# Continuity proof

• February 1st 2008, 08:31 AM
WWTL@WHL
Continuity proof
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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• February 1st 2008, 08:52 AM
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.
• February 1st 2008, 10:05 AM
WWTL@WHL
Quote:

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! :D

$f(x) - f(c) < \frac{f(c)}{2}$ 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. :o

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?
• February 1st 2008, 10:50 AM
Plato
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.
• February 1st 2008, 11:13 AM
WWTL@WHL
Excellent. Thanks so much, Plato. :)

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