2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.