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I am trying to follow the accepted answer here in order to prove that for $1 \leq p < q \leq \infty$ we have $||x_n||_q < ||x_n||_p$ and $\ell^q \subset \ell^p$.

Start with taking $(x_n)_{n \in \mathbb{N}} \in \ell^p$. Since $\sum |x_n|^p < \infty$ then $\exists_M \forall_{m>M}$ we have $|X_m|^p < 1$. But $q > p$ so $|X_m|^q < |X_m|^p$. That implies any sequence from $\ell^p$ will be summable with power of $q$ too and thus in $\ell^q$. How do I conclude the norm inequality?

If we skipped the first $M$ elements in the sequence we would get $$ ||x_n||^q_q \backsimeq \left((\sum_{i=M+1}^{\infty} |x_i|^q)^{\frac{1}{q}}\right)^q = \sum_{i=M+1}^{\infty} |x_i|^q \leq \sum_{i=M+1}^{\infty} |x_i|^p = \left((\sum_{i=M+1}^{\infty} |x_i|^p)^{\frac{1}{p}}\right)^p \backsimeq||x_n||^p_p $$ (no equality sign since I skipped M first elements in the norm)

Now my questions:

  1. Can I just "skip" some elements like that and give $=$ instead of $\backsimeq$? What if skipped elements summed to something bigger on the left side than on the right side?
  2. Would $||x_n||^q_q \leq ||x_n||^p_p$ imply $||x_n||_q < ||x_n||_p$? If yes, how?
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  • $\begingroup$ What is that $n$ in $x_n$ mean? Do you mean $x =(x_n)_{n\in \mathbb N}$? Also you are summing in m, but the expression has no m? And what is $\hat{x_n}$? $\endgroup$ Oct 25, 2020 at 8:33
  • $\begingroup$ @ArcticChar I have edited the question. $\endgroup$
    – blahblah
    Oct 25, 2020 at 8:42

1 Answer 1

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  1. No, you cannot skip elements like that.
  2. No, it does not imply that.

What you want to prove is what is sometimes called Jensen inequality, and even if only finitely many terms are nonzero it is not completely trivial.

Here is how a possible proof goes: You can divide the proof into 2 steps.

  1. Show the implication $\lVert x\rVert_p\le1\implies\lVert x\rVert_q\le1$.
  2. Show that this implication implies $\lVert x\rVert_q\le\lVert x\rVert_p$.

For step 1: This can be shown similarly as you did: Note that the proof does not run into your above two difficulties because

  1. If $x=(x_n)_n$ satisfies $\lVert x\rVert_p\le1$ then $\lvert x_n\rvert\le1$ for every $n$ (so you do not have to exclude some $n$).
  2. To prove $\lVert x\rVert_q\le1$ it is sufficient to show that $\lVert x\rVert_q^r\le1$ for some power $r>0$ (because $1^r=1$), so there is no problem with the powers.

For step 2: Let $x\in\ell_p$. In case $x=0$, the inequality is trivial. In case $x\ne0$, there is a number $\lambda>0$ such that $y=\lambda x\in\ell_p$ satisfies $\lVert y\rVert_p=1$. Now apply step 1 to obtain that $\lVert y\rVert_q\le1=\lVert y\rVert_p$. Dividing this inequaliy by $\lambda>0$ gives the result.

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