Continuous function on compact interval $[a,b]$ with non-negative values

Question

let $f:[a,b]\longrightarrow[0, \infty)$ be a continuous function satisfying the following: $f(\frac{a+x}{2})+f(\frac{2b+a-x}{2})=f(x), \forall x \in [a,b]$. Then the only function that satisfies these conditions is $f\equiv 0$? Setting $x=a$, we obtain $f(a)+f(b)=f(a)$ thus $f(b)=0$. Next, for $x=b$: $f(\frac{a+b}{2})+f(\frac{a+b}{2})=f(b)$, hence $f(\frac{a+b}{2})=0$. How can one use the continuity to conclude that $f\equiv 0$? Thank you!

Answer 1

First, we can assume, without loss of generality, that $a=0$ and $b=1$. A homothetic transformation transforms one problem into the other.

Thus, let $f:[0,1]\to\mathbb R^+$ such that

$f(x/2) + f(1-x/2)=f(x)$, $\forall x\in[0,1]$

As you noticed, putting $x=0$ and $x=1$ in the above equation gives $f(1)=f(1/2)=0$. Now, consider the following hyopthesis $(H_n)$ :

For all $k\in\{1,2,\dots,2^n\}$, $f(k/2^n)=0$.

We proved that $(H_0)$ and $(H_1)$ are true. Assume that $(H_n)$ is true for some $n\in\mathbb N$, and let $k\in\{1,2,\dots,2^n\}$. Then

$$ f(k/2^{n+1})+f(1-k/2^{n+1})=f(k/2^n)=0, $$

and since $f\geq 0$, we get that

$$ f(k/2^{n+1})=f(1-k/2^{n+1})=0. $$

Using the fact that

$$ \{1,2,\dots,2^n,2^n+1,\dots,2^{n+1}\}=\{1,2,\dots,2^n,2^{n+1}-(2^n-1),\dots,2^{n+1}\} $$

and since $f(2^{n+1}/2^{n+1})=f(1)=0$, $(H_{n+1})$ is proved.

Finally, $D=\{k/2^n,n\in\mathbb N,k\in\{1,\dots,2^n\}\}$ is dense in $[0,1]$, and $f$ being continuous on $[0,1]$ and $0$ on $D$, it is $0$ everywhere.