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I have some questions regarding the following theorems:

Theorem Let $A$ be an open subset of a Banach space $W$, let $I$ be an open interval in $\mathbb{R}$, and let $F$ be a continuous mapping from $I \times A$ to $W$ which is locally uniformly Lipschitz in its second variable. Then for any point $$ in $I \times A$, for some neighborhood $U$ of $\alpha_0$ and for any sufficiently small interval $J$ containing $t_0$, there is a unique function $f$ from $J$ to $U$ which is a solution of the differential equation passing through the point $$.

the solution function is proved to be the fixed point of the mapping $K: f\rightarrow g$, where $f:J \rightarrow A$ is any continuous mapping and $g:J \rightarrow W$ is defined by $$ g(t)=\alpha_0+\int^t_{t_0}F(s,f(s))ds$$

the proof starts by choosing a neighborhood $L \times U$ of $$ on which $F$ is bounded and Lipschitz in $\alpha$ uniformly over $t$.

  • Question 1: why not choose $I \times A$ instead of the neighborhood $L \times U$. Is the purpose to make $F$ bounded?

A lemma to the above theorem: Let $g_1$ and $g_2$ be any two solutions of $d\alpha/dt=F(t,\alpha)$ through $$. Then $g_1(t)=g_2(t)$ for all t in the intersection $J=J_1 \cap J_2$ of their domains.

Proof by contradiction. Otherwise there is a point $s$ in $J$ such that $g_1(s) \neq g_2(s)$. Suppose that $s>t_0$, and set $C=\{t:t>t_0 \mbox{and} g_1(t)\neq g_2(t)\}$ and x = glbC. The set $C$ is open, and therefore $x$ is not in $C$. That is $g_1(x)=g_2(x)$

  • Question 2: why is the set $C$ open?

Call this common value $\alpha$ and apply the theorem to $$. With $r$ such that $B_r(\alpha) \subset C$ [Edit (T.B.): This should be $B_r(\alpha) \subset A$], we choose $\delta$ small enough so that the differential equation has a unique solution $g$ from $(x-\delta,x+\delta)$ to $B_r(\alpha)$

  • Question 3: why $B_r(\alpha)$ has to be a subset of $C$? Why is this possible?

The above lemma allows us to remove the restriction on the range of $f$

  • Question 4: Can you elaborate this? and why removing the restriction useful?

Thanks

Edit (T.B.): This is taken from Section 6 of Lynn H. Loomis, Shlomo Sternberg, Advanced Calculus, Jones and Bartlett Publishers, 1990.

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    Hello user11869, welcome to this site! It would be nice if you could give the name of the source you're reading. This might make answering a bit easier. Here's a very quick answer: Q1: The purpose is to make $F$ bounded and *uniformly Lipschitz* on $L \times U$ as opposed to locally bounded and locally uniformly Lipschitz. Q2: Because $g_1$ and $g_2$ are continuous (see Q1). Q3. Because $C$ is open. I'm not sure I understand what the sentence right before question 4 means. But that's why I asked about a reference.2011-06-07
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    Thank you. This is from Chapter 6 of Advanced Calculus by Loomis.2011-06-07
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    Thanks! By the way: If everybody put that much effort into questions here, this site would be much more interesting!2011-06-07
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    For Q3, why is it impossible that there exists some x' in $B_r(\alpha)$ such that $g_1(x')=g_2(x')$2011-06-07
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    I'm sorry but I'm a bit confused. Are you sure that the ball $B_r(\alpha)$ in Q3 should be a subset of $C$ and not of $A$ (to apply the theorem)? If I understand what you write correctly, then it should rather be $A$, no? Then you can simply apply the uniqueness assertion of the theorem.2011-06-07
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    Finally, for question 4 don't you ultimately want to go to a uniqueness statement without restricting that the values of $f$ are confined to a small neighborhood $U$ of $\alpha_0$ but rather $A$ itself?2011-06-07
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    You are right, it should be $B_r(\alpha) \subset A$. The book I'm reading has a typo2011-06-07
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    Are your questions now settled or are there still things you'd want to have clarified?2011-06-08

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