# Real Analysis Continuous Functions

• Oct 6th 2008, 12:36 PM
ynn6871
Real Analysis Continuous Functions
Let f:[a, b] -> R be continuous at c of [a,b] and suppose that f(c) > 0. Prove that there exist a positive number m and an interval [u,v] is a subset of [a,b] such that c of [u,v] and f(x) >(or equal) for all x of [u,v].

I don't know where to begin with this problem!!
• Oct 6th 2008, 12:43 PM
Plato
Use the basic definition of continuity at c.
Use $\varepsilon = \frac{{f(c)}}{2}=m$ and find the $\delta$ that goes with it.
Then let $u = c - \frac{\delta }{2}\,\& \,v = c + \frac{\delta }{2}$.
• Oct 7th 2008, 06:47 AM
ynn6871
ok, so we know that f is continuous so,

$|f(x) - f(c)|< \varepsilon$ and $|x-c|< \delta$.

So to prove my thing do I go something like $|f(x) - f(c)|< \frac{{f(c)}}{2} = m$ with $c$ $\in [u,v]$ with $u = c - \frac{\delta }{2}\,\& \,v = c + \frac{\delta }{2}$

You told me to find $\delta$ and I found it to be $\delta = v-u$

I know this may sound stupid but I have starred at that since last night and I still can't figure out how to get to $f(x) \geq m$ $\forall x \in [u,v]$
• Oct 7th 2008, 07:16 AM
Plato
Well, that was not a full proof but rather an outline more or less.
There are several careful adjustments to make to get u & v .
In the definition of continuity at c let $\varepsilon = \frac{{f(c)}}{2} > 0$.
Now corresponding to that epsilon $\left( {\exists \delta > 0} \right)\left[ {\left| {x - c} \right| < \delta \Rightarrow \left| {f(x) - f(c)} \right| < \frac{{f(c)}}
{2}} \right]$
.
Remove the absolute value: $- \frac{{f(c)}}{2} < f(x) - f(c) < \frac{{f(c)}}{2} \Rightarrow \quad \frac{{f(c)}}{2} < f(x)$.
That means that every x within a distance of $\delta$ of c has the property $\frac{{f(c)}}{2} < f(x)$.
Let $u = \max \left\{ {a,c - \frac{\delta}{2} } \right\}\,\& \,v = \min \left\{ {b,c + \frac{\delta}{2} } \right\}$.