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how can I find out the general solution of Jensen's functional equation?

The Jensen's functional equation is : $$ \frac{f(x)+f(x^{\prime})}{2}= f(\frac{x+x^{\prime}}{2})$$

I need the general solution, i found something on $\mathbb{Q}$ but i need general, so i need te proof

daw
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Maria Pop
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1 Answers1

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This is discussed, inter alia, in Evan Chen's "Introduction to functional equations". Briefly:

$\begin{align*} \frac{f(x) + f(y)}{2} &= f\left(\frac{x + y}{2}\right) \end{align*}$

Clearly, $f(x) = k x + c$ works. So it makes sense to try $g(x) = f(x) - f(0)$:

$\begin{align*} \frac{g(x) + g(y) + 2 f(0)}{2} &= f\left(\frac{x + y}{2}\right) \\ \frac{g(x) + g(y)}{2} &= f\left(\frac{x + y}{2}\right) - f\left(\frac{0 + 0}{2}\right) \\ \frac{g(x) + g(y)}{2} &= g\left(\frac{x + y}{2}\right) \end{align*}$

Same equation, but we know $g(0) = 0$. Now we have both:

$\begin{align*} g(x) + g(0) &= 2 g\left(\frac{x}{2}\right) \\ g(x) &= 2 g\left(\frac{x}{2}\right) \\ g(x) + g(y) &= 2 g\left(\frac{x + y}{2}\right) \\ &= g(x + y) \end{align*}$

This is Cauchy's functional equation.

First off, $2 g(0) = g(0)$, so $g(0) = 0$; $g(x) + g(-x) = g(0) = 0$, so $g(-x) = - g(x)$. If $n$ is a positive integer, by repeating the equation $n$ times we get:

$\begin{align*} g(n) &= n g(1) \end{align*}$

Take a rational $p / q$ ($p, q$ integers). Repeating the function $q$ times we get:

$\begin{align*} q \cdot g\left(\frac{p}{q}\right) &= g\left(q \cdot \frac{p}{q}\right) \\ g\left(\frac{p}{q}\right) &= \frac{g(p)}{q} \\ &= \frac{p}{q} \cdot g(1) \end{align*}$

But this does not define the values of $g$ at irrational $x$, we can take e.g. $g(\sqrt{2})$ to be anything, and the equation allows us to extend this to, for rational $r$:

$\begin{align*} g(r \sqrt{2}) &= r g(\sqrt{2}) \end{align*}$

But, again by the equation, for rational $r_1$ and $r_2$, we can write:

$\begin{align*} g(r_1 + r_2 \sqrt{2}) &= g(r_1) + g(r_2 \sqrt{2}) \\ &= r_1 g(1) + r_2 g(\sqrt{2}) \end{align*}$

Same for any other irrational that is linearly independent over $\mathbb{Q}$... this gives a whole truckload of different solutions, all endearingly discontinuous.

And it looks so innocent...

vonbrand
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