Let us consider the function defined as $$ F: l^4 \rightarrow l^6 \\ (x_1, \dots,x_n, \dots) \mapsto (x_1^{20}, \dots, x_n^{20}, \dots) $$ I am asked to prove whether this function is continuous or not. Let us try proving the continuity, then using the $\epsilon - \delta$ definition I need to find, given $\epsilon > 0$ a suitable $\delta$ such that $$ \lvert \lvert (x_1, \dots,x_n, \dots) - (y_1, \dots,y_n, \dots) \rvert \rvert_{l^4} = \left(\sum_{n \in \mathbb{N}}\left| x_n-y_n \right|^{4}\right)^\frac{1}{4} < \delta $$ Implies $$ \lvert\lvert F(x_1, \dots,x_n, \dots) - F(y_1, \dots,y_n, \dots)\rvert\rvert_{l^{6}} = \left(\sum_{n \in \mathbb{N}}\left| x_n^{20}-y_n^{20} \right|^{6}\right)^\frac{1}{6} < \epsilon $$ Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.


1 Answer 1


Probably not the best proof, but the best I can think of right now.

For sequences, $\|\boldsymbol{x}\|_q\le\|\boldsymbol{x}\|_p$ whenever $p\le q$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+\cdots+y^{19}|\le c|x-y|.$$ Hence $$\|(x_n^{20})-(y_n^{20})\|_6\le c\|x-y\|_6\le c\|x-y\|_4<c\delta$$

To justify the constant $c$, note that \begin{align*}\left|\sum_{i=0}^{19}x_n^iy_n^{19-i}\right|&\le\sum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\\ &\le\sum_{i=0}^{19}\frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\\ &= 20(|x_n|^{19}+|y_n|^{19})\\ &\le 20(\|\boldsymbol{x}\|_\infty^{19}+\|\boldsymbol{y}\|_\infty^{19})=c \end{align*}

  • $\begingroup$ What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something. $\endgroup$
    – JCF
    Jan 31, 2019 at 10:52

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