Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $<t_0,\alpha_0>$ 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 $<t_0, \alpha_0>$.

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 $<t_0,\alpha_0>$ 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 $<t_0,\alpha_0>$. 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 $<x, \alpha>$. 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?


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

share|cite|improve this question
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. – t.b. Jun 7 '11 at 20:18
Thank you. This is from Chapter 6 of Advanced Calculus by Loomis. – user11869 Jun 7 '11 at 20:30
Thanks! By the way: If everybody put that much effort into questions here, this site would be much more interesting! – t.b. Jun 7 '11 at 20:35
For Q3, why is it impossible that there exists some x' in $B_r(\alpha)$ such that $g_1(x')=g_2(x')$ – user11869 Jun 7 '11 at 20:37
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. – t.b. Jun 7 '11 at 21:15

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.