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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.

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    $\begingroup$ 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. $\endgroup$ Jan 6 at 9:59
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    $\begingroup$ @SarveshRavichandranIyer I tried to prove enough context in order to receive an answer $\endgroup$ Jan 6 at 10:17
  • $\begingroup$ 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. $\endgroup$ Jan 6 at 10:25
  • $\begingroup$ @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. $\endgroup$ Jan 6 at 10:33
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    $\begingroup$ 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 $\endgroup$ Jan 6 at 10:34

1 Answer 1

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

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

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