How to show this $\ell^p$ inequality

Question

I have the following problem: Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $\ell^p (\mathbb{N})$ for some $p \in (0,2]$. Let $q \geq p$, $N \in \mathbb{N}$. Show: There exists a finite index set $I \subseteq \mathbb{N}$ of size $N$, such that $$\left(\sum_{n \in \mathbb{N} \setminus I} (a_n)^q \right)^{1/q} \leq || (a_n)_{n \in \mathbb{N}}||_{\ell^p} (N+1)^{1/q - 1/p}.$$ So the obvious choice for $I$ would be to pick the biggest values of $a_n$, since the sum converges these values cannot become arbitrary big. I know of a similar inequality regarding finite measure sets, see e.g. Inequalities in $l_p$ norm or An inequality about $L^p$ norm? , but this is a little bit different. What can I do here?

Answer 1

I will use an approach very similar to the one in the accepted answer by Syl.Qiu of An inequality about $L^p$ norm? which you had linked in your question. You also had the right idea to consider the biggest values of $a_n$, as this is what I'll be doing.

Let $I_0 := \emptyset$ and $i_1 := \min\{i \in \mathbb{N} \mid |a_i| = \sup |a_n|\}$, then define recursively $I_N := \{i_1 , \dots, i_N\}$ and $$i_{N+1} := \min\left\{i \in \mathbb{N} \setminus I_N \mid |a_i| = \sup_{n \in \mathbb{N}\setminus I_N} |a_n|\right\}$$ This is just to formalise the process of "picking the biggest values" by choosing an ordering.

Then $I_N$ is of size exactly $N$, even for $N = 0$, and we have: $$\begin{split} \left(\sum_{n \in \mathbb{N} \setminus I_N} |a_n|^q \right)^{\frac1q} &= \left(\sum_{n \in \mathbb{N} \setminus I_N} |a_n|^p |a_n|^{q-p} \right)^{\frac1q}\\ &\leq \left(\sum_{n \in \mathbb{N} \setminus I_N} |a_n|^p \right)^{\frac1q} \left(\sup_{n \in \mathbb{N} \setminus I_N}|a_n|^{q-p}\right)^{\frac1q}\\ &\leq \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1q} \bigg(\underbrace{\sup_{n \in \mathbb{N} \setminus I_N}|a_n|}_{=\, |a_{i_{N+1}}|}\bigg)^{(q - p) \cdot \frac1q \cdot \frac p p}\\ &\leq \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1q} \left(|a_{i_{N+1}}|^p\right)^{\frac{q-p}{pq}}\\ &\leq \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1q} \left(\frac{1}{N+1}\sum_{k=1}^{N+1}|a_{i_k}|^p\right)^{\frac1p - \frac1q}\\ &\leq \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1q} \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1p - \frac1q} (N+1)^{\frac1q - \frac1p}\\ &\leq \left(\sum_{n \in \mathbb{N}} |a_n|^p \right)^{\frac1p} (N+1)^{\frac1q - \frac1p}\end{split}$$ where we used the fact that $|a_{i_{N+1}}|^p \leq |a_{i_k}|^p$ for $k \leq N$, so that $(N+1) |a_{i_{N+1}}|^p \leq \sum_{k = 1}^{N+1} |a_{i_k}|^p$.