# Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$

Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$.

We have $f(0) = 0$ and $f(x) = 2f(2x) - x$, but I am not sure how to convert this functional equation into something that is easier to solve. Maybe using induction may work, but I don't see an easy way to induct since we only have one variable.

• $f(x)=\frac{x}{3}$ works. – vadim123 Apr 15 '16 at 20:30

Set $g(x)=f(x)-ax$; then $f(x)=g(x)+ax$ and $$g(x)+ax+x=2(g(2x)+2ax)=2g(2x)+4ax$$ and we can choose $a=1/3$, so the equation becomes $$g(x)=2g(2x)$$ For $x/2$, we get $$g(x)=\frac{1}{2}g(x/2)=\frac{1}{4}g(x/4)=\dots=2^{-n}g(2^{-n}x)$$ Since $g$ is continuous, we have $$g(x)=\lim_{n\to\infty}2^{-n}g(2^{-n}x)=0$$

• How is it true that $g(x)=\lim_{n\to\infty}2^{-n}g(2^{-n}x)=0$? – user19405892 Apr 22 '16 at 19:16
• @user19405892 $\lim_{n\to\infty}2^{-n}=0$ and $g(0)$ is continuous at $0$, so you get $0\cdot g(0)$. – egreg Apr 22 '16 at 19:32

We prove that there is only one function $f:\mathbb R\to\mathbb R$ that satisfies $2f(2x)=f(x)+x$ and \begin{align*} \lim_{t\to0}tf(xt)=0 \end{align*} for all $x\in\mathbb R$, namely $f(x)=x/3$. Note that if $f$ is continuous, it satisfies this limit condition (why?)!

By iterating $f(x)=\frac{1}{2}f(x/2)+x/4$ one finds the formula \begin{align*} f(x)=\frac{1}{2^n}f\left(\frac{x}{2^n}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k,\quad n\geq 1,x\in\mathbb R.\qquad (\star) \end{align*} This can also be shown by induction: For $n=1$, this is just the definition of $f$. Assume that $(\star)$ has been proven for some $n\geq 1$. Then, again by definition, \begin{align*} f\left(\frac{x}{2^n}\right)=\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}}, \end{align*} and if we put this into our formula (induction!), we obtain \begin{align*} f(x)= \frac{1}{2^n}\left(\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k= \frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k= \frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}\sum_{k=0}^{n}4^k, \end{align*} which proves $(\star)$.

Now, if we fix $x\in\mathbb R$ and take the limit $n\to\infty$ in $(\star)$ we obtain due to the limit condition that $f(x)=x/3$.

Try with a polynomial of degree $n$ $$\sum_{k=0}^{k=n}a_kx^k$$ We have $$\sum_{k=0}^{k=n}a_kx^k+x=\sum_{k=0}^{k=n}2^{k+1}a_kx^k$$ It follows $$\begin {cases}a_k=2^{k+1}a_k;\space k=2,3,.....,n\\a_1x+x=4a_1x\\a_0=2a_0\end{cases}$$

Hence the only polynomial satisfying the property is $$f(x)=\frac x3$$

Similarly with an homography $$\frac{ax+b}{cx+d}$$ one gets at once $c=0$ and $d=3a$ which comes to $f(x)=\frac x3$

It is doubtful that the property is worth for rational or transcendental function. I stop here.

Exists an elegant and general proof to show that $$f(x)=\frac{x}{3}$$ is the only one solution.

## Elegant proof

Using complex variable, rewrite the functional equation as $$2f(2z)-f(z)=z$$. Since $$2z$$ and $$z$$ are complex-differentiable at $$z\in\mathbb{C}$$ it follows that $$f(z)$$ is necessarily complex-differentiable for all $$z\in\mathbb{C}$$ and hence necessarily continuous. Moreover, since $$2$$, $$2z$$, $$-1$$ and $$z$$ are finite-grade polynomials then $$f(z)$$ too and

$$\deg(f) = \max( \frac{\deg(z) - \deg(2)}{\deg(2z)} , \frac{\deg(z) - \deg(-1)}{\deg(z)} ) = \max(1,1) = 1.$$

As a consequence, the general solution is $$f(z) = c_0+zc_1$$ where $$c_0 = 0$$ and $$c_1 = \frac{1}{3}$$ by replacing in the functional equation. Therefore, $$f(z) = \frac{z}{3}$$ is the only one solution.

## On a general proof

Of course, $$f(z)$$ is necessarily continuous because it is necessarily complex-differentiable. Note that the fact that being a complex-differentiable function is an even stronger property than being a continuous function. Hence if you have a functional equation

$$A(z) f(U(z)) + B(z) f(V(z)) = P(z),$$

where $$A(z)$$, $$B(z)$$, $$U(z)$$, $$V(z)$$ and $$P(z)$$ are complex-differentiable in an open $$\Omega\subset\mathbb{C}$$, then $$f(z)$$ is also complex-differentiable for all $$z\in\Omega$$ (HINT: you can use the Cauchy-Riemann equations to prove this). Therefore, in order to get a general solution you can always compare the Taylor's coefficients of both members of the functional equation.

In particular, if you have that $$A(z)$$, $$B(z)$$, $$U(z)$$, $$V(z)$$ and $$P(z)$$ are finite order polynomials, as in your case, then $$f(z)$$ is complex-differentiable for all $$z\in\mathbb{C}$$ and its Taylor series exists, is unique and converges for all $$z\in\mathbb{C}$$. Moreover, for these cases it can easily be shown that $$f(z)$$ will always be a finite polynomial. In fact,

$$\deg f = \max( \frac{\deg P - \deg A}{\deg U} , \frac{\deg P - \deg B}{\deg V} ),$$

that you can use to avoid comparing infinite coefficients.