Arzela-Ascoli Theorem and Applications

The Arzela-Ascoli Theorem is a rather formidable-sounding theorem that gives a necessary and sufficient condition for a sequence of real-valued continuous functions on a closed and bounded interval to have a uniformly convergent subsequence.

Statement: Let (f_n) be a uniformly bounded and equicontinuous sequence of real-valued continuous functions defined on a closed and bounded interval [a,b]. Then there exists a subsequence (f_{n_k}) that converges uniformly.

The converse of the Arzela-Ascoli Theorem is also true, in the sense that if every subsequence of (f_n) has a uniformly convergent subsequence, then (f_n) is uniformly bounded and equicontinuous.

Explanation of terms used: A sequence (f_n) of functions on [a,b] is uniformly bounded if there is a number M such that |f_n(x)|\leq M for all f_n and all x\in [a,b]. The sequence is equicontinous if, for all \epsilon>0, there exists \delta>0 such that |f_n(x)-f_n(y)|<\epsilon whenever |x-y|<\delta for all functions f_n in the sequence. The key point here is that a single \delta (depending solely on \epsilon) works for the entire family of functions.

Application

Let g:[0,1]\times [0,1]\to [0,1] be a continuous function and let \{f_n\} be a sequence of functions such that f_n(x)=\begin{cases}0,&0\leq x\leq 1/n\\    \int_0^{x-\frac{1}{n}}g(t,f_n(t))\ dt,&1/n\leq x\leq 1\end{cases}

Prove that there exists a continuous function f:[0,1]\to\mathbb{R} such that f(x)=\int_0^x g(t,f(t))\ dt for all x\in [0,1].

The idea is to use Arzela-Ascoli Theorem. Hence, we need to show that (f_n) is uniformly bounded and equicontinuous.

We have

\begin{aligned}|f_n(x)|&\leq |\int_0^{x-\frac{1}{n}} 1\ dt|\\    &=|x-\frac{1}{n}|\\    &\leq |x|+|\frac{1}{n}|\\    &\leq 1+1\\    &=2    \end{aligned}

This shows that the sequence is uniformly bounded.

If 0\leq x\leq 1/n,

\begin{aligned}|f_n(x)-f_n(y)|&=|0-f_n(y)|\\    &=|\int_0^{y-\frac{1}{n}} g(t,f_n(t))\ dt|\\    &\leq |\int_0^{y-\frac{1}{n}} 1\ dt|\\    &=|y-\frac{1}{n}|\\    &\leq |y-x|    \end{aligned}

Similarly if 0\leq y\leq 1/n, |f_n(x)-f_n(y)|\leq |x-y|.

If 1/n\leq x\leq 1 and 1/n\leq y\leq 1,

\begin{aligned}|f_n(x)-f_n(y)|&=|\int_0^{x-1/n} g(t,f_n(t))\ dt-\int_0^{y-1/n}g(t,f_n(t))\ dt|\\    &=|\int_{y-1/n}^{x-1/n}g(t,f_n(t))\ dt|\\    &\leq |\int_{y-1/n}^{x-1/n} 1\ dt|\\    &=|(x-1/n)-(y-1/n)|\\    &=|x-y|    \end{aligned}

Therefore we may choose \delta=\epsilon, then whenever |x-y|<\delta, |f_n(x)-f_n(y)|\leq |x-y|<\epsilon. Thus the sequence is indeed equicontinuous.

By Arzela-Ascoli Theorem, there exists a subsequence (f_{n_k}) that is uniformly convergent.

f_{n_k}(x)\to f(x)=\int_0^x g(t,f(t))\ dt.

By the Uniform Limit Theorem, f:[0,1]\to\mathbb{R} is continuous since each f_n is continuous.

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