# What are the necessary conditions for $\delta =\min \left\{ f^{-1}\left( x_{0}-\varepsilon \right) ,f^{-1}\left( x_{0}+\varepsilon \right) \right\}$?

I was doing several exercises in which I was asked to find a delta, given a particular continuous function. Soon I started thinking for a general way to find the delta. I've come up with something, but I'm not sure if I can always apply it or not. So here is my question:

Let $f:D\rightarrow \mathbb{R}$ be a continuous bijective function on $D$ and $x,x_0 \in D$
If we want to find a $\delta$ such that $$\forall \varepsilon > 0 : \left| x_0 - x\right| \leq \delta \Rightarrow \left| f(x_0) - f(x)\right| \leq \varepsilon$$
what are the necessary conditions for always being able to set $\delta$ to $$\delta =\min \left\{ f^{-1}\left( x_{0}-\varepsilon \right) ,f^{-1}\left( x_{0}+\varepsilon \right) \right\}$$?

• What was your intuition for coming up with that condition? It doesn't seem very applicable. Commented Dec 7, 2016 at 14:48

A necessary condition is $f(x_0)=0.$

Assume $f(x_0)>0$ WLOG. Since $f$ is continuousthere exists $r>0$ such that $$|x-x_0|<r\implies |f(x)-f(x_0)|<\frac{f(x_0)}{3}.$$ Thus, for $\epsilon<r$ we have that

$$\min\{f(x_0-\epsilon),f(x_0+\epsilon)\}>\frac{2f(x_0)}{3}.$$

Thus $\delta >\frac{2f(x_0)}{3},$ which is not possible for $\epsilon$ small enough.

If $D$ is an interval, then by the IVT, $f$ is monotone, and so we can let $\delta=\min\{|x_0-f^{-1}(f(x_0)-\epsilon)|,|x_0-f^{-1}(f(x_0)+\epsilon)|\}$. This isn't your condition, but I'm not sure what we can do with yours...

• I see. Yes, you're right, my formula for choosing the delta wasn't very applicable indeed.
– Zyx
Commented Dec 7, 2016 at 14:54
• @Abdallah Hammam I'm sorry that your answer got downvoted, but it wasn't me.
– Zyx
Commented Dec 7, 2016 at 15:40

You write $|f(x)-f(x_0)|=|x-x_0|g(x)$

and you add on $x$, the condition

$|x-x_0|<\delta_0$ and find $K>0$ such that

$g(x)\leq K$.

then you take $\delta=\min(\delta_0,\frac{\epsilon}{K})$.