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Let $\left \| x \right \|_{p}= \Bigg( \sum_{i \in \mathbb{N}} \left | x_{i} \right |^{p} \Bigg)^{\frac{1}{p}} $ be the norm of $\ell_p$ for $1 \leq p < \infty$, show that when $p\neq q$ that the norms of $\ell_p$ and $\ell_q$ are not equivalent on the linear space $f $.

So far I've considered the sequences $g_{n} \in f$ with $g_{1} = (1,0,0,....), g_{2} = (1,1,0,....), ...$ I get that $\left \| g_{n} \right \|_{p}=n^\frac{1}{p}$

Next I assume that $\left \| .\right \|_{p}$ and $\left \| .\right \|_{q}$ are equivalent hoping for a contridiction and get the inequality $an^\frac{1}{p} \leq n^\frac{1}{q} \leq bn^\frac{1}{p}$ for $a,b$ constants $n \in \mathbb{N}$ and $p,q>1$ Letting n=1 give $a \leq 1 \leq b$ and I can also assume wlog that $p$ greater than $q$ to get more information about the relative sizes of the powers, but I'm still not really sure how to dispove this inequality for a contradiction.

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You're almost there. Multiplying your inequality through by $n^{-1/q}$ gives $$ an^{\frac{1}{p} - \frac{1}{q}} \leq 1 \leq b n^{\frac{1}{p}-\frac{1}{q}}, \quad \forall n \geq 1 $$ for some $a, b > 0$. Can you see why this is impossible whenever $p \neq q$?

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  • $\begingroup$ I think so. Let $q>p$ then $\frac{1}{p}-\frac{1}{q}>0$ So $n^{\frac{1}{p}-\frac{1}{q}}\rightarrow \infty $ as $n \rightarrow \infty$ so there can't be a constant $a$ such that the inequality is true. $\endgroup$ – Roger Oct 6 '18 at 16:10

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