If $f(\frac{x+f(x)}{2})=x$, for all $x \in \Bbb R$, Prove that $f(x)=x$ for all $x\in \Bbb R$.

Question

Let $f: \Bbb R \rightarrow \Bbb R$ be a strictly increasing function such that $f(\frac{x+f(x)}{2})=x$, for all $x \in \Bbb R$. Prove that $f(x)=x$ for all $x\in \Bbb R$.

I don't understand this proof:

Suppose that for some $x\in\Bbb R$, $f(x)\neq x$. Then either $f(x)>x$ or $f(x)<x$. In the first case $\frac{f(x)+x}{2}>x$, but since $f$ is strictly increasing, it follows that $f(\frac{f(x)+x}{2})>f(x)>x$, which is a contradiction with the assumption. Analogously for $f(x)<x$.

In particular I don't understand this part

In the first case $\frac{f(x)+x}{2}>x$...

I can't see how they got that from $f(x)>x$? Does it mean $f(x)=\frac{f(x)+x}{2}$? If so, why?

Answer 1

Note that if $a<b$ then $a < {1 \over 2} (a+b) < b$.

We are given that $f$ is strictly increasing.

If $f(x) > x$ then $f(x) > {1 \over 2} (f(x)+x) > x$ and applying $f$ to all of these gives $f(f(x)) > f({1 \over 2} (f(x)+x)) > f(x)$.

Since we have assumed that $f(x) >x$, this shows that $f({1 \over 2} (f(x)+x)) > f(x) > x$ which contradicts the fact that $f({1 \over 2} (f(x)+x)) = x$.

Similarly for the other side.

Answer 2

$f(x)>x$ implies $f(x)+x>2x$, so $\frac{f(x)+x}{2}>\frac{2x}{2}=x$.