Bounded function - Proving $f(x)=0$ for all $x$ Let $f$ be a bounded function on $\mathbb R$
such that
$f(x) = \frac{1}{4}(f(\frac{x}{2})+f(\frac{x+1}{2}))$ for all $x$
Prove that $f(x)=0$ for all x
I let $|f(x)|≤M$ where $M$ is fixed then showed $\frac{M}{2^k}$ is a bound and this tends to $0$ as $k$ tends to $\infty$. But I wanted to ask if there is an easier way/alternative way.
 A: Instead of the limiting process you can choose $M$ as the
least upper bound:
$$A := \{ | f(x) | : x \in \mathbb R \} \, , \quad M :=  \sup A \, .$$
The supremum exists because every non-empty bounded set of
real numbers has a supremum (https://en.wikipedia.org/wiki/Least-upper-bound_property). 
Then for all $x \in \mathbb R$
$$
|f(x)| = \left| \frac{1}{4}(f(\frac{x}{2})+f(\frac{x+1}{2})) \right|
\le \frac{1}{4}(M + M) = \frac {M}2
$$
so that $\frac {M}2$ is also an upper bound of $A$.
Since $M$ is the least upper bound (supremum) it follows
that
$$
M \le \frac {M}2
$$
and therefore $M = 0$.
A: Another way: 
If $M = \sup \{|f(x)|: x \in \mathbb R\} > 0$, there is some $x$ for which
$|f(x)| > M/2$. But 
$$\max\left(\left|f\left(\frac{x}{2}\right)\right|, \left|f\left(\frac{x+1}{2}\right)\right|\right) \ge \frac{1}{2} \left( \left|f\left(\frac{x}{2}\right)\right|+ \left|f\left(\frac{x+1}{2}\right)\right|\right) \ge \frac{1}{2} \left| f\left(\frac{x}{2}\right)+ f\left(\frac{x+1}{2}\right)\right| = 2 f(x) > M $$ 
contradiction.
