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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!

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  • $\begingroup$ Yes, this absolutely fine ;) The neighbourhood would be $(-f(x_0)/2,f(x_0)/2)$ when $f(x_0)>0$. $\endgroup$ – weirdo May 24 at 16:24
  • $\begingroup$ 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. $\endgroup$ – Dayton May 24 at 16:24
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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 $$ B_\epsilon (x_0)$$

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