# An exercise in Lorentz sequence spaces

I'm trying to solve an exercise about Lorentz sequence spaces. Below is the text.

It is known that the Lorentz sequence space, denoted by $$\ell\left(p,q\right)$$ is the set of all sequences $$a = \left\lbrace a_n\right\rbrace$$ such that the functional $$\Vert a\Vert_{pq} < \infty,$$ where

$$\Vert a\Vert_{pq} =\begin{cases} \left(\sum_{n=1}^{\infty} \left(n^{\frac{1}{p}} a_{n}^{\ast}\right)^{q} n^{-1}\right)^{\frac{1}{q}}\\ \sup_{n\geq 1} n^{\frac{1}{p}} a^{\ast}_{n} \end{cases}$$

if $$0 and $$0 respectively. Moreover $$a^{\ast} = \left\lbrace a^{\ast}_{n}\right\rbrace$$ is the sequence $$\left\lbrace\vert a_{n}\vert\right\rbrace$$ permutated in a decreasing order.

Now consider the two-dimensional Lorentz sequence space $$\ell^{\left( 2\right)}\left(p,q\right),$$ that is the space of all sequences $$a = \left\lbrace a_{1}, a_{2}\right\rbrace$$ with the quasi-norm $$\Vert a\Vert_{pq} = \left( a_{1}^{\ast^{q}} + 2^{\frac{q}{p} -1} a_{2}^{\ast^{q}}\right)^{\frac{1}{q}}.$$

Prove that $$\Vert\cdot\Vert_{pq}$$ cannot be a norm for $$\ell^{\left( 2\right)}\left(p, q\right)$$ when $$p < q.$$

Hint: Consider the unit ball in $$\ell^{\left(2\right)}\left(p, q\right)$$ for different values of $$p$$ and $$q.$$

It's clear that it can be shown as done for the standard Lorentz spaces, but (about me) the author chooses $$\ell^{\left( 2\right)}\left(p,q\right)$$ because, since the dimension is $$2,$$ the unit ball can be drawn in the plane. My idea is to prove that, in the case $$p < q,$$ the unit ball is not convex so $$\Vert\cdot\Vert_{pq}$$ fails to be a norm. Is that right what i said until now?

For $$p = q = 1,$$ we have $$\Vert a\Vert_{11} = \vert a_1\vert + \vert a_2\vert \leq 1,$$ that is a rhombus, that is convex and then $$\Vert\cdot\Vert_{11}$$ is a norm, as expected. I would like to prove that, for example, for $$p = 1, q = \infty,$$ the $$\Vert\cdot\Vert_{1\infty}$$ is not a norm, but i don't know how to express the norm in this case, i.e. what is the value of $$\Vert a\Vert_{1\infty}?$$

I expect to draw a not covex figure in the plane, but i don't know how.

I hope everyone could help me! Thank you!

• I can at least verify the result is correct graphically on Desmos desmos.com/calculator/f8vn8we6h7. Coming to the result caveman style is another thing... – Calvin Khor Oct 12 '19 at 8:25
• PS I'd expect in analogy with the full sequence norm, $$\|a\|_{1\infty} = \max\{ a^*_1, 2^{1/p}a^*_2\}$$ for this space, $(1/2,1)$ and $(1/2,1)$ are in the (closed) unit ball, but their average $(3/4,3/4)$ is not. – Calvin Khor Oct 12 '19 at 8:36
• I'm sorry but i thought I'd solve it like a caveman. You offered a very good way to solution. – C. Bishop Oct 13 '19 at 8:11
• I think the method will work eventually, but I don't consider the problem solved. I would like to see the full "caveman" answer :) or maybe someone has a slick method that we are missing...so I don't agree with accepting my answer. – Calvin Khor Oct 13 '19 at 8:30

## 1 Answer

(the below gets the result for $$p\ll q$$. I don't have time for the full result, but hope this is helpful anyway...)

In analogy with the full sequence norm, I'd say that $$|a|_{p\infty} := \max\{ a^*_1, 2^{1/p}a^*_2\}.$$ For this space, $$(1/2^{1/p},1)$$ and $$(1/2^{1/p},1)$$ are in the (closed) unit ball, but their average is not. Indeed, the average has norm $$\frac{1/2^{1/p}+1}2 |(1,1)|_{p\infty} = \frac{2^{1/p}(1/2^{1/p}+1)}2 = \frac{2^{1/p} + 1}2 > 1.$$

Hoping that this works naively for other $$q$$. Let $$p. This translates to $$2^{q/p-1}>1$$. Set $$a\le 1$$, and consider $$x_1 = (1,a), \quad x_2 = (a,1).$$ Their norms are $$|x_1|_{pq} =|x_2|_{pq} = (1+a^q2^{q/p-1})^{1/q}$$ Setting $$x_3 = (x_1 + x_2)/2 = (\frac{1+a}2,\frac{1+a}2) = \frac{1+a}2(1,1)$$, $$|x_3|_{pq} = \frac{1+a}2 |(1,1)|_{pq} = \frac{1+a}2(1+2^{q/p-1})^{1/q}$$ Convexity of the (closed) ball of radius $$|x_1|_{pq}$$ (or equivalently any ball of radius 1) would of course imply that $$|x_3|_{pq} \le \frac12 |x_1|_{pq} + \frac12 |x_2|_{pq} = (1+a^q2^{q/p-1})^{1/q}$$

So the question is - is it true that for $$a\in[0,1]$$, $$\frac{1+a}2(1+2^{q/p-1})^{1/q} \overset{\Huge ?}{\le} (1+a^q2^{q/p-1})^{1/q}$$ Setting $$a = 2^{1/q - 1/p}<1$$ in analogy with the $$p\infty$$ case. This inequality is

$$\frac{1+2^{1/q - 1/p}}2(1+2^{q(1/p-1/q)})^{1/q} \overset{\Huge ?}{\le} 2^{1/q}$$

As noted, this inequality is false for $$q\to\infty$$. Its true at $$p=q$$. It looks likely to work on the graph(see https://www.desmos.com/calculator/qg2cpfocrj), but I don't have the time now to persue this further.