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Math Help - Real Analysis Continuous Functions

  1. #1
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    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!!
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  2. #2
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    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}.
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  3. #3
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    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]
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  4. #4
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    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)}}<br />
{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\}.
    Last edited by Plato; October 7th 2008 at 08:47 AM.
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