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Math Help - Real Analysis continuous functions

  1. #1
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    Real Analysis continuous functions

    Given that h is a continuous func. on the compact set K such that h(x) \neq 0 on K, and also given that (g_n) is a sequence of functions such that they converge uniformly to g on K, prove \left(\frac{g_n}{h}\right) converges uniformly to \frac{g}{h} on K.

    Not sure. Apparently its a very hard proof.
    Last edited by mr fantastic; October 6th 2009 at 12:36 AM. Reason: Corrected a word in post title
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    From the topic title, is it safe to assume that these are function defined on the real numbers to the real numbers?
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    Quote Originally Posted by Plato View Post
    From the topic title, is it safe to assume that these are function defined on the real numbers to the real numbers?
    Mmhmm.
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    Quote Originally Posted by Ideasman View Post
    Mmhmm.
    I will take that as an "yes you can".

    Because h is never zero and K compact we may assume that |h| has a minimum positive value in K. That is: \left( {\exists x_0  \in K} \right)\left( {\forall y \in K} \right)\left[ {\left| {h(y)} \right| \ge \left| {h(x_0 )} \right| > 0} \right]. Moreover, \left( {\exists x_0  \in K} \right)\left( {\forall y \in K} \right)\left[ {\left| {\frac{1}{{h(y)}}} \right| \le \left| {\frac{1}{{h(x_0 )}}} \right|} \right].

    Now you can get: \left| {\frac{{g_n (y)}}{{h(y)}} - \frac{{g(y)}}{{h(y)}}} \right| \le \frac{{\left| {g_n (y) - g(y)} \right|}}{{\left| {h(x_0 )} \right|}}.
    So if \varepsilon  > 0 by the uniform convergence of the g_n choose N such that n \ge N\quad  \Rightarrow \quad \left| {g_n (y) - g(y)} \right| < \varepsilon \left| {h(x_0 )} \right|.
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