1

Which of the following metric spaces are complete?
(a) The space $C^1[0, 1]$ of continuously differentiable real-valued functions on $[0, 1]$ with the metric $d(f, g) = \max_{t∈[0,1]}|f(t) − g(t)|$

(b) The space $C[0, 1]$ of continuously differentiable real-valued functions on $[0, 1]$ with the metric $d(f, g) = \max_{t∈[0,1]}|f(t) − g(t)|$

(c) The space $C[0, 1]$ with the metric $d(f, g) =∫_0^1 |f(t) − g(t)| dt$.

(d)The space $C^1[0, 1]$ with the metric $d(f, g) =∫_0^1 |f(t) − g(t)| dt$.

my attempts: from the Weierstrass Approximation Theorem option a), option b) are true and option C) and option D) are incorrect ........ as

IS my answer is correct or not ? pliz verified and tell me the solution if u have a time ...i would be more grateful....

Thanks in advance

  • 1
    In b), you only have §continuous", not "continuously differentiable". Also note that this is the only difference between a) and b) - don't you think that this may suggest that they require different answers? – Hagen von Eitzen Jan 01 '18 at 14:12
  • ok@ Hagen Von Eitzen ,,,,pliz provide counter example ..im not getting –  Jan 01 '18 at 14:17

1 Answers1

1

a) is NO from Weierstrass: you can approximate all continuous functions uniformly by polynomials (which are in $C^1([0,1])$). Also functions that are continuous and non-differentiable. Such an approximation sequence is Cauchy in the uniform metric on $C^1$ but has no limit in $C^1([0,1])$ etc. Also, this uniformly convergent sequence will be Cauchy in the integral metric too (simple estimates show this), and so will give a negative answer to d) as well.

The uniform limit of continuous functions is continuous so b) does hold. (would still need a proof, depending on what you covered in class).

For c) look and follow links here, to find a Cauchy sequence of continuous functions without a continuous limit.

Henno Brandsma
  • 242,131