# are they complete metric spaces?

I need to know whether the following spaces are complete or not

1. Space of all continuos real valued functions with compact support with supnorm metric

2. The space $C^1[0,1]$ with metric $d(f,g)=max_{t\in[0,1]}|f(t)-g(t)|$

• Generally speaking, do you know the completion of the space of continuous real functions with compact support with regards to the supremum norm? – Ayman Hourieh Jan 6 '13 at 11:22
• @AymanHourieh, I have no idea. – Marso Jan 6 '13 at 11:24
• OK, do you know that the limit of a uniformly convergent sequence of continuous functions is also continuous? – Ayman Hourieh Jan 6 '13 at 11:27
• @AymanHourieh Yes That I know :) – Marso Jan 6 '13 at 11:28

1. Consider the sequence functions $$f_n(x)=\max\left\{0,\frac1{1+x^2}-\frac1n\right\}$$
2. Let $f(x)=\left\vert x-\frac12\right\vert$. Can you find functions $f_n\in C^1([0,1])$ with $\max_{t\in[0,1]}|f_n(t)-f(t)|<\frac1n$? What does that impliy?
1. is complete. A proof you can find in baby Rudin or Stokey Lucas Prescott. For 2. consider the sequence of functions $f_n(x) = \left\{ \begin{array}{lr} -x & : x <-\frac{1}{n}\\ \frac{1}{2n}+\frac{n}{2} x^2 & : -\frac{1}{n}\leq x \leq\frac{1}{n}\\ x & : x>\frac{1}{n} \end{array} \right.$
• Presumably you mean the space of continuous functions on a compact space in 1., which is indeed a complete space. In general, this is not the same as the space of functions with compact support (the support of a function is the closure of the set of points where $f$ takes nonzero values). The functions $f_n\colon \mathbb{R} \to \mathbb{R}$ from part 1. of Hagen von Eitzen's answer have compact support and they form a Cauchy sequence, yet they do not converge to a function with compact support. – Martin Jan 6 '13 at 12:47
• The answer to 2 is almost okay. You "smoothen out" the corner of the absolute value at zero and the sequence converges uniformly to the absolute value which fails to be differentiable at $0$. The only minor point is that the problem asked about $C^1[0,1]$, not $C^{1}[-1,1]$, but that's not a big deal, just replace $f_n(x)$ by $f_{n}(x-1/2)$. – Martin Jan 6 '13 at 13:54