How can I solve the following functional equation? $$f(2x) = \frac{f(x)+x}{2},$$ for $x \in \mathbb{R}$ with $f$ being a continuous function.

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    $\begingroup$ Where is the problem from? $\endgroup$ – Jonas Meyer Jan 20 '15 at 5:36

Here's an alternative proof. Notice that by substituting in $\frac{1}{2^{n+1}}x$ for $x$, we can obtain

$$f\left(\frac{1}{2^n}x\right) = \frac{1}{2}f\left(\frac{1}{2^{n+1}}x\right) + \frac{x}{2^{n+2}}$$

Using this expression, we find that

\begin{align*}f(x) &= \frac{1}{2}f\left(\frac{1}{2}x\right) + \frac{x}{4}\\ &= \frac{1}{2}\left(\frac{1}{2}f\left(\frac{1}{4}x\right) + \frac{x}{8}\right) + \frac{x}{4}\\ &= \frac{1}{4}f\left(\frac{1}{4}x\right) + \frac{x}{4}+\frac{x}{16}\\ \\&\vdots\\ &= \frac{1}{2^n}f\left(\frac{1}{2^n}x\right) + \sum_{k=1}^{n}\frac{x}{4^k} \end{align*}

Let's look at the behavior of this last term as $n\to \infty$. We have that $f\left(\frac{1}{2^n}x\right) \to f(0)$, which is finite, so $\frac{1}{2^n}f\left(\frac{1}{2^n}x\right) \to 0$. It follows that $$f(x) = \sum_{k=1}^{\infty}\frac{x}{4^k} = \frac{x}{3}$$

  • $\begingroup$ I understand this solution, but I still don't see how I should have thought of this myself. Could you recommend some material that teaches how to go about solving a functional equation? $\endgroup$ – user62029 Jan 20 '15 at 11:06
  • $\begingroup$ Well, the main idea of my solution was to construct a sequence $x_n$ converging to $0$ for every $x$, so I could relate all of the values $f(x)$ to $f(0)$ and so $f(x_n)$ was related to $f(x)$ somehow, so I could take the limit to relate $f(x)$ to $f(0)$. $\endgroup$ – user88319 Jan 20 '15 at 21:13
  • $\begingroup$ As far as material goes, there isn't really a unifying theory of solving functional equations, so lecture notes can be hard to find. I found a couple here and here which have some good tips and examples, but seem to be working on much harder (IMO) problems than you posted. This book also looks alright, but I haven't read it, so I can't necessarily 'recommend' it. $\endgroup$ – user88319 Jan 20 '15 at 21:16

Let $f(x)=ax$. Then it's easy to show that $a=1/3$.

Next prove uniqueness.

Claim: The only function $h(x)$ which is continuous at $x=0$ and satisfying $h(2x)=h(x)/2$ is $h(x)=0$.


It's easy to show that $h(0)=0$. Suppose $h(b)\ne0$ for some non-zero $b$. Then $h(b/2^N)=2^Nh(b)$. For all $\epsilon$, $\delta$, there exists $N$ such that $|b/2^N|<\delta$ but $|h(b/2^N)|>\epsilon$, contradicting the continuity of $h(x)$ at $0$.

Now suppose there are two functions $f(x)$ and $g(x)$ both continues at $x=0$ and satisfying $f(2x)=\frac{f(x)+x}{2}$ and $g(2x)=\frac{g(x)+x}{2}$. Define $h(x)=f(x)-g(x)$ which is continuous at $x=0$. Then $h(2x)=\frac{h(x)}{2}$ and hence $f(x)=g(x)$.


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