# Real Analysis - Sequences #2

• Jan 20th 2009, 01:21 PM
ajj86
Real Analysis - Sequences #2
Hey, I have been having trouble ever since I took part 1 of this class. So, here is question 2:

Assume that (fn) converges uniformly to f on A and that each fn is uniformly continuous on A. Prove that f is uniformly continuous on A.

Any help would be greatly appreciated. Thanks.
• Jan 20th 2009, 02:57 PM
TheEmptySet
Quote:

Originally Posted by ajj86
Hey, I have been having trouble ever since I took part 1 of this class. So, here is question 2:

Assume that (fn) converges uniformly to f on A and that each fn is uniformly continuous on A. Prove that f is uniformly continuous on A.

Any help would be greatly appreciated. Thanks.

Here is the idea of the proof.

Since $\displaystyle f_n \to f$ uniformly there exits an $\displaystyle N \in \mathbb{N}$ such that for $\displaystyle n > N$$\displaystyle |f_n-f|< \frac{\epsilon}{3}$ Now fix $\displaystyle M > N$

Since each $\displaystyle f_n$ is uniformly continous there exists a delta such that $\displaystyle |x-y|< \delta \implies |f_M(x)-f_M(y)|< \frac{\epsilon}{3}$

Now if $\displaystyle |x-y|<\delta$ then

$\displaystyle |f(x)-f(y)|= |f(x)-f_M(x)+f_M(x)-f_M(y)+f_M(y)-f(y)|$

$\displaystyle <|f(x)-f_M(x)|+|f_M(x)-f_M(y)|+|f_M(y)-f(y)|< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsil on}{3}=\epsilon$