I was asked to prove/disprove the following:
If $f:\mathbb{R} \to \mathbb{R}$ continuous function, so for all $x \in \mathbb{R}$ and $\epsilon > 0$, exists $\delta > 0$ s.t. if $y \in \mathbb{R}$ and $|y-x|< \epsilon \to ֻֻ|f(y)-f(x)|< \delta$
So my intuition told me it was wrong for some reason (maybe because the requirement looks a bit like the definition of uniform continuity) so here is what I did:
Let's consider $f:\mathbb{R} \to \mathbb{R}$ defined as $f(x)=x^2$
So it's continuous:
Let $\epsilon >0$. We need to show that we can find $\delta_{x_0} >0$ such that $|x-x_0|<\delta_{x_0}$ implies $|f(x)-f(x_0)|<\epsilon$ $ \ \ $ $\forall x,x_0 \in \mathbb{R}$
We have $$|f(x)-f(x_0)|<|x^2-x_0^2|=|x+x_0||x-x_0|$$ And let's choose $0<\delta_{x_0}<1$ so $|x-x_0|<\delta_{x_0}$ implies $|x+x_0|<1+2|x_0|$ so: $$|f(x)-f(x_0)|<|x^2-x_0^2|=|x+x_0||x-x_0|<(1+2|x_0|)\delta_{x_0}$$ So we can choose $\delta_{x_0}<\frac{\epsilon}{1+2|x_0|}$ to get what we want.
Now I want to show it doesn't satisfy the claim:
Let $\epsilon >0$ and $x \in \mathbb{R}$ we need to try and find $\delta >0$ s.t. $|y-x|<\epsilon \to \ \ |f(y)-f(x)|<\delta$ . So:
$$|f(y)-f(x)|=|y+x||y-x|<(\epsilon-2|x|)\epsilon$$
But now I can just choose $\delta >(\epsilon-2|x|)\epsilon $ and the claim will hold.
So am I looking at this wrong? Or maybe the function I chose wasn't right?
A function that is nowhere continuous and for which the given property holds is $$f(x)=\begin{cases}1&\text{if }x\in\mathbb Q\\0&\text{if }x\notin \mathbb Q\end{cases} $$
Given $x\in \mathbb R$ and $\epsilon>0$ you can pick $\delta=2$ and have that for all $y\in \mathbb R$ with (or without) $|x-y|<\epsilon$ we have $|f(x)-f(y)|<\delta$.
On the other hanbd, any $f\colon \mathbb R\to\mathbb R$ with your property is also continuous. This is because continujos $f$ is bounded on the compact interval $[x-\epsilon,x+\epsilon]$.