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.

  • $\begingroup$ 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

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.


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