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May I ask if $p < q$ (where $q$ may be $\infty$), then is $A|\mathbf{x}|_q \leq |\mathbf{x}|_p \leq B|\mathbf{x}|_q$, where $A,B$ are some positive constants with respect to $\mathbf{x}$ (ie. depending on $n,p,q$ is fine), for all $\mathbf{x} \in \mathbb{R}^n$?

I've tried to find the answer in other posts on Stack Exchange such as this one but I couldn't understand.

Follow-up Question: If this is true, then I think it should be true that convergence in $p$-norm for one value of $p$ is equivalent to convergence in $p$-norm for any $p$, right?

HIH
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  • Yes, it's true. There is a general result that all norms defined on a finite dimensional normed space are equivalent to each other. In particular it follows that convergence in one norm implies convergence in all other norms on the same space. This is not true for infinite dimensional normed spaces though. – Mark Oct 11 '22 at 21:19
  • Have a look at this answer and the comment that follows it for infinite dimensional spaces (which is not what you are after). – Jean Marie Oct 11 '22 at 21:19

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Yes, because the $p$-norm dominates the $q$-norm when $p \lt q$. The proof is just a matter of algebra. The key is to apply Hölder's inequality: $$ \sum_{i=1}^n |x_i|^p \le \left( \sum_{i=1}^n |x_i|^q \right)^{p/q} = |x|_q^p $$ and $$ \sum_{i=1}^n |x_i|^q \ge \left( \sum_{i=1}^n |x_i|^p \right)^{q/p} = |x|_p^q. $$

Amirreza Hashemi
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