As part of a proof of the Picard–Lindelöf theorem, I am using the following space:
$X = \{ u \in C([0,T]) : u(0) = \alpha , || u - \alpha || \leq K\}$
where $K \in \mathbb{R}_{> 0} , \ \alpha \in \mathbb{R}$ and the norm is defined as follows:
$|| u || = ^{\text{sup}}_{t \in (0,T)} |u(t)|$.
I have read and understood the proof just fine, but no version I have read so far has provided a proof for whether or not $X$ is actually complete, which is necessary to make use of the Banach Fixed Point Theorem.
I have shown in previous exercises that such $X$ is complete under the norm:
$||u|| = ^{\text{sup}}_{t \in (0,T)} ( |u(t)| + |u'(t)| ) $
but I cannot figure out how to do so without the derivative included in the norm.
Here is my progress so far:
Let $u_{n}$ be a Cauchy sequence in $X$. I have shown that there is a function, $u$, to which $u_{n}$ converges pointwise, and thus $u$ is continuous. I have also shown that $||u|| \leq K$. However, I have not figured out how to show that $u$ is differentiable, and thus inside $X$.
I would like to prove the following claim:
if $u_{n}$ is a Cauchy sequence in $X$, then $u_{n}'$ is also Cauchy in $X$.
I would surely be done then. Can someone guide me as to how I can prove this claim? Or if by any chance I am completely mistaken here and $X$ is in fact not complete at all? In that case, how does one make use of such a set to prove the Picard–Lindelöf theorem?
Thank you very much.
Your $X$ does not consist of differentiable functions, but continuous ones (obviously all differentiable functions are in there). There is no need to even talk about $u'$. More to the point your space $X$, with that norm, is only "interested" in continuous functions.
You state that $u_n$ converges to $u$ pointwise, thus $u$ is continuous, but this is not true. You need to show that $u_n$ converges to $u$ uniformly, and only then can you conclude that $u$ is continuous. However this follows quite simply from the fact that $u_n$ must converge to $u$ in the $\sup$ norm because $u_n$ is Cauchy in the $\sup$ norm and converges pointwise to $u$.