# functions satisfying $f(x)=2f(2x)$

How can I prove every smooth real function f satisfying $f(x)=2f(2x)$ is of the from $f(x)=k/x$ where $k$ is a constant? I have tried this for integers that can be divided by $2^n$, but then I cannot proceed anymore. Can anybody give me some help???

I presume your function $f:\mathbb{R}^+\mapsto \mathbb{R}^+$.

Define $g:\mathbb{R}^+\mapsto \mathbb{R}^+$ by $g(x)=x{f(x)},\quad \forall x\in\mathbb{R}^+.$ Then, you have $g(x)=g(2x).$ Now using this post to show that $g$ is constant. Consequently, you will arrive at what your wanted.

NB: You can generalize your question as follows:

Given a>1. Then a continuous function $f$ (not necessary smooth) satisfying $$f(x)=af(ax)$$ implies $f(x)=k/x$ for some constant $k$.

• I believe you meant $g(x)=xf(x)$. – user1337 Jun 10 '14 at 15:28
• Exactly @user1337. I've edited there. Thanks :D – Jlamprong Jun 10 '14 at 15:34

Hint: Set $$r(x)=\dfrac{1}{f(x)}$$

then

$$r(2x)=\dfrac{1}{f(2x)}=\dfrac{2}{f(x)}=2r(x)$$

From that you will get $r(nx)=nr(x)$ and you will find $r(x)=cx$.

• What happens to this change of variables at the zeros of $f$? – user1337 Jun 10 '14 at 15:27
• @user1337: if $f$ is constant and equal to $0$ then $f(x)=2f(2x)$ is satisfied and it is also smoot which show that this question has not uniqe answer under this assumptions. Now, if $f$ is not constant, above equation must true for all nonzero values of $f$ and it turns out to be all nonzero values of $f$ lies on $k/x$ and by using smoothness we can conclude. – mesel Jun 10 '14 at 15:33

We note that one solution is $g(x) = 1/x$. Suppose that another solution is $h(x)$. Then $$g(x)/h(x) = 2g(2x)/2h(2x) = g(2x)/h(2x)$$ But looking at it the other way around, we have $$g(x)/h(x) = g(x/2)/h(x/2)$$ continuing this inductively we have $$g(x)/h(x) = g(x/2)/h(x/2) = g(x/4)/h(x/4) = g(x/8)/h(x/8) = g(x/16)/h(x/16) = \dots$$ Thus we have for all $a$ and $b \in R$ $$g(a)/h(a) = \lim_{n\to\infty} g(a/2^n)/h(a/2^n) = \lim_{t\to 0} g(t)/h(t) = \lim_{n\to\infty} g(b/2^n)/h(b/2^n) = g(b)/h(b)$$

Thus $g(x)/h(x)$ evaluates to a constant for all $x$, which implies that $h(x)=kg(x)=k/x$ for some $k\in R$

I hope this helps. :)