This post shows how to prove $C^1 [0, 1]$ is incomplete in the uniform norm. But I want to get a deeper understanding, specifically how to come up with an example. Here's my understanding:
I know $C^0[0,1]$, the set of continuous functions on $[0,1]$, is indeed complete in the uniform norm. So if $(f_n)$ is a Cauchy sequence of $C^1$ functions, then it must converge to some $C^0$ function $f$, so the reason $C^1$ is incomplete must be that $f$ may not be continuously differentiable. In other words, the problem lies in the fact that differentiable functions may not converge to another differentiable function.
Now the problem is reduced to finding an example. Imagine I do not know any standard example, I may do:
Step 1: Choose the simplest non-differentiable continuous function I could think of as a candidate limit, eg. $f(x)=|x-\frac{1}{2}|$.
Step 2: Think of a sequence of $C^1$ functions $g_n$ (eg. polynomials) which uniformly converges to it. I know the graphs of $x^t$ on $[0,1]$ is getting flatter as $t \to \infty$, so I reverse the process and observe the graphs of $x^{1+\frac{1}{n}}$ as $n \to \infty$ is getting closer to the graph of $y=x$.
Step 3: I adjust the graph to match $|x-\frac{1}{2}|$ and get the example sequence $f_n(x)=|x-\frac{1}{2}|^{1+\frac{1}{n}}$.
Since $C^0$ is a normed space in the uniform norm and hence a metric space, $f_n$'s limit is unique. Now $f_n$ is Cauchy for the uniform norm and its unique limit is $C^0$ but not $C^1$. So we proved $C^1[0,1]$ is incomplete.
May I ask is my understanding detailed enough?
