# Expanding a norm over a given space

Let $1 \leq p<q \leq \infty$ (p an q are not related)

Let $\Phi$ be the vector space of all sequences with at most finitely many nonzero elements, meaning $\Phi=\{\{x_n\}_{n=1}^\infty|$ there is $n_o$ such that $x_n=0$ whenever $n\leq n_0\}$

Prove that $\|x\|_q \leq \|x\|_p$ for all $x \in \Phi$ and there is no constant $C$ such that $\|x\|_p \leq C\|x\|_q$ for all $x \in \Phi$

How can I show that $\sup_{0\neq x\in \Psi} \frac{\|x\|_p}{\|x\|_q}=\infty$? this proves both inequations but I cant manage to epand and solve the sup function. What is the system here? is it at all a good direction?

Hint: (I assume $q>p$)
1. You can assume that $||x||_p = 1$. Then each $|x_i| \le 1$. So $|x_i|^q \le |x_i|^p$. Now sum this up.
2. Try the sequence $u_n = \frac{1}{n^{1/p}}$
• Your assumption is true and i have corrected the post. why can i simply assume $\|x\|_p=1$? what if it is not true? Commented May 13, 2016 at 10:38
• Well, it happens because of normalization. Any element is of the form $tx$ where $t \in \mathbb R$ and $||x||_p = 1$. Now in order to prove $||tx||_q \le ||tx||_p$, you just have to prove $||x||_q \le ||x||_p$. Clear? Commented May 13, 2016 at 10:43
• I'm not sure. you are saying that if i prove it for the case $\|x\|_p=1$ i have proved it for any other norm $\|t\|$ since i can denote it $\|tx\|$? Commented May 13, 2016 at 10:56
• $t$ is a constant. It comes out and gets cancelled from both sides. Commented May 13, 2016 at 10:57
• Assume you have proven it for any $x$ with $||x||_p = 1$. Take some other $y$. Write it as $||y||_p (\frac{y}{||y||_p})$. Let $t = ||y||_p$. Then $y = tx$ for some $x$ with $||x||_p = 1$. Commented May 13, 2016 at 11:00