# If $f$ is continuous, then $\exists \delta > 0$ so that $\vert f(x) \vert \geq \frac{\vert f(x_0) \vert}{2} > 0$

Problem:

Let $$\, f : \mathbb{R} \rightarrow \mathbb{R}$$ be continuous at the point $$x_0$$ and $$f(x_0) \ne 0 \ \forall x \in \mathbb{R}$$. Show that $$\exists \delta > 0$$ such that $$\vert f(x) \vert \geq \ \frac{\vert f(x_0) \vert}{2}>0 \ \forall x \in \mathbb{R}$$ for $$\vert x-x_0 \vert < \delta$$

Attempted solution:

My idea so far has been to find $$\epsilon > 0$$ so that I could derive the inequality from $$\vert f(x)-f(x_0) \vert < \epsilon \ \forall x \in \mathbb{R}, \vert x-x_0 \vert < \delta$$, where the existence of such $$\delta$$ is guaranteed by the definition of continuity.

I tried quite a few different values of $$\epsilon$$, one of which was $$\epsilon=\vert f(x_0) \vert >0$$. Using this value, I arrived at $$\frac{\vert f(x)-f(x_0) \vert}{\vert f(x_0) \vert}<1 \iff 0 but this isn't helpful at all. Any ideas?

• Then how about $\varepsilon = |f(x_0)|/2$? – xbh Nov 12 '18 at 18:00

As xbh said, we can find $$\delta>0$$ such that $$|f(x)-f(x_0)|<\frac{|f(x_0)|}{2} \hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta)$$ since $$f$$ is continuous at $$x_0$$ and $$f(x_0)\neq 0$$ so $$|f(x_0)|>0$$. Then by triangular inequaltiy, $$|f(x_0)|-|f(x)|\leq|f(x)-f(x_0)|<\frac{|f(x_0)|}{2}\hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta)$$ so, $$\frac{|f(x_0)|}{2}=|f(x_0)|-\frac{|f(x_0)|}{2}< |f(x)| \hspace{3mm}\forall x\in (x_0-\delta,x_0+\delta).$$