# Let $f(x)$ be continous in $x_0$ and $f(x_0)\ne 0$. Prove that $\exists C>0$ and neighbouhood of $x_0$ such that $|f(x)| \ge C$

Let a function $$f(x)$$ be continuous in some point $$x_0$$ and $$f(x_0) \ne 0$$. Prove that there exists a number $$C > 0$$ and neighbourhood of $$x_0$$ such that forall $$x$$ in that neighbourhood the following inequality holds: $$|f(x)| \ge C$$

The problem statement says that $$f(x_0) \ne 0$$, so it is either greater than $$0$$ or less than $$0$$. Let's consider the case for $$f(x_0) > 0$$. In such case by continuity of $$f(x)$$ we know: $$\lim_{x\to x_0} f(x) = f(x_0) > 0$$

Or in other words: $$\forall \epsilon > 0\ \exists \delta_\epsilon > 0\ \forall x: |x-x_0| < \delta_\epsilon \implies |f(x) - f(x_0)| < \epsilon$$

Since $$f(x_0) \ne 0$$ we may let $$\epsilon = {f(x_0)\over 2}$$. In such case: $$|f(x) - f(x_0)| < \epsilon \\ |f(x) - f(x_0)| < { f(x_0)\over 2 } \\ -{ f(x_0)\over 2 } < f(x) - f(x_0) < { f(x_0)\over 2 }\\ 0<{f(x_0)\over 2} < f(x) < {3f(x_0)\over 2}$$

So we may now choose any $$C \in \left(0; {f(x_0)\over 2}\right)$$, which would imply $$f(x) \ge C$$.

Similar reasoning is applied to the case when $$f(x_0) < 0$$.

I'm not sure my reasoning above is valid, so I would like to kindly ask for verification. Or for a correct proof if the above makes no sense. Thank you!

• Yes, this absolutely fine ;) The neighbourhood would be $(-f(x_0)/2,f(x_0)/2)$ when $f(x_0)>0$. – weirdo May 24 at 16:24
• This looks correct. You have $f(x) \geq C$ for all $|x-x_0|\leq\delta_\epsilon$. The reason this works is because of continuity, i.e. the statement is for all $\epsilon >0$..., the the particular choice of $\epsilon = f(x_0)/2$ works. – Dayton May 24 at 16:24

Yep, the above proof is valid. But for a couple suggestions in terms of proof writing, in the last line you said that $$f(x_0) \geq C$$, but this is only true for all $$x$$ such that $$|x-x_0|<\delta$$. Just make sure you include that in your last line so that people don't think you are saying that you are proving it for every $$x$$.
Have you been introduced to the neighbourhood notation? It is basically $$B_r(x_0) := \{|x-x_0| < r \quad| \quad \forall x \text{ in your domain.}\}$$ This way you can say $$|f(x) - f(x_0)|<\epsilon$$ as $$B_\epsilon (x_0)$$