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I'm being introduced into the Picard Existence Theorem. I am fairly comfortable with math terminology but not great at it. In your answer I would appreciate a mathematicians answer, sparsed with some actual english please, to help explain things =).

The statement is defining a Lipschitz continuous function. Here is the statement.

A function $f: U \times [t_0;t_0 + T ] \rightarrow \mathbb{R}^n, U \subset \mathbb{R}^n$, is Lipschitz continuous in $U$ if there exists a constant $L$ such that $\| f(y,t) - f(x,t) \| \le L\|x - y\|$ for all $x,y \in U$ and $t \in [t_0;t_0 + T ]$. If $U = \mathbb{R}^n$, $f$ is called globally Lipschitz.

What I don't get:

1) The $\times$ after the first $U$. Is that saying the cartesian product of $U$ and $[t_0;t_0 + T ]$? I get the underlying meaning of this statement...I have a function that is mapping $U$ into $\mathbb{R}^n$ but I don't get the specifics of that first part.

2) The second statement I understand the math this is the part I'm most confused about, but don't understand the meaning of it. I mean what is this constant $L$ in the first place? I just don't see where this statement comes from or what it even means.

The rest of the statement I understand, thanks for the help!

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up vote 3 down vote accepted

1) The times is just the cartesian product. Your function takes an $x \in U$ as one input and a $t \in [t_0, t_0 + T]$ as the other input.

2) This statement of Lipschitz tells you that if you take any two points in your space, $x$ and $y$, and use them as inputs, then there is some $L$ that makes it so that the difference in function values ( $||f(x,t)-f(y,t)||$ ) is smaller than in the original space ($||x-y||$) but with a multiplying factor of $L$ on the original space. Note that this $L$ must work for any choice of $x$ and $y$.

An example would be a contraction mapping. If $L<1$, that would mean that if you apply the function $f$ to the whole set $U$ (with some fixed $t$), the image would be in some sense smaller than it was before.

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thanks for the good explanation =) – Spaderdabomb Aug 21 '13 at 20:25

It's easier to figure out why Lipschitz continuous functions are important if we look at the one dimensional case. When you get the underlying mechanics, it's easy to generalize to more complex cases.

So, let $I$ be an open interval, $f:I \rightarrow \mathbb{R}$ is Lipschitz continous function if there exists a constant $L$ so $|f(x) - f(y)| \leq L |x - y|$ for all $x, y \in I$.

If f is differentiable (and in your case, it surely is because of Picard's theorem), it isn't hard to show, dividing by $|x-y|$, that L-continuity is equivalent to $f'$ being bounded on $I$.

So, the function is limited in how fast it can change. This is the ''gist'' of L-continuity. Try finding a few examples of differentiable functions which are L-continuous and few which aren't.

Also worth noting is that L-continuity implies uniform continuity.

EDIT: In your case, all this talk about $f'$ being bounded translates nicely to $\frac{\delta f}{\delta t}$ being bounded... So the function is limited in how fast it can change in second variable.

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After reading Andrew W's answer, this answer makes more sense to me now than the first time I read it haha =p thanks – Spaderdabomb Aug 21 '13 at 20:27

Begin with a function $f:\ U\to{\mathbb R}^n$ defined on some reasonable set $U\subset{\mathbb R}^n$. Such a function is continuous when inputting an $x$ near the actually meant point $x_0$ produces a value $f(x)$ near the actual value $f(x_0)$, for each individual $x_0\in U$. (Of course the exact definition is more complicated.)

Now Lipschitz continuity is a very simple quantitative realization of this idea: The function $f$ is $L$-Lipschitz continuous on $U$ if for all $x$, $x_0\in U$ we have $$|f(x)-f(x_0|\leq L|x-x_0|\ ,\tag{1}$$ i.e., if the error in the output is at most $L$ times the error in the input. The existence of such an $L>0$ is a stronger condition than the general notion of continuity, but Lipschitz-continuity is extremely handy in practice.

Now in your setup you have not only one function $f:\ U\to{\mathbb R}^n$, but a whole family $\bigl(f_t\bigr)_{t_0\leq t\leq t_0+T}$ of such functions (which is the same thing as a function $f:\ U\times\ [t_0,t_0+T]\to{\mathbb R}^n$), and you require that each member $f_t$ of this family satisfies $(1)$, resp. $$|f_t(x)-f_t(y)|\leq L|x-y|\qquad \forall x\ \forall y\in U\ ,$$ with the same $L$ for all $t\in[t_0,t_0+T]$.

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Excuse me instead of your request I'll be succinct: Yes for the first question and for the second $L$ is called the Lipschitz constant and it appears in the inequality called Lipschitz inequality: it's just the definition.

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