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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.

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  • $\begingroup$ 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? $\endgroup$ Jan 5 '20 at 16:24
  • $\begingroup$ 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 $x \in E$. $\endgroup$ Jan 5 '20 at 16:33
  • $\begingroup$ Are you assuming that $\lvert \lvert \lvert \cdot \rvert \rvert \rvert$ makes $E$ a Banach space? $\endgroup$ Jan 5 '20 at 16:50
  • $\begingroup$ Yes, I think so $\endgroup$ Jan 5 '20 at 16:59
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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 $\mathbb{R}^k$.

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  • $\begingroup$ How do you apply the OMT to show $\vert \vert x \vert \vert_E \leq c \vert \vert \vert x \vert \vert \vert $? $\endgroup$ Jan 5 '20 at 17:19
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    $\begingroup$ 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. $\endgroup$ Jan 5 '20 at 17:27
  • $\begingroup$ 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 $? $\endgroup$ Jan 5 '20 at 18:15
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    $\begingroup$ 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. $\endgroup$ Jan 5 '20 at 18:18
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    $\begingroup$ 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. $\endgroup$ Jan 5 '20 at 19:22

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