0
$\begingroup$

I need to show that the following function is constant: $f:(0,\infty)\to\mathbb{R}$, $f(2^x)=f(3^x)$ and $f$ is continuous. I need to find the value of $f(x)$.

So I did something but I do not know how to continue. Let $x\to \log_2 x$ and we get that $f(x)=f(3^{\log_2 x)}$. Now, let $x\to 3^{\log_2 x}$ and we get that $f(3^{\log_2 x)}=f(3^{\log_2 3^{\log_2 x})}$. And so on. I think I need to find a rule. Do you have any idea?

  • 0
    Hint: One way to approach the problem would be to take the derivative! – 2017-01-09
  • 0
    @cool.coolcoolcool this will, however, only cover the derivable functions... – 2017-01-09
  • 0
    I do not know how to use derivates yet. I need to learn them. – 2017-01-09
  • 0
    Corollary: The only continuous functions $f:(0,\infty)\to\mathbb{R}$ such that $f(x)=f(x^a)$ for every $x$ in $(0,\infty)$, for some $a\ne\pm1$, are constant. – 2017-01-09
  • 0
    How do you know that? – 2017-01-09

2 Answers 2

2

Let $y= 3^x$ and $z = 2^x$, then you can write that $z = y^{\log_3 2}$. The functional equation on $f$ becomes $f(y^{\log_3 2}) = f(y)$.

Let $a>0$, $b>0$ and $f(a)\ne f(b)$. Consider the sequences $a_n = (a_{n-1})^{\log_3 2}$, $a_1=a$ and similarly $b_n$.

Clearly, $$\forall n\quad f(a_n)=f(a),\quad f(b_n)=f(b)$$ $$\lim a_n =1,\quad \lim b_n = 1,$$ which leads to a contradiction - two sequences both converge to $1$, yet the values of $f$ on these sequences do not converge simultaneously to $f(1)$.

  • 0
    I do not understand what you do after you find f(y). – 2017-01-09
  • 0
    @OvyOvy you have $f(y) = f(y^{\log_3 2}) = f((y^{\log_3 2})^{\log_3 2})=\ldots $. The sequences formalize the chain of identites. – 2017-01-09
  • 0
    I tried to do something like this as I said in description. Do you have any idea of that is correct? – 2017-01-09
0

First observe that the two functions $f(2^x)$ and $f(3^x)$ are continuous in the variable $x$, and both are defined on the whole real line. Let's say $f(1)=k$. We want to show that $f(x)=k$ for all $x\in (0,\infty)$. We use epsilon-delta arguments to show this.

By continuity at $x=0$ of the two functions $f(2^x)$ and $f(3^x)$, we have for any $\varepsilon\gt0$, there exists $\delta_1,\delta_2\gt0$ such that whenever $\lvert x\rvert\lt \delta_1$, we have $\lvert k-f(2^x)\rvert\lt \varepsilon$, while whenever $\lvert x\rvert\lt \delta_2$, we have $\lvert k-f(3^x)\rvert\lt \varepsilon$. Choose $\delta:=\min\{\delta_1,\delta_2\}$ so that whenever $\lvert x\rvert\lt \delta$, we have simultaneously $\lvert k-f(2^x)\rvert\lt \varepsilon$ and $\lvert k-f(3^x)\rvert\lt \varepsilon$.

Provided that all $x\in(-\delta,\delta)$ satisfies $\lvert k-f(3^x)\rvert\lt \varepsilon$ and $\lvert k-f(2^x)\rvert\lt \varepsilon$, we have that all $x\in(-\delta\log_2 3,\delta\log_2 3)$ also satisfies $\lvert k-f(3^x)\rvert\lt \varepsilon$ and $\lvert k-f(2^x)\rvert\lt \varepsilon$ because of the equality $f(2^x)=f(3^x)$ $\color{red}{(1)}$, and this process of multiplying the interval by $\log_2 3$ can be repeated indefinitely so that we obtain that every $x\in\Bbb R$ satisfies $\lvert k-f(3^x)\rvert\lt \varepsilon$. This corresponds to that every $x\in(0,\infty)$ satisfies $\lvert k-f(x)\rvert\lt \varepsilon$.

This is true for every $\varepsilon\gt 0$, so that $f(x)=k$ for all $x\in(0,\infty)$.

Edit$:$ You may want to see a proof for $\color{red}{(1)}$. We rewrite "for all $x\in(-\delta,\delta)$, $\lvert k-f(3^x)\rvert\lt \varepsilon$" as $f(3^{({-\delta},{\delta})})\subseteq(k-\varepsilon,k+\varepsilon)$ (note that two statements are equivalent). Then we have $3^{({-\delta},{\delta})}=2^{(-\delta\log_2 3,\delta\log_2 3)}$, so that $f(2^{(-\delta\log_2 3,\delta\log_2 3)})=f(3^{({-\delta},{\delta})})\subseteq(k-\varepsilon,k+\varepsilon)$, i.e. for all $x\in(-\delta\log_2 3,\delta\log_2 3)$, $\lvert k-f(2^x)\rvert\lt \varepsilon$. By the equality $f(2^x)=f(3^x)$, we also have for all $x\in(-\delta\log_2 3,\delta\log_2 3)$, $\lvert k-f(3^x)\rvert\lt \varepsilon$.