# Real Analysis, Folland Problem 5.3.32 The Baire Category Theorem

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.

• See this. – Pedro Mar 20 '16 at 18:24
• 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. – Wolfy Mar 20 '16 at 18:27
• Do you want to prove it without the open mapping theorem? – Pedro Mar 20 '16 at 18:31
• I guess so maybe I do not understand the open mapping theorem well cause I don't know how it could be applied here. – Wolfy Mar 20 '16 at 18:32
• 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. – 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.