1
$\begingroup$

Problem 5.3.32 - Let $\lVert x \rVert_{1}$ and $\lVert x \rVert_{2}$ be norms on the vector space $\mathscr{X}$ such that $\lVert x \rVert_{1}\leq \lVert x \rVert_{2}$. If $\mathscr{X}$ is complete with respect to both norms, then the norms are equivalent.

I know that $0\in \mathscr{X}$ and also I know that the norms are equivalent if there exists a $C_1,C_2 > 0$ such that $$C_1\lVert x \rVert_{1} \leq \lVert x \rVert_{2} \leq C_2\lVert x \rVert_{1}$$ but I am not sure how to show this, any suggestions is greatly appreciated.

|cite|improve this question
$\endgroup$
  • $\begingroup$ See this. $\endgroup$ – Pedro Mar 20 '16 at 18:24
  • $\begingroup$ Thanks for the link but I am still pretty lost, I just want to know if there is another of proving it other than that method. $\endgroup$ – Wolfy Mar 20 '16 at 18:27
  • $\begingroup$ Do you want to prove it without the open mapping theorem? $\endgroup$ – Pedro Mar 20 '16 at 18:31
  • $\begingroup$ I guess so maybe I do not understand the open mapping theorem well cause I don't know how it could be applied here. $\endgroup$ – Wolfy Mar 20 '16 at 18:32
  • $\begingroup$ Note that this exercise is in the section of the open mapping theorem. So, the open mapping theorem is probably supposed to be used in the expected solution. $\endgroup$ – Pedro Mar 20 '16 at 18:39
2
$\begingroup$

As you have said, you have to prove that there exist $C_1,C_2 > 0$ such that $$C_1\lVert x \rVert_{1} \leq \lVert x \rVert_{2} \leq C_2\lVert x \rVert_{1}$$

By hypothesis, we have $$C_1\lVert x \rVert_{1} \leq \lVert x \rVert_{2}\tag{1},$$ where $C_1=1$. So, it's enough to prove that there exists $C_2 > 0$ such that $$\lVert x \rVert_{2} \leq C_2\lVert x \rVert_{1}$$

In other words, you have to prove that the mapping $F:(\mathscr{X},\|\cdot\|_1)\to (\mathscr{X},\|\cdot\|_2)$ given by $F(x)=x$ is continuous.

Note that the mapping $F$ is the inverse of the mapping $T:(\mathscr{X},\|\cdot\|_2)\to (\mathscr{X},\|\cdot\|_1)$ given by $T(x)=x$.

Because of $(1)$, $T$ is continuous. So, by Corollary 5.11 of Folland, $F$ is continuous as desired.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.