# Show the norms of $\ell_p$ and $\ell_q$ are not equivalent on the linear space $f$

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.

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 $$p \neq q$$?
• 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. – Roger Oct 6 '18 at 16:10