5
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Where is the problem from? $\endgroup$ – Jonas Meyer Jan 20 '15 at 5:36
4
$\begingroup$

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 $$f(x) = \sum_{k=1}^{\infty}\frac{x}{4^k} = \frac{x}{3}$$

$\endgroup$
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
17
$\begingroup$

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 $h(x)=0$.

Proof:

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 $f(x)=g(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.