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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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    Are your questions now settled or are there still things you'd want to have clarified?2011-06-08

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Question 1: $F$ is only locally uniformly Lipschitz in the statement of the theorem, but you need it to be Lipschitz in the proof. That's why they have to pass to some small enough neighborhood so its actually Lipschitz.

For Question 2: $C$ is open because $C = A \cap B$ where $A = \{t \in R: t > t_0\}$ and $B = \{t \in R: g_1(t) \neq g_2(t)\}$. $A$ is obviously open, and $B = (g_1 - g_2)^{-1}(R -\{0\})$. So $B$ is also open since $g_1 - g_2$ is continuous and $R - \{0\}$ is open.