$\forall x\exists \epsilon>0$ such that $\forall 0

Question

$f(0)=1$, for any $x$ there exists $\epsilon>0$ such that $\forall 0<y<\epsilon$, $f(x+y)=f(x)$. Is it possible to prove that $f(1)=1$?

I am so sorry that I forgot to include that $y$ must be greater than zero! Otherwise the solution is obvious and does not involve any real analysis.

It seems really obvious: intuitively by induction, $f(nc)=1$ for any $n\in\mathbb N$, and some $c$. However, I am unsure that if $\epsilon$ is infinitesimal, then there might not exist a $n$ such that $n\epsilon=1$.

@DustanLevenstein not really pure math background with little understanding of compactness. — High GPA, Oct 19 '19 at 15:33
It doesn't follow that $f(n\epsilon)=1$ for all $n\in\Bbb N$, because $\epsilon$ depends on $x$. (It would follow if the proposition started with $\exists\epsilon\forall x$.) — TonyK, Oct 19 '19 at 15:36
Can you confirm that the hypothesis is only $\forall y < \epsilon$, i.e. $y$ can be negative or $\forall 0 \le y < \epsilon$... — mathcounterexamples.net, Oct 19 '19 at 15:37
I would consider the quantity $\sup{ c : f(x) = f(0) \text{ for all } x \in [0, c]}$. — Sera Gunn, Oct 19 '19 at 15:38
@mathcounterexamples.net note that the claim is false if you require $\forall 0 \le y < \epsilon$... — Dustan Levenstein, Oct 19 '19 at 15:39
My guess is that you meant $|y|<\epsilon$. The claim is true for that assumption, see this question — lulu, Oct 19 '19 at 15:46
@lulu It might take me a while to understand that question but I would surely try. — High GPA, Oct 19 '19 at 15:51

Hagen von Eitzen · Accepted Answer · 2019-10-19T15:46:43.320

We are given $$\tag1 \forall x\;\exists \epsilon>0\;\forall y<\epsilon\;f(x+y)=f(x).$$ Specialize to $x=1$: $$ \exists \epsilon>0\;\forall y<\epsilon\;f(1+y)=f(1).$$ So for suitable positive $\epsilon$, $$ \forall y<\epsilon\;f(1+y)=f(1).$$ Specialize to $y=-1$ (which certainly is $<\epsilon$): $$ f(0)=f(1).$$

After an edit of the question: We are given $$\tag2 \forall x\;\exists \epsilon>0\;\forall 0<y<\epsilon\;f(x+y)=f(x).$$ Then we have a counterexample, given by $$f(x)=\lfloor x\rfloor+1.$$ Then $(2)$ holds as we can pick $\epsilon=1>0$ when $x\in \Bbb Z$, and $\epsilon=\lceil x\rceil -x>0$ otherwise. However, $f(1)=2\ne f(0)$.

So let's assume additionally that $f$ is continuous on $[0,1]$. If $f$ is constant on $[0,1]$ we are done. So assume it is not and let $S=\{\,x\in[0,1]\mid f(x)\ne f(0)\,\}\ne\emptyset$ and $a=\inf S$. By continuity, $f(a)=f(0)$ and hence $f(x)=f(0)$ for $x\in[0,a]$. So if $a=1$ we are done. But if $a<1$, there exists $\epsilon>0$ with $f(x)=f(a)$ for $a<x<a+\epsilon$. It follows that $[0,a+\epsilon)\cap S=\emptyset$ and hence $a\ge a+\epsilon$, contradiction.

Upvoted though I have to say sorry for not be very clear of the question. $y>0$ and at least some real analysis must be used in solution — High GPA, Oct 19 '19 at 15:38

Robert Z · Answer 2 · 2019-10-19T15:55:15.233

Consider the function
$$f(x) = \begin{cases} 1 & \text{ if } x <1 \\ 0 &\text{ if } x \geq 1\end{cases}$$ It satisfies your conditions but $f(1)\not=1$.

In fact, $f(0)=1$, and for any $x$ there exists $\epsilon>0$ given by $$\epsilon(x) = \begin{cases} 1-x & \text{ if } 0<x <1, \\ 1 &\text{otherwise.} \end{cases}$$ such that $f(x+y)=f(x)$, $\forall y\in (0,\epsilon)$.

The given statement holds if we assume that $f$ is continuous on $[0,1]$.

score 1 · Answer 3 · answered Oct 19 '19 at 15:44

If the question is: $\forall x\exists \epsilon>0$ such that $\forall 0<y<\epsilon$, $f(x+y)=f(x)$. Show $f(1)=f(0)$, then the claim is false. Consider the function $$f(x) = \begin{cases}0 & \text{ if } x < 0.5 \\ 1 &\text{ otherwise.}\end{cases}$$ When $x$ is smaller than $0.5$, take $\epsilon = 0.5-x$. When $x \ge 0.5$, $\epsilon$ can be anything.

If the question is: $\forall x\exists \epsilon>0$ such that $\forall y, |y|<\epsilon$, $f(x+y)=f(x)$. Show $f(1)=f(0)$:

Hint: What is the derivative of $f$?

Mirko · Answer 4 · 2019-10-19T15:57:53.453

Are you assuming that $f$ is continuous? If not, a counter example would be $f(x)=1$ for $0\le x<\frac12$ and $f(x)=2$ for $\frac12\le x\le1$.

This is really a variation of the proof that $[0,1]$ is connected. I would assume something weaker than what you state, namely I would assume that :

$f(0)=1$, for any $x$ there exists $\epsilon>0$ such that whenever $0<y<\epsilon$, we have $f(x+y)=f(x)$. (Well apparently that was now fixed in the statement.)

Let $A=\{x\in[0,1]:f(x)=1\}$. Then $0\in A$. (I will finish when you confirm that $f$ is assumed continuous. Well, it was already done in another answer, let me include it anyway.) So, assume that $f$ is continuous and let $a=\sup(A)$. By continuity we have $f(a)=1$. If $a=1$ then we are done. If $a<1$ then $f(a+y)=1$ for some $\varepsilon>0$ and all $0<y<\varepsilon$, so then $a\ge a+\varepsilon$, a contradiction.

$\forall x\exists \epsilon>0$ such that $\forall 0

4 Answers4