Is ${\cal C}^\infty$ complete?

Question

Given $a \in \mathbb R \backslash \{ 0 \}$, is the metric space $ X \equiv\{ f \in {\cal C}^{\infty} ([-a, a]) \}$ equipped with the uniform norm $||.||_\infty \displaystyle := \underset {X}{\sup}|f| $ complete? If not, does it depend exclusively on the choice of the norm? Or is it due to something else?

Answer 1

This space is not complete with the supremum norm because it doesn't control the norm of the derivatives. For instance : $$\forall x \in [-a, a] \quad |\sqrt{x^2 + \frac1n} - |x| | = \frac{\frac1n}{\sqrt{x^2 + \frac1n} + |x|} \leq \frac1{\sqrt{n}}$$ Thus $x \mapsto \sqrt{x^2 + \frac1n}$ is a sequence of a $C^{\infty}$ functions wich converges uniformly to $x \mapsto |x|$ wich is not $C^1$.

To be complete $C^{\infty}$ may be equiped with a family of seminorms to control all the derivatives. This is called a Fréchet space (https://en.wikipedia.org/wiki/Fr%C3%A9chet_space). Here the derivative at $x = 0$ equals $n$ wich diverges.