Real Analysis continuous functions

**Ideasman** · November 29th 2007, 12:01 PM

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.

**Plato** · November 29th 2007, 12:23 PM

From the topic title, is it safe to assume that these are function defined on the real numbers to the real numbers?

**Ideasman** · November 29th 2007, 12:25 PM

Originally Posted by Plato

From the topic title, is it safe to assume that these are function defined on the real numbers to the real numbers?

Mmhmm.

**Plato** · November 29th 2007, 12:46 PM

Originally Posted by Ideasman

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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