# Equivalent norms on Banach spaces

For $$n \in \mathbb{N} \ \$$ let $$(E_1, \vert \vert * \vert \vert_{E_1}), ..., (E_n, \vert \vert * \vert \vert_{E_n})$$ be Banach spaces. Define $$E:= \prod_{j=1}^{n}E_n$$, $$\vert \vert * \vert \vert_E : E \rightarrow \mathbb{R}, \ \ (x_1, ..., x_n) \mapsto \vert \vert (\vert \vert x_1 \vert \vert_{E_1}, ..., \vert \vert x_n \vert \vert_{E_n} )\vert \vert_p \ \$$ for $$\ \ p \in [1, \infty].$$

Let $$\vert \vert \vert * \vert \vert \vert$$ be another norm on $$E$$, such that for all $$k \in \{1, ..., n\}$$ the function $$p_k: (E, \vert \vert \vert * \vert \vert \vert ) \rightarrow (E_k, \vert \vert * \vert \vert_{E_k} ), \ \ (x_1, ..., x_n) \mapsto x_k$$ is continuous. Prove that the norms $$\vert \vert * \vert \vert_{E}$$ and $$\vert \vert \vert * \vert \vert \vert$$ are equivalent.

• If i get it right both $(E, \vert \vert \vert * \vert \vert \vert)$ and $(E, \vert \vert * \vert \vert_{E})$ are Banach spaces. Is it so? Jan 5 '20 at 16:24
• So I have to show that there are $c_1, c_2 >0$ such that $c_1 \vert \vert \vert x \vert \vert \vert \leq \vert \vert x\vert \vert_E \leq c_2 \vert \vert \vert x \vert \vert \vert$ for every $x \in E$. Jan 5 '20 at 16:33
• Are you assuming that $\lvert \lvert \lvert \cdot \rvert \rvert \rvert$ makes $E$ a Banach space? Jan 5 '20 at 16:50
• Yes, I think so Jan 5 '20 at 16:59

If we assume that $$\lvert \lvert \lvert \cdot \rvert \rvert \rvert$$ makes $$E$$ a Banach space then by the Open mapping theorem (or Banach's isomorphism theorem or whichever your favourite equivalent form of the OMT is), it will suffice to see that $$\|x\|_E \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$$.
There are many ways to see this. One way is to recall that $$\|\cdot\|_E$$ induces the product topology on $$E$$ so that if $$Y$$ is a topological space $$f:Y \to (E, \|\cdot\|_E)$$ is continuous iff $$p_k \circ f$$ is continuous for each $$k$$. In particular, it is immediate that $$\operatorname{Id}: (E, \lvert \lvert \lvert \cdot \rvert \rvert \rvert) \to (E, \|\cdot\|_E)$$ is continuous which is precisely that $$\|x\|_E \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$$ as desired.
Alternatively, continuity of $$p_k$$ means that $$\|x_k\|_{E_k} \leq c_k \lvert \lvert \lvert x \rvert \rvert \rvert$$. Hence there is a $$c$$ such that $$\max_k \|x_k\| \leq c \lvert \lvert \lvert x \rvert \rvert \rvert$$. To conclude exploit the fact that the $$p$$-norms are equivalent on $$\mathbb{R}^k$$.
• How do you apply the OMT to show $\vert \vert x \vert \vert_E \leq c \vert \vert \vert x \vert \vert \vert$? Jan 5 '20 at 17:19
• The identity map is a continuous bijection from $(E, \lvert \lvert \lvert \cdot \rvert \rvert \rvert \rvert) \to (E, \|\cdot \|)$. Applying e.g. Banach's isomorphism theorem then tells you that the its inverse (which is the identity map with the norms the other way) is also continuous which gives the inequality. Jan 5 '20 at 17:27
• Ok. Could you please explain why continuity of $p_k$ means $\vert \vert x_k \vert \vert_{E_k} \leq c_k \vert \vert \vert x \vert \vert \vert$? Jan 5 '20 at 18:15
• This is just the statement that $p_k$ is a bounded linear operator for $\lvert \lvert \lvert \cdot \rvert \rvert \rvert$ since $p_kx = x_k$. It is a standard fact that a linear map is bounded iff it is continuous. Jan 5 '20 at 18:18
• At that point, you know that $\|(\|x_i\|)_{i=1}^k\|_\infty \leq \lvert \lvert \lvert x \rvert \rvert \rvert$ where $\|\cdot\|_\infty$ is the $\infty$-norm on $\mathbb{R}^k$. Since the $p$-norms on $\mathbb{R}^k$ are equivalent, the desired result is immediate. Jan 5 '20 at 19:22