1

enter image description here

This is an exercise problem given by my TA in my Real Analysis class. However, I am having quite a bit of problem with just understanding the problem. In particular, I am not sure how to show (1) and the motivation for defining such $\phi_n(t)$

For (3), I suppose I need to express $\phi_n(t)$ as a sequence of real-valued continuous function in order to get the result.

For (4), I thought the function is defined as $\phi_n(t)=y_0+\int_0^t F(\phi_n(s-\frac{1}{n}),s)$ already. The question makes me a bit confused.

In summary, I would like to hear anything that can help me understand the question better as well as proceeding with the answers. The textbook I am using is Analysis II by Tao

I have also found a comparable proof where I was only able to understand part of it. Proving Peano's Existence Theorem by approximating with $C^{\infty}$ functions using Weierstrass' Theorem.

Rico
  • 303
  • Why is part of (3) blurred? –  Mar 13 '19 at 06:47
  • It basically says that the Arzela-Ascoli Theorem is the Corollary 3.7 in the book. There are some website information in there that I would not like to share (such as my school). – Rico Mar 13 '19 at 06:49
  • In a nutshell, the idea of the proof is to approximate a solution of $y'(t)=F(y(t),t)$, $y(0)=y_0$, by functions satisfying $y'(t)=F(y(t-\tfrac1n),t)$ (plus the IC). Such functions are just $\phi_n$ given by the formula in (1). Now, by the Arzela-Ascoli theorem, the sequence $(\phi_n)_{n=1}^{\infty}$ has a uniformly convergent subsequence, and the limit is just a solution looked for. – user539887 Mar 13 '19 at 07:02
  • @user539887 Then how may I define a function as (1)? I am not quite clear with the motivation of how to define such function. (or how to define such function as the hint) – Rico Mar 13 '19 at 07:05
  • Observe that in the integrand in $(1)$ for $t\in[0,\tfrac1n]$ you need only to know the values of $\phi_n$ on $[-\tfrac1n,0]$ (which is $y_0$), for $t\in[\tfrac1n,\tfrac2n]$ you need only to know the values of $\phi_n$ on $[0,\tfrac1n]$ (which you know already, by the first step), and so on. – user539887 Mar 13 '19 at 13:37

0 Answers0