1. ## Find all functions

Find all functions f : IR ---> IR : ∀ (x,y) ∈ IRē :

2. If we replace $f(t) ~~$ by $\frac{1}{g(t)}$

Then it becomes

$\frac{1}{g(\frac{x+y}{2})}[ g(y) + g(x) ] = 2$

$g(x) + g(y) = 2 g( \frac{x+y}{2})$

It implies that $g(t) = t ~~$ so $f(t) = \frac{1}{t}$

Is there another function satisifying the requirement ?

3. How about $f(t)=\frac1{at+b}$ for constants $a$, $b$ not both zero?

4. The equation $g\left(\frac{x+y}2\right)=\frac{g(x)+g(y)}2$ that occurs in simplependulum's reply is known as Jensen's equation.

To solve it, put $y=0$ and then $g\left(\frac x2\right)=\frac{g(x)+b}2$ where $b=g(0)$.

Put $x+y$ in this latter equation: $g\left(\frac{x+y}2\right)=\frac{g(x+y)+b}2$.

And now we see that $g(x+y)+b=g(x)+g(y)$.

Finally, put $h(x)=g(x)-b$. Then $h(x+y)=h(x)+h(y)$. This is the well-known Cauchy's equation.

The most general continuous solution of Cauchy's equation is $h(x)=ax$ where $a$ is constant. Therefore $g(x)=ax+b$ is the most general continuous solution of Jensen's equation.

However, it must be said that there exist solutions of Cauchy's equation on $\mathbb R$ which are not continuous, and they turn out to be extremely pathological. For example, if $h$ is one of these solutions and $I=(a,b)$ is any open interval then $h(I)$ is dense in $\mathbb R$.

So it makes sense in many cases to ask for conditions on the solution, such as continuity, or boundedness on a finite interval, or monotonicity, anything which would avoid these weird functions.