lfdahl said:
The function, $f$, is defined on the interval $[0;1]$, and satisfies the following conditions:
(a). $f(0) = f(1) = 0$.
(b). For any $a,b \in [0;1]$: $f\left ( \frac{a+b}{2} \right ) \leq f(a)+f(b)$.
Prove, that the equation: $f(x)=0$ has infinitely many solutions.
Give an example of such a function, that is not identically zero on any subinterval of $[0;1]$.
[sp]Putting $a=0$ and $b=1$, you see that $f\bigl(\frac12\bigr) \leqslant 0+0 = 0$.
Now suppose that $f\bigl(\frac12\bigr) = k < 0$. Then (putting $a=0$ and $b= \frac12$), $f\bigl(\frac14\bigr) \leqslant 0+k = k$. Also ((putting $a=1$ and $b= \frac12$), $f\bigl(\frac34\bigr) \leqslant 0+k = k$. Then putting $a=\frac14$ and $b = \frac34$, you see that $f\bigl(\frac12\bigr) \leqslant k+k = 2k$. But $f\bigl(\frac12\bigr) = k$, so that $k\leqslant 2k$. That implies that $k\geqslant0$, contradicting the assumption that $k<0$. It follows that $f\bigl(\frac12\bigr)$ must be $0$.
Next, put $a=0$ and $b=\frac12$ to see that $f\bigl(\frac14\bigr) \leqslant 0+0 = 0$. Also, exactly as in the previous paragraph, if $f\bigl(\frac14\bigr) < 0$ then we get a contradiction. So $f\bigl(\frac14\bigr) = 0$, and similarly $f\bigl(\frac34\bigr) = 0$.
Continuing to bisect the interval in this way, you see that $f(x) = 0$ for every dyadic rational $x$. Thus $f(x) = 0$ has infinitely many solutions.
A nontrivial example of such a function is $f(x) = \begin{cases}0&\text{if }x\text{ is rational,}\\1&\text{if }x\text{ is irrational.}\end{cases}$ For that function, if $a$ and $b$ are both rational then so is $\frac{a+b}2$, and so the inequality $f\left ( \frac{a+b}{2} \right ) \leqslant f(a)+f(b)$ becomes $0\leqslant 0+0$, which is certainly true. Otherwise, if at least one of $a$, $b$ is irrational then the right side of the inequality will be at least $1$, and the left side will be at most $1$. So again the inequality is satisfied.
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