# Extending inequality for $\ell^p$ from integer $p$ to real $p$

Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are two decreasing sequences of positive numbers such that $a_1<b_1$ and $$\sum_{k=1}^\infty a^p_k\leq \sum_{k=1}^\infty b^p_k<\infty$$ for $p=2,3,\cdots.$

Does it follow that $$\sum_{k=1}^\infty a^p_k\leq \sum_{k=1}^\infty b^p_k$$ for all real $p\geq 2$?

$\textbf{Remark 1:}$ Its transparent that the latter inequality cannot be violated by a non-integral sequence $(r_n)$ that runs to infinity. Otherwise by continuity argument $a_1=\max a_n=\displaystyle\lim_{n\to\infty}\|a_n\|_{r_n}\geq\lim_{n\to\infty}\|b_n\|_{r_n} =\max b_n=b_1$!!

• I think it is sufficient to show it for $p=m/2$ with odd $m\ge3$. By re-scaling, we then get it for every $p=q/2^n$ (with $q>2^n, q\in\mathbb N)$ and by continuity for all $p>1$. – Wolfgang Nov 27 '15 at 17:07
• @Wolfgang Unfortunately there was a typo and I corrected it which may or may not change the answer :-( . All I know at the moment ,is that it holds for all integers greater than or equal to 2. – BigM Nov 27 '15 at 17:10

All we have to do is solve some equations. If I did it right, then:

Let $q \approx 2.45$ satisfy $204q^4-580q^3+867q^2-4028=0$. Let $$r = \frac{578}{4323}q^3+\frac{2465}{12969}q^2-\frac{8056}{12969}q+\frac{41905}{2593}\approx 3.20 .$$ Let $$a_1 :=3 ,\qquad a_2 := q,\qquad a_3 := \frac{1}{2},$$

$$b_1 := r,\qquad b_2 := 2,\qquad b_3 := 1.$$

Then $f(p) := (b_1^p+b_2^p+b_3^p) - (a_1^p+a_2^p+a_3^p)$ is negative for $2\lt p\lt 3$, zero for $p=2,3$ and positive for $p\gt 3$ and $0 < p < 2$. In particular, $f(p) \ge 0$ for all natural numbers $p$.

I did this by solving for $q,r$ in the two equations $f(2)=0, f(3)=0$.

• Well seems the case is closed. I appealed to Mathematica to plot the graph of your example. Guess that was a wishful thinking. Thanks – BigM Nov 28 '15 at 0:29

This is a counterexample for the original question that had $p \ge 1$ instead of $p \ge 2$.

Consider a case where $(a_1, a_2, a_3) = (3,2,1)$. We'll take $$(b_1, b_2, b_3) = \left(2 + \sin(\theta) + \frac{\cos(\theta)}{\sqrt{3}}, 2 - \sin(\theta) + \frac{\cos(\theta)}{\sqrt{3}}, 2 - \frac{2\cos(\theta)}{\sqrt{3}}\right)$$ and $b_k = a_k$ for $k \ge 4$. Note that $\sum_k b_k = \sum_k a_k$ and $\sum_k b_k^2 = \sum_k a_k^2$. For $0 < \theta < \pi/3$ we have $b_1 > b_2 > b_3 > 0$, with $b_k = a_k$ for $\theta = \pi/6$ and $b_1 > a_1$ for $\pi/6 < \theta < \pi/3$. Taking e.g. $\theta = \pi/4$, we have $\sum_k b_k^{3/2} < \sum_k a_k^{3/2}$, while $\sum_k b_k^n > \sum_k a_k^n$ for all $n \ge 2$.

I suspect this can be modified to produce an example where $\sum_k a_k^p - \sum_k b_k^p = 0$ for $p = 2$ and $3$, $> 0$ for $2 < p < 3$ and $< 0$ for $p > 3$.

EDIT: For a numerical example, take $(a_1, a_2, a_3) = (5,4,1)$ and $(b_1, b_2, b_3) \approx (5.4, 2.551364169, 2.516056612)$.

• Let's for the sake of argument assume both inequalities are proper, then intuitively it shouldn't be exceptional cases (since $\|.\|)_p$ is monotone deceasing). – BigM Nov 27 '15 at 19:46