Define a map $T:\ell^\infty \to \ell^2$ as $$T(a)=\{a_1,\frac{a_2}{2},\frac{a_3}{3}...\}.$$ The map is clearly linear and continuous as one can see $$T(a+b)=\{a_1+b_1,\frac{a_2+b_2}{2},\frac{a_3+b_3}{3}...\}=T(a)+T(b) $$And, $$||T(a)||_2=||\{a_1,\frac{a_2}{2},\frac{a_3}{3}...\}||_2=\bigg(\sum_{k=1}^\infty |\frac{a_k}{k}|^2 \bigg)^{1/2}\le \sup_{k\in\Bbb N} |a_k|\bigg(\sum_{k=1}^\infty \frac{1}{k^2} \bigg)^{1/2}=\frac{\pi}{\sqrt6}||a||_\infty $$ So $T$ is bounded, therefore continuous. But I am stuck to prove or disprove whether it is invertible or uniformly continuous or both? My intuition is that it must be both invertible and uniformly continuous but I want some proper rigorous arguments in this direction. Kindly help me.
uniform continuity and invertiblity of a map on $\ell^p$ -spaces?
2
$\begingroup$
real-analysis
functional-analysis
lp-spaces
-
1The boundedness proof is not correct. You have to prove that $\|T(a)\|_2\leq C\|a\|_\infty$ for some $C>0$, not $\|T(a)\|_2\leq C\|a\|_2$, as the norm equipped on the domain is $\|\cdot\|_\infty$. – 2017-01-18
-
0I corrected sir thank you! – 2017-01-18
-
1$T$ is not surjective. Consider the sequence $\{b_n\}$ defined by $b_n=n^{-\frac{3}{4}}$. Clearly $\{b_n\}$ is in $\ell^2$. However if there is some $a$ with $T(a)=b$, then by the formula of $T$ we get the explicit formula of $a$ is $a_n=n^{\frac{1}{4}}$, which is clearly not a bounded sequence. Thus $\{b_n\}$ has no pre-image. – 2017-01-18
1 Answers
4
$T$ is not surjective. Consider the sequence $\{b_n\}$ defined by $b_n=n^{-\frac{3}{4}}$. Clearly $\{b_n\}$ is in $\ell^2$. However if there is some $a$ with $T(a)=b$, then by the formula of $T$ we get the explicit formula of $a$ is $a_n=n^{\frac{1}{4}}$, which is clearly not a bounded sequence. Thus $\{b_n\}$ has no pre-image.
For the uniformly continuity part, observe that $$\|T(a)-T(b)\|_2=\|T(a-b)\|_2\leq\frac{\pi}{\sqrt{6}}\|a-b\|_\infty.$$ A bounded linear operator automatically satisfies the Lipschitz condition, thus is uniformly continuous.