Let $\mathcal{A}$ be an arbitrary Banach space with norm $\|\cdot\|_{\mathcal{A}}$ and let $\mathcal{A}^{n}$ be its Cartesian product. I came across the following statement and wonder whether it is true or not:

All norms $\|\cdot\|_{\mathcal{A}^{n}}$ on $\mathcal{A}^{n}$ such that $\|(0,...,0,\cdot,0,...,0)\|_{\mathcal{A}^{n}}=\|\cdot\|_{\mathcal{A}}$ for all $i=1,...,n$ (i.e. the inclusion is an isometry) are equivalent.

It is clear that for every $x=(x_{1},...,x_{n})\in\mathcal{A}^{n}$ it holds:

\begin{equation} \|x\|_{\mathcal{A}^{n}}\leq\sum_{i=1}^{n}\|x_{i}\|_{\mathcal{A}}=:\|x\|_{1} \end{equation}

Therefore, $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{\mathcal{A}^{n}}$.

What what about the other direction? Is it true?

Thanks for your help.

Let $\mathcal{A}$ be an arbitrary (typically infinite-dimensional) Banach space with norm $\|\cdot\|_{\mathcal{A}}$ and let $\mathcal{A}^{n}$ be its Cartesian product. I came across the following statement and wonder whether it is true or not:

All norms $\|\cdot\|_{\mathcal{A}^{n}}$ on $\mathcal{A}^{n}$ such that $\|(0,...,0,\cdot,0,...,0)\|_{\mathcal{A}^{n}}=\|\cdot\|_{\mathcal{A}}$ for all $i=1,...,n$ (i.e. the inclusion is an isometry) are equivalent.

It is clear that for every $x=(x_{1},...,x_{n})\in\mathcal{A}^{n}$ it holds:

\begin{equation} \|x\|_{\mathcal{A}^{n}}\leq\sum_{i=1}^{n}\|x_{i}\|_{\mathcal{A}}=:\|x\|_{1} \end{equation}

Therefore, $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{\mathcal{A}^{n}}$.

What what about the other direction? Is it true?

Thanks for your help.