$2f(x)=f(y) \Rightarrow 2f(tx)=f(ty)$

Question

Find all continuous and strictly monotonic function $f:[0,\infty)\to \Bbb R$ such that:

• If there is a pair $(x,y)\neq (0,0)$ such that $2f(x)=f(y)$ then $2f(tx)=f(ty)$ for all $t>0$;

• There is at least one pair $(x,y)$ where the above condition hold.

What I got so far:

• Check that $f(x)=ax^b$ is a solution. I want to discover if there is any other;

• $t\to 0$ then $2f(0)=f(0) \Rightarrow f(0)=0$;

• We can prove that $y> x$;

• $t:=t/x$ and then $2f(t)=f(tk)$, for all $t>0$, with $k=y/x >1$;

• $f(tk^n)=2^nf(t)$ and setting $t=1$ we get $f(k^n)=2^nf(1)$. It give us that, if $f(1)>0$ then $f$ is increasing, otherwise, $f$ is decreasing.

Answer 1

Actually, there are many such $f$.

Here I assume $f$ is positive, increasing. (If you have decreasing $f$, try to consider $-f$.)

Take $a<b$ s.t. $2f(a)=f(b)$. Let $d$ be $b/a$.

The restricted function $g=f|_{[a,b]}$ should be continuous, strictly increasing on $[a,b]\subset(0,\infty)$ and should satisfy $2g(a)=g(b)>0$.

Now, the main claim is that we can uniquely extend any such $(g,[a,b])$ to a function $f$ satisfying all the properties you asked.

(Uniqueness) As you mentioned above, $f(0)=0$. Choose any $t>0$. $\exists k\in\mathbb{Z},c\in[a,b]$ s.t. $t=d^kc$. Then, $f(t)=2^kf(c)$.

Therefore $f$ is unique on $[0,\infty)$.

(Existence) Define $f$ as follows.

$$f(t)=\begin{cases}0,t=0\\ 2^kg(c),t=d^kc,k\in\mathbb{Z},c\in[a,b]\end{cases}$$

Now it is left to check that such $f$ is positive, continuous and strictly increasing.

$f$ is positive by its definition.

$\forall p,q$ s.t. $0<p<q$, $f(p)<f(q)$ since $p=d^{k_1}c_1,q=d^{k_2}c_2,k_i\in\mathbb{Z},c\in[a,b)\Rightarrow k_1<k_2\,or\,k_1=k_2,c_1<c_2$, and if $k_1<k_2$, $f(p)=2^{k_1}f(c_1)<2^{k_1}f(b)\le \frac{1}{2}\cdot 2^{k_2}f(b)=2^{k_2}f(a)\le 2^{k_2}f(c_2)=f(q)$, and if $k_1=k_2,c_1<c_2$, $f(p)=2^{k_1}f(c_1)<2^{k_2}f(c_2)=f(q)$.

Thus $f$ is strictly increasing.

$f$ is continuous on each $(d^ka,d^kb)$ since $g$ is continuous. At $d^ka=d^{k-1}b$, $f$ is right continuous since so was $g$ at $b$, and is left continuous since so was $g$ at $a$.

Now we check that $f$ is continuous at $0$.

$\forall\epsilon>0$, we can take $N$ s.t. $2^{-N}\cdot f(a)<\epsilon$, then $\forall x\in(0,d^{-N}a)$, $f(x)<f(d^{-N}a)=2^{-N}f(a)<\epsilon$.

Then, since $f$ is positive, $\lim_{x\rightarrow 0}f(x)=0$ and this means $f$ is continuous at $0$.

-So we are done.

P.S. You can also rewrite this in the following more closed form.

$f$ is answer iff $$f(t)=\begin{cases}0,t=0\\ 2^kg(c),t=d^kc,k\in\mathbb{Z},c\in[1,d]\end{cases}$$ where $d>1$ and $g:[1,d]\rightarrow(0,\infty)$ is a continuous strictly monotonic function s.t. $g(d)=2g(1)$.

Answer 2

We assume $f$ is increasing (take $-f$ if it is not.) By the strictly increasing condition, $f(x)>0$ for all positive $x$. We now define a new function $g:$ $$g(\ln(x)) = \ln (f(x)),$$ for all positive $x$. Let $a= \ln(x)$, $b= \ln(y)$, $c= \ln(2)$. The condition becomes $g(t+a) +c = g(t+b) \: \forall t \in \mathbb{R}.$ Furthermore, note that $g$ is still increasing and continuous. Let $S$ be the set of functions $h: \; [0, b-a] \rightarrow [0,c]$ such that $h$ is strictly increasing, continuous, and satisfies $h(0)=0, h(b-a)=c$. It is clear that all solutions to $g$ are given by $g(n(b-a)+t) = h(t) + cn + d$ for any $h \in S$, all integers $n$, all $0 \leq t < b-a$ and any constant $d$. Using $f(x) = e^{g(\ln (x))}$, and then remembering that $f$ was possibly negated at the start, yields all possible $f$.

Answer 3

Calling $y = x+\delta$ with $\delta > 0$ we have if

$$ 2f(x)-f(x+\delta) = 0\Rightarrow 2f\left(\frac{x}{\delta}\right)-f\left(\frac{x}{\delta}+1\right) = 0 $$

now calling $u = \frac{x}{\delta}$ we have the recurrence functional equation

$$ 2f(u) - f(u+1) = 0 $$

with solution

$$ f(u) = 2^u \Phi(u) $$

where $\Phi(u)$ is any periodic function with period $1$,

and also choosing $\Phi(u) = C_0$

$$ f\left(\frac{x}{y-x}\right)=C_02^{\frac{x}{y-x}} $$

for $x \ne y$ and such that $2f(x) = f(y)$

Answer 4

I found another solution and I think it is insteresting to post here. It was inspired in change of variable sugested by @Yuri Negometyanov.

Let's set $tx=2^z$ and $ty=2^w$, where $z,w \in \Bbb R$ and $t>0$. We also call $g:\Bbb R \to \Bbb R$ such that $g(s)=f(2^s)$, and then we get $$2f(tx)=f(ty)\to 2f(2^z)=f(2^w)\to 2g(z)=g(w)$$

but $\frac{ty}{tx} =\frac{y}{x}=2^{w-z}$ and then $w=z+\log_2 (y/x)=z+d$. So, $$2g(z)=g(z+d)$$ wich has solution $$g(z)=2^{z/d}\phi(z)$$ where $\phi(z)=\phi (z+d)$, it means that $\phi$ is a periodic function with period $d$. . But $\phi$ has to follow some constrains. For example, $\phi(z)=\sin^2\left(z\frac{\pi}{d}\right)+10$ works and $\phi(z)=\sin^2\left(z\frac{\pi}{d}\right)$ don't. So, suppose that $f$ is increasing (if it is decreasing we work with $-f$), then $g$ is increasing, which means that for $p>q$ we have $g(p)>g(q)$, and $$2^{p/d}\phi(p)>2^{q/d}\phi(q)\to 2^{(p-q)/d}>\frac{\phi(q)}{\phi(p)}.$$