# Proving $f(x) = 0$ everywhere

I'm stuck on this problem, I've given it considerable effort and tried using the bounded nature of continuous functions on closed, bounded intervals but I just can't solve it. I think I might need to pick a clever sequence somewhere but I can't see exactly what to do.

Suppose that $f:[0,1] \to\mathbb{R}$ is continuous and $f(0) = f(1) = 0$. Also, suppose further that for all $x \in (0,1)$ there exists a $0 < d < \min\{x,1-x\}$ such that: $$f(x) = \frac12\Big(f(x-d) + f(x+d)\Big)\;.$$ Prove that $f(x)=0$ everywhere.

Thank You.

-
You changed the problem! Earlier you had $d = \min\{x, 1-x\}$. Can you please tell us where this problem is from? – Aryabhata Mar 3 '12 at 1:17
yes, I had it written down wrong! It's a question mid way through problem sheet 2 of the Cambridge 1st year Analysis course. – user27182 Mar 3 '12 at 1:33
Do you have a link to the problem sheet which we can look at. It seems to be an interesting course! – Aryabhata Mar 3 '12 at 1:33
Thanks........... – Aryabhata Mar 3 '12 at 1:44
Perhaps you should try to construct a proof by contradiction. Suppose $f(a) = b$ for some $a \in (0,1)$ and $b > 0$, and suppose $f$ is continuous. Then choose an $\epsilon > 0$, such that for each $\delta$ you can apply the rule for $f$ sufficiently many times to get sufficiently small intervals, and arrive at some contradiction with the definition of continuity. – TMM Mar 3 '12 at 1:50

For the current case

Suppose $f$ attains the maximum value $M \gt 0$, at $c$.

Now consider the set $S = \{ x: f(x) = M\}$.

Since $2f(c) = f(c-d) + f(c+d)$, we must have that $f(c-d) = f(c+d) = M$.

This implies that $\inf S = 0$.

Thus there is a sequence $c_n \to 0$ such that $f(c_n) = M$. Thus $M=0$.

Similarly, we can show that the minimum value is $0$.

For the earlier case

(Though the above proof still carries over, just having this here because of the curious sequence we get).

You basically get the sequence:

$x_{n+1} = 2x_n$ if $\ 0 \le x_n \le \frac{1}{2}$ and

$x_{n+1} = 2x_n - 1$ if $\frac{1}{2} \le x_n \le 1$

Consider $x$ which has a finite decimal representation in base $2$.

For $x_1 = x$, for such $x$, we can show that the sequence converges to $0$ or $1$ (see what happens to the digits before and after the decimal point).

Since such $x$ are dense, and $f$ is continuous, you are done.

-
Also see Bernoulli map. – TMM Mar 3 '12 at 1:32
@TMM: Thanks for the link! Very interesting. – Aryabhata Mar 3 '12 at 1:35
By showing that f(c-d)=f(c+d)=M, you can say that S is non empty set(is that was your purpose?, and you deduced that infS=inf{x\in(0,1) | f(x)=m }=0. I – Salech Alhasov Mar 3 '12 at 17:13
@SalechAlhasov: $S$ is non-empty because the maximum exists and is attainable because of continuity. inf S is zero because if $c \in S$, then $c -d \in S$. – Aryabhata Mar 3 '12 at 17:23

Suppose that $f$ is not identically $0$. Then $f$ attains a non-zero extremum at some point $c\in(0,1)$. Let $A=\{x\in[0,1]:f(x)=f(c)\}$; since $f$ is continuous, $A$ is closed and has a least element $a$; clearly $a>0$. There is a positive $d<\min\{a,1-a\}$ such that $$f(a)=\frac12\Big(f(a-d)+f(a+d)\Big)\;,$$ and since $f$ attains its maximum at $a$, $f(a)=f(a-d)$, which is impossible.

HINT for an earlier version of the problem in which $d=\min\{x,1-x\}$:

$\quad1.$ Given $f(0)$ and $f(1)$, what is $f\left(\frac12\right)$?

$\quad2.$ Given $f\left(\frac12\right)$ and the earlier points, what are $f\left(\frac14\right)$ and $f\left(\frac34\right)$?

$\quad3.$ Given $f\left(\frac14\right)$ and $f\left(\frac34\right)$ and the earlier points, what are $f\left(\frac18\right),f\left(\frac38\right),f\left(\frac58\right)$, and $f\left(\frac78\right)$?

$\quad4,5,\dots$

$\quad\omega.$ A continuous function is completely determined by its values on a dense set.

-
I think that you should have changed 5 to $\omega$, after all the fourth step is infinitely long! :-) – Asaf Karagila Mar 3 '12 at 1:23
@Asaf: I wanted to, but the automatic numbering system overrode it, and I was too lazy to work around it. – Brian M. Scott Mar 3 '12 at 1:34

Suppose that $f$ is not identically $0$. Without loss of generality, we may assume that $M:=\max\{f(x):0<x<1\}>0$. Define $x_0:=\inf\{x\in(0,1): f(x) =M\}$. Then $x_0\in(0,1)$ and $f(x_0)=M$, but the averaging property hypothesized will yield another point $x_1\in(0,x_0)$ with $f(x_1)=M$, and a contradiction.

-

I have not worked it out completely, but I think that if you look at the numbers $x = 0/k, 1/k, \ldots, k/k$ for some positive integer $k$ and use the aforementioned property of $f$, you get equations implying $f(i/k) = 0$ for all $i = 0, \ldots, k$. Doing this for every $k$ then implies that $f(x) = 0$ for all $x \in (0,1) \cap \mathbb{Q}$, and since $f$ is continuous this should imply $f(x) = 0$ for all $x \in (0,1)$.

-