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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 $d(f,g)=max_{t\in[0,1]}|f(t)-g(t)|$

Please help, I have no clue. Thank you.

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

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  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.$
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  • $\begingroup$ 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. $\endgroup$ – Martin Jan 6 '13 at 12:47
  • $\begingroup$ you are right, I obviously missed that point. Hope my answer to 2. is ok tho. $\endgroup$ – mathemagician Jan 6 '13 at 13:44
  • $\begingroup$ 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 $f_{n}(x-1/2)$. $\endgroup$ – Martin Jan 6 '13 at 13:54

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