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.

  • $\begingroup$ Your $X$ is a subset of the normed space $C([0,T])$, but it is linear space itself. $\endgroup$ – daw Mar 19 at 12:33

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 $u$.

  • $\begingroup$ Hi K. Power, Thank you very much! I realise now that I simply misunderstood the notation of C([0,T])... This makes my job much easier indeed. Thank you. $\endgroup$ – David Hughes Mar 19 at 8:47
  • $\begingroup$ Hi K.Power, if you wouldn't mind me asking, I'm wondering where I can find help for the last part of this question. I have shown that a solution, u, to the Integral Equation is unique in X, and I would like to show that it is unique in all C([0,T]). $\endgroup$ – David Hughes Mar 20 at 17:19
  • $\begingroup$ @DavidHughes Are you asking where to find a proof of the Picard-Lindelöf Theorem? We have shown $X$ is complete so you can use it in the proof $\endgroup$ – K.Power Mar 20 at 17:23
  • $\begingroup$ I am asking about one final point of the Picard Lindelof theorem. I understand how u is unique in X, which is a subset of C([0,T]) where the functions are bounded. I do not know how to show that this uniqueness carries over to all C([0,T]) $\endgroup$ – David Hughes Mar 20 at 17:24
  • $\begingroup$ @DavidHughes I'd recommend just asking a new question here on MSE, and linking to this post. Make sure to give context to the question and make clear exactly what you aren't clear about. $\endgroup$ – K.Power Mar 20 at 17:28

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