# Thread: Real Analysis continuous functions

1. ## 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.

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

3. 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.

4. 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 $n \ge N\quad \Rightarrow \quad \left| {g_n (y) - g(y)} \right| < \varepsilon \left| {h(x_0 )} \right|$.