Completeness of a Normed Space of Smooth, Bounded Functions

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 $X$ is a subset of the normed space $C([0,T])$, but it is linear space itself. – 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 $$u$$.
• @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 – K.Power Mar 20 at 17:23