# Question on “Proving $f(x) = 0$ everywhere”

I've some questions on Brian M Scott's proof for Proving $f(x) = 0$ everywhere. Please do combine my thread with the original which might be easier to read. Thank you.

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

My version of Brian M Scott's Proof (by Contradiction) $^{\huge{2}}$:

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 $a$. $^{\huge{3}}$ $^{\huge{4}}$ By definition of $A$ , $a \in (0,1]$.
Then there exists $0 < d<\min\{a,1-a\}$ such that $f(a)=\frac12\Big(f(a-d)+f(a+d)\Big).$

But since $f$ attains its maximum at $a$, $f(a)=f(a-d)$ $^{\huge{5}}$, which is impossible.

Question 1: How can I understand this complicated definition of $f(x)$ intuitively? This might help me with better understanding this problem.

Question 2: Is there a direct way to show this? Why choose to prove by contradiction?

Question 3: I don't understand why we need $A$ to be closed?

Question 4: On top of Question 3, wouldn't Extreme Value Theorem here be enough? I'd just say: "Since $f$ is continuous, then by Extreme Value Theorem, $A$ has a least element $a$."

Question 5: $max f(x) = f(a)$ means $f(a) \geq f(x)$ for all $x \in domf$. How does this imply $f(a) = f(a - d)$?

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Question 2 is a good one i don't see a direct proof for this one, did a left another question open ? –  Dominic Michaelis Feb 17 '13 at 15:00
Sorry, what did you mean by "did a left another question open?" Thank you for your help so far. –  dresserse Feb 17 '13 at 15:28
If all questions are answered, would you be so kind to accept an answer? –  Dominic Michaelis Feb 17 '13 at 16:44
Thank you very much Dominic Michaelis! Yes, of course! I stepped away. –  dresserse Feb 17 '13 at 17:05

Mh how to descripe the function the best, the function has the property, that it is the mean of two symmetric positioned values.
$A$ is closed because it is a preimage of a continuous function,
The extrem value theorem only works for compact sets.
Well since the Definition of your function you know $$f(a)=\frac{1}{2} (f(a-d)+f(a+d))$$ because of $f(a-d)\leq f(a)$ and $f(a+d)\leq f(a)$ it follows $f(a-d)=f(a+d)=f(a)$

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Thank you. Some follow-ups: $\Large{\text{Question 1}}$ : What does “Mh mean”? $\Large{\text{Question 3}}$: Are you using this result: “f is continuous iff if the inverse image of every closed set is a closed set”? Is there something easier? $\Large{\text{Question 4}}$: Isn’t A closed and bounded? $A := {x \in [0,1] : … }$ $\Large{\text{Question 5}}$: Sorry, still don’t see this. How does $$A \leq C \text{ and } B \leq C \text{ and } C = \frac{1}{2}(A + B)$$ imply $A = B = C$? –  dresserse Feb 17 '13 at 15:31
Mh means i am not sure if there isn't a better explanation, to question 3 yes that is exactly what i am using, to question for $A$ is bounded but not necessary bounded, taking for example the function $$g(x)=\left\{ \begin{array}{rl} \sin\left( \frac{1}{x}\right) & x\neq 0 \\ 1 & x=0\\ \end{array} \right.$$ the set $A$ would be not open nor closed. ok Taking $$C=\frac{1}{2} (A+B)$$ we know $$C\leq \max\{A,B\}.$$ But we even know $$C\geq \max\{A,B\}$$, so $$C=\max\{A,B\}$$. But since $A=B$ we know $A=B=C$ –  Dominic Michaelis Feb 17 '13 at 15:33
@Dominic.Michaelis: Thank you very much. $\Large{\text{Question 3}}$: Do you mean $A$ is bounded in this question, but $A$ doesn't have to be bounded in general? I'm just a bit confused since you wrote "$A$" is bounded but not necessarily bounded"? $\Large{\text{Question 5}}$: How do we know $A = B$? –  dresserse Feb 17 '13 at 16:20
oh my brain trolled me :D i mean $A$ is always bounded, but in general it doesn't need to be closed (can't correct the comment anymore). It is closed cause $f$ is continuous. –  Dominic Michaelis Feb 17 '13 at 16:21
@Dominic.Michaelis: Great! Thanks. My last follow-up is: $\large{\text{Question 5:}}$ How do we know $f(a - d) = f(a + d$? The function $f(x)$ doesn't have to be symmetric about $x = a$? –  dresserse Feb 17 '13 at 16:25