Continuity proof

**WWTL@WHL** · February 1st 2008, 08:31 AM

Hi, I'm stuck on the following question. Any help will be fantastic, thanks.

Question:

$f:\Re \to \Re$ is a continuous function. Prove that if for some $c \in \Re$ , $f(c)>0$ , then there exists a $\delta > 0$ such that $\forall x \in ( c - \delta, c + \delta)$ , $f(x) > 0$

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I tried a contradiction proof but didn't really get anywhere. Please help! Thank you.

**Plato** · February 1st 2008, 08:52 AM

$f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}} {2}$

Expand the last inequality. Add f(c) to all three parts. See what you get.

**WWTL@WHL** · February 1st 2008, 10:05 AM

Originally Posted by Plato

$f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}} {2}$

Expand the last inequality. Add f(c) to all three parts. See what you get.

Thanks for the reply, Plato!

Following your help, I get...

$f(x) - f(c) < \frac{f(c)}{2}$ if $f(x)-f(c) > 0$
and
$f(c) - f(x) < \frac{f(c)}{2}$ if $f(x)-f(c) < 0$

So $f(x) > \frac{3}{2} f(c)$ and $f(x) > \frac{1}{2} f(c)$ ...each of which is greater than 0 since f(c)>0

Is this what you meant? I don't see where the third inequality comes from so I don't think I've followed your advice properly.

Also, if a function is continuous at c, by definition, $\forall \epsilon > 0$ , there exists a $\delta > 0$ s.t. $\forall x \in \Re$ such that $|x-c|< \delta$ , $|f(x) - f(c)|< \epsilon$ .

As far as I can see, you've let $\epsilon = \frac{f(c)}{2}$ which is fixed. So I don't see how this covers the 'for all epsilon' bit. I'm probably being very dull here though.

EDIT: Thinking about this again, I think you might've been suggesting a proof by contradiction - which I totally missed. If you let f(x) < 0, then we have a contradiction. Is that what you wanted me to do?

**Plato** · February 1st 2008, 10:50 AM

First of all, there is absolutely no reason why we should for “all epsilon”.
The problem says that f is positive is some neighborhood of c.
Use the definition of continuity with $\varepsilon = \frac{{f(c)}}{2}> 0$ .
We know that $\exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)} }{2}$
We also know that $\left| {x - c} \right| < \delta \quad \Leftrightarrow \quad x \in (c - \delta ,c + \delta )$ which is a neighborhood of c. For any x there we have
$\begin{gathered} \left| {f(x) - f(c)} \right| < \frac{{f(c)}} {2} \hfill \\ - \frac{{f(c)}} {2} < f(x) - f(c) < \frac{{f(c)}} {2} \hfill \\ 0 < \frac{{f(c)}} {2} < f(x) < \frac{{3f(c)}} {2} \hfill \\ \end{gathered}$ .

So the function is positive in that neighborhood of c.

**WWTL@WHL** · February 1st 2008, 11:13 AM

Excellent. Thanks so much, Plato.

I was focusing on the definition of continuity more than the question, me thinks.

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