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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?

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1 Answer 1

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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 $u_n = \frac{1}{n^{1/p}}$

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  • $\begingroup$ Your assumption is true and i have corrected the post. why can i simply assume $\|x\|_p=1$? what if it is not true? $\endgroup$
    – Maxim
    Commented May 13, 2016 at 10:38
  • $\begingroup$ 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? $\endgroup$
    – Shubhodip Mondal
    Commented May 13, 2016 at 10:43
  • $\begingroup$ 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 $\|tx\|$? $\endgroup$
    – Maxim
    Commented May 13, 2016 at 10:56
  • $\begingroup$ $t$ is a constant. It comes out and gets cancelled from both sides. $\endgroup$
    – Shubhodip Mondal
    Commented May 13, 2016 at 10:57
  • $\begingroup$ 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 $||x||_p = 1$. $\endgroup$
    – Shubhodip Mondal
    Commented May 13, 2016 at 11:00

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