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I have been searching for an answer to the following question:

Given a normed linear space $V$ and two norms that are not equivalent, but $\exists K\in\mathbf{R}$ such that $\|v\|_1\leq K\|v\|_2$ for all $v\in V$. We know that $(V,\|\cdot\|_2)$ is complete, i.e. a Banach space. Is $(V,\|\cdot\|_1)$ also complete?

Some related questions posted so far are:

(i) Example of two norms on same space, non-equivalent, with one dominating the other

(ii) If two normed spaces are Lipschitz equivalent, then one if complete iff the other is

In (i) there is a good example from julien, but I am unable to relate it to Cauchy sequences.

Could someone please shine some light on this, please?

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user136475
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  • From the answers you got in (i) you can produce an example. Take $C[0,1]$. Take $||\cdot||_1$ to be the $L^2$ norm, and $||\cdot||_2$ to be the uniform norm. $C[0,1]$ with the uniform norm is complete. But the completion of it in the $L^2$ norm is $L^2[0,1]$. – OR. Mar 19 '14 at 11:23
  • Maybe to make it more explicit consider a sequence of continuous function on $[0,1]$ converging, in the $L^2$ norm, to $\chi_{[1/2,1]}$ (the function that is equal to $1$ on $[1/2,1]$ and zero otherwise). – OR. Mar 19 '14 at 11:25
  • @ABC Thank you that is clear to me. I was also thinking about the reverse - Is it possible to find an example where $V$ is complete with respect to both norms? – user136475 Mar 19 '14 at 11:29
  • Use the open mapping theorem with the identity function. It will imply the norms are equivalent. – OR. Mar 19 '14 at 11:34
  • @ABC The identity function $I:(V,|\cdot|_1)\to(V,|\cdot|_2)$ is continuous but the inverse $I^{-1}:(V,|\cdot|_2)\to(V,|\cdot|_1)$ is discontinuous. So, using the Open Mapping Theorem, how do I relate it to Cauchy sequences that does not converge? – user136475 Mar 19 '14 at 11:40
  • The open mapping theorem says that if a continuous linear map between Banach spaces is surjective, then it is open. This means that the inverse is continuous. Because the inverse is continuous and linear, the inverse is also bounded. Since the inverse is also the identity, the definition of bounded gives you the missing inequality to say that the norms are equivalent. – OR. Mar 19 '14 at 11:43
  • @ABC I get that part, but we don't have a linear map between Banach spaces since one of them is not complete, or am I missing something? My biggest problem is that I have until now mostly proven completeness from Cauchy sequences, and I have a problem relating the above to the theory of converging sequences. – user136475 Mar 19 '14 at 11:53
  • I will put all in an answer so you understand. – OR. Mar 19 '14 at 11:57

1 Answers1

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Part 1: (There are examples in which is complete under $||\cdot||_2$ and not under $||\cdot||_1$)

Take $V:=C[0,1]$, $||f||_1:=\int_{0}^{1}f^2$, and $||f||_2:=\sup|f|$. Then $||\cdot||_1\leq||\cdot||_2$. While $V$ is complete under $||\cdot||_2$ we can get a sequence of continuous functions converging to $\chi_{[1/2,1]}$ under $||\cdot||_1$.

Part 2: (It is not possible for $V$ to be complete under $||\cdot||_1$)

Assume that $V$ were complete under $||\cdot||_1$. Consider the identity function $I:(V,||\cdot||_2)\rightarrow (V,||\cdot||_1)$. Since $||I(f)||_1=||f||_1\leq K||f||_2$, we have that $I$ is a continuous linear map. Since it is also surjective (and we are assuming $(V,||\cdot||_2)$ and $(V,||\cdot||_1)$ to be Banach) we get, by the open mapping theorem, that $I$ is open.This means that $I^{-1}:(V,||\cdot||_1)\rightarrow (V,||\cdot||_2)$ is continuous. Since $I^{-1}$ is linear, then is is also bounded: $$||f||_2=||I^{-1}(f)||_2\leq C||f||_1$$ for some constant $C>0$.

Therefore $\frac{1}{C}||\cdot||_2\leq||\cdot||_1\leq K||\cdot||_1$. This means that the norms would be equivalent. But this is a contradiction with the fact that they are not. Therefore $V$ cannot be complete under $||\cdot||_1$.

OR.
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