Uniformly Cauchy

[Imum Coeli]

Q: If \(\displaystyle (f_n) \) is uniformly Cauchy on D, prove that there is a function \(\displaystyle f : D \to \mathbb{R} \) such that \(\displaystyle (f_n) \) convereges uniformly to \(\displaystyle f \) on D.

Definitions:

• Uniformly Cauchy: For \(\displaystyle D \subseteq \mathbb{R} \), a sequence of functions \(\displaystyle f_n : D \to \mathbb{R} \) is said to be uniformly Cauchy if for each \(\displaystyle \epsilon > 0 \; \exists\; N \in \mathbb{N}\) such that \(\displaystyle |f_n(x)-f_m(x)|<\epsilon \) whenever \(\displaystyle m,n \geq N \; \forall\; x \in D\).
• Uniform Convergence: A sequence \(\displaystyle (f_n)_{n=1}^\infty\) of functions \(\displaystyle D \to \mathbb{R} \) converges uniformly to \(\displaystyle f : D \to \mathbb{R} \) if \(\displaystyle \forall \epsilon > 0\; \exists\: N \in \mathbb{N} : |f_n(x)-f(x)|<\epsilon \) if \(\displaystyle n \geq N\; \forall\: x \in D \).

Can anyone point out how to go about proving this, or even just how to start...

 

[HallsofIvy]

An obvious first step is to define f. Saying that \(\displaystyle \{f_n\}\) is "Uniformly Cauchy" means, first, that the sequence \(\displaystyle \{f_n(x)\}\) is Cauchy for every x in D so that it converges for every x in D. We can define f by "for every \(\displaystyle x\in D\), \(\displaystyle f(x)= \lim_{n\to\infty} f_n(x)\). Now show that the convergence of \(\displaystyle \{f_n\}\) to f is uniform.

 

[Imum Coeli]

Thanks heaps.

I came up with this...

Since \(\displaystyle (f_n)\) is Cauchy then we can choose \(\displaystyle N \in \mathbb{N}\) such that \(\displaystyle |f_n(x)-f_m(x)|<\epsilon \; \forall\; x\in D \) if \(\displaystyle m,n>N\)

Define \(\displaystyle \forall x\in D\), \(\displaystyle f(x) := \lim_{m\to\infty} f_m(x)\)

Then \(\displaystyle f_m(x)-f_n(x) = f_m(x)-f(x)-(f_n(x)-f(x))\)

So \(\displaystyle f_n(x)-f(x)= f_m(x)-f(x)+(f_n(x)-f_m(x))\)

Taking absolute values and applying the triangle inequality \(\displaystyle |f_n(x)-f(x)|\leq |f_m(x)-f(x)|+|f_n(x)-f_m(x)|\)

Suppose that \(\displaystyle n \geq N\) and \(\displaystyle x \in D\). Then for every \(\displaystyle m>N\)

\(\displaystyle |f_n(x) - f(x)| < |f_m(x) - f(x)| + \epsilon \)

But since \(\displaystyle f_m(x) \to f(x)\) as \(\displaystyle m \to \infty\) we can choose \(\displaystyle m>N\) such that \(\displaystyle |f_m(x) - f(x)| < \epsilon\)

Then if \(\displaystyle n>N \)

\(\displaystyle |f_n(x) -f(x)| < 2 \epsilon \)

Since \(\displaystyle \epsilon \) is arbitrary \(\displaystyle (f_n) \) converges uniformly to \(\displaystyle f \) on \(\displaystyle D \).