# Is this operator between $\ell^{25}$ and $\ell^{12}$ continuous?

Problem: Let us define $$\ell^p$$ as the space of sequences $$(x_n)_{n \in \mathbb{N}}$$ such that $$\sum\limits_{n \in \mathbb{N}}|x_n|^p < +\infty$$ with the usual norm $$\|x\|_p = \big( \sum\limits_{n \in \mathbb{N}}|x_n|^p \big)^{\frac 1 p}$$ Define $$T:\ell^{25} \to \ell^{12}$$ in the following way: $$T(x_1,\dots,x_n, \dots) = (x_1^{2018},\dots,x_n^{2018}, \dots)$$ I am asking if the map $$T$$ is continuous.

Attempt: Let us fix $$x \in \ell^{25}$$ and let us study what happens for $$y \in \ell^{25}$$ with $$\|y-x\|_p \leq 1$$. We have that $$|y_n-x_n| \leq 1$$ for all $$n \in \mathbb{N}$$ and thus $$|x_n^{2018}-y_n^{2018}| \leq |x_n - y_n| K$$ for $$K > 0$$ depending only on $$\|x\|_p$$ which has been fixed. This is because $$a^n-b^n = (a-b)p(a, b)$$ where $$p$$ is a polynomial on $$a$$ and $$b$$.

Thus $$\|Tx-Ty\|_{12}^{12} \leq K^{12} \|x-y\|_{12}^{12}$$ but this does not help me to conclude because we cannot control the $$\mathcal{l}^{12}$$ with the $$\mathcal{l}^{25}$$ norm.

Trivial remark: the operator is not linear and thus we have no hope to prove that $$T$$ is Lipschitz continuous.

• Wasn't this question asked recently and received quite a few votes? It may have lacked context : is that the reason for reasking? +1. The other question is here. Jan 6 at 9:59
• @SarveshRavichandranIyer I tried to prove enough context in order to receive an answer Jan 6 at 10:17
• Thanks, Filippo, this is a very nice question, given that if it has a nice answer, one could use it as an example of working with non-linear operators in function/sequence spaces. Jan 6 at 10:25
• @SarveshRavichandranIyer I haven't seen it yet. Thank you so much. The fact is that in that case $4<6$ but in our case $25>12$. The spaces are in some way inverted. If it was $T:\mathcal{l}^{12} \to \mathcal{l}^{25}$ the proof would be exactly the same. Jan 6 at 10:33
• Sorry, I just removed the post you referred to above, because I thought I'd put the wrong link, when in fact I'd put the right one. I'll just put it up , it's here @Filippo Jan 6 at 10:34

Note that for all $$n \in \mathbb{N}$$ and for all $$a,b \in \mathbb{R}$$ we have $$a^n-b^n = (a - b) \ ( \ a^{n-1} \ + \ a^{n-2}b \ + \ a^{n-3}b^2 \ + \ \ldots \ + \ b^{n-1} \ )$$. Let $$x \in \mathcal{l}^{25}$$ be fixed and let $$y \in \mathcal{l}^{25}$$ be such that $$\|y-x\|_{25} \leq 1$$. Thus for all $$n$$ we have $$|x_n| \leq \|x\|_{25}$$ and $$|y_n| \leq \|y-x\|_{25}+\|x\|_{25} \leq 1+\|x\|_{25}$$.

Let us compute \begin{align*} \|Tx-Ty\|_{12}^{12}&= \sum\limits_{n \in \mathbb{N}}|x_n^{2018}-y_n^{2018}|^{12} \\ &= \sum\limits_{n \in \mathbb{N}}|(x_n - y_n) \ ( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12} \\ &= \sum\limits_{n \in \mathbb{N}}|(x_n - y_n) |^{12} |( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12} \\ &\leq \left( \sum\limits_{n \in \mathbb{N}}|(x_n - y_n)|^{12 \cdot 3} \right)^{\frac 1 3} \left( \sum\limits_{n \in \mathbb{N}} |( x_n^{2017 } + x_n^{2018-2}y_n + x_n^{2018-3}y_n^2 + \ldots + y_n^{2018-1} )|^{12 \cdot \frac 3 2} \right)^{\frac 2 3} \end{align*}

Now $$\big( \sum\limits_{n \in \mathbb{N}}|(x_n - y_n)|^{12 \cdot 3} \big)^{\frac 1 3}=\|x-y \|_{36}^{12} \leq \|x-y\|_{25}^{12}$$.

Finally: $$\big( \sum\limits_{n \in \mathbb{N}} |( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12 \cdot \frac 3 2} \big)^{\frac 2 3} \leq \big( \sum\limits_{n \in \mathbb{N}} |x_n|^{2017 \cdot 12 \cdot \frac 3 2} \ + \ |x_n|^{2016 \cdot 12 \cdot \frac 3 2}|y_n|^{12 \cdot \frac 3 2} \ + \ |x_n|^{2015 \cdot 12 \cdot \frac 3 2}|y_n|^{2 \cdot 12 \cdot \frac 3 2} \ + \ \ldots \ + \ |y_n|^{2017 \cdot 12 \cdot \frac 3 2} \ )|^{12 \cdot \frac 3 2} \big)^{\frac 2 3} \leq C(\|x\|_{25})$$ since $$y \in B_{\mathcal{l}_{25}}(x,1)$$. Thus we found an estimate as:

$$\|Tx-Ty\|_{12} \leq \big( C(\|x\|_{25}) \big) ^{\frac{1}{12}} \|x-y\|_{25}$$

for all $$y \in B_{\mathcal{l}_{25}}(x,1)$$

• Nice use of Holder's inequality to bump up the exponent. Jan 6 at 18:17
• Excellent work! I am happy that things worked out. +1, and I'll bookmark this one. Jan 6 at 18:23
• @SarveshRavichandranIyer thank you Jan 6 at 20:41