or try showing there is NO such function which is continuous at $(0, +\infty)$ by contradiction.

As I understand it: $$f(x) = f(x^2)\quad \Rightarrow$$ $$f(x^\frac{1}{2}) = f(x)\quad \Rightarrow$$ $$f(x) = f(x^\frac{1}{2^n})_{n \in N}$$
take $a,b \in (0,+\infty)$, then:

$$f(a) =\lim_{n\rightarrow\infty} f(a^\frac{1}{2^n})=f(1)$$ $$f(b) =\lim_{n\rightarrow\infty} f(b^\frac{1}{2^n})=f(1)$$ $$f(a)=f(b)\Rightarrow_{def}\mathtt {f\quad is\quad constant}$$

Is it correct? Is it rigorous?
My thought process was: continuity has limit in its definition, so I probably have to use lim in writing my proof.

• when we have $f(x)=f(x^{\frac{1}{2n}})$, we can take limit in one side only ? Commented Mar 23, 2013 at 18:37
• @Idonknow, you take the limit of both sides, but the limit on the left-hand side is trivial because $n$ does not appear there. Commented Mar 23, 2013 at 18:37
• @AntonioVargas: ooh, i see. Thx Commented Mar 23, 2013 at 18:42

Try proving the statement using contradiction. Suppose that $f(x^2) = f(x)$ for all $x$ and that there is a point $a$ such that $f(a) \neq f(1)$. Prove that $f$ cannot be continuous at $x=1$.

• I donno how to proceed Commented Mar 23, 2013 at 15:14
• @Idonknow: Given the function is continuous. Contradict that function $f(a) \neq f(1)$ isn't continuous at $x=1$. Commented Mar 23, 2013 at 15:17
• @Inceptio, try using the sequence definition of continuity. What can you say about the values of $f$ at the points of the sequence $x_n = a^{1/2^n}$? Commented Mar 23, 2013 at 15:20
• ooh, so the proof is like this? The sequence $(x_n) \rightarrow 1$ and since $f$ is continuous, we have $lim_{n}{f(x_n)}=f(1)=f(x)$ for all $x$. Hence , $f$ is a constant function. Commented Mar 23, 2013 at 18:28
• @Idonknow, yes, exactly. Commented Mar 23, 2013 at 18:35

First it's useful to find the constant. That's easy, it would be $C = f(1)$.

Now assume that $f(x) = D$ for some $x > 1$ and some $D \ne C$. Then there is a smallest $x_0 > 1$ with this property. What is $f( \sqrt{x_0})$? Can you get to a contradiction from there? How do you handle the case $x < 1$?

• It's worth noting that that $x_0>1$ must exist because $f$ is continuous (in general, sets of reals needn't have a least element). Commented Mar 23, 2013 at 15:07
• how do we know the constant is $f(1)$? Commented Mar 23, 2013 at 15:11
• Just set $C = f(1)$ and then prove that $f(x) \ne C$ for some $x > 0$ leads to an contradiction. Commented Mar 23, 2013 at 15:14
• To understand the argument that I sketched above (it's NOT a proof, not even an informal one), draw a graph. Commented Mar 23, 2013 at 15:15
• @Idonknow, one reason for looking at what happens near $x=1$ is that it is the only point in your domain where $x=x^2$. One might suspect it would be important because of that. Commented Mar 23, 2013 at 15:23