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$C^1[0,1]$ is the space of continuous functions $f:[0,1]\to\Bbb R$ with a continuous derivative $f':[0,1]\to\Bbb R$. Show that with the norm $||f||:=||f||_{\infty}+||f'||_{\infty}$ the space $C^1[0,1]$ is complete.

My attempt:

enter image description here

Now, define $f_n$ as a sequence of functions. Clearly $f_n$ is cauchy with respect to the sup norm since $||f_n-f_m||_{\infty}=|| ||f_n-f_m||_{\infty}+||f_n'-f'_m||_{\infty} ||_{\infty} <\epsilon$ for every $\epsilon>0$. So by the above theorem, I can state that $\exists f\in C^1[0,1]$ such that $f_n\to f$ and $f'_n\to f'$. Now, by definiton of a complete space, $C^1[0,1]$ is complete. The definition is given below:

enter image description here

Would this be sufficient for this proof?

  • What is the meaning of the expression $$|| ||f_n-f_m||{\infty}+||f'||{\infty} ||_{\infty}$$ ? – Jack M May 11 '20 at 17:56
  • @JackM the sup norm of $||f||$ which is $||f||:=||f||{\infty}+||f'||{\infty}$ –  May 11 '20 at 17:59
  • "The sup norm of $||f||$" doesn't make sense as a phrase - the sup norm is something defined for a function, but $||f||$ is a real number. And in any case, what is $f$? There isn't an "$f$" in the expression I quoted above, just $f_n$ and $f_m$. – Jack M May 11 '20 at 18:05
  • @JackM sorry a lot for the mistake, i edited it –  May 11 '20 at 18:26
  • I think what you meant to write with that equation is $$||f_n-f_m||=||f_n-f_m||{\infty}+||f_n'-f'_m||{\infty}$$ What you've written doesn't make sense because you're using the sup norm on the left hand side when I think you mean the "sum norm" introduced in the exercise, and on the right hand side you're taking the sup norm of a real number, which doesn't make sense. – Jack M May 11 '20 at 18:30
  • .You may look at this. – Bora Doğan May 14 '20 at 14:17
  • @BoraDoğan ah, okay. This is actually what i needed –  May 14 '20 at 14:20
  • Is there any way a moderator can cancel my bounty? As this is actually a duplicate –  May 14 '20 at 14:21
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    As I did recently, I can save your bounty points by providing a fake answer. If you’ll award it with the bounty then I’ll return your points by awarding any your answer at MSE. – Alex Ravsky May 14 '20 at 14:59
  • @AlexRavsky sure, let's do it –  May 14 '20 at 15:04
  • I think you can choose your answer which you want to be awarded. – Alex Ravsky May 14 '20 at 15:08
  • (This link)[https://math.stackexchange.com/questions/214218/uniform-convergence-of-derivatives-tao-14-2-7 ] will help you. – N.Quy May 14 '20 at 20:58

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