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Thread: Unif. Continuity Again

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    Unif. Continuity Again

    If $\displaystyle f,g$ are unif. cont. on an interval $\displaystyle [a,b]$ and $\displaystyle g(x) \not= 0$ for $\displaystyle x \in [a,b]$, then $\displaystyle f/g$ is unif. cont. on $\displaystyle [a,b]$

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    Quote Originally Posted by Math2010 View Post
    If $\displaystyle f,g$ are unif. cont. on an interval $\displaystyle [a,b]$ and $\displaystyle g(x) \not= 0$ for $\displaystyle x \in [a,b]$, then $\displaystyle f/g$ is unif. cont. on $\displaystyle [a,b]$
    The interval $\displaystyle [a,b]$ is closed so any continious function is uniformally continuous.
    First prove that $\displaystyle \frac{1}{g}$ is continuous.
    Without loss of generally, we can assume that $\displaystyle g$ is positive on $\displaystyle [a,b]$.
    Therefore, $\displaystyle \left( {\exists c \in [a,b]} \right)\left( {\forall x \in [a,b]} \right)\left[ {0 < g(c) \leqslant g(x)} \right]$, the low point theorem.
    Now we have a bound on $\displaystyle \frac{1}{g}$.
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