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Let $l^{\infty}=\{(a_{n}):a_{n}\in\mathbb{C},sup_{n}|a_{n}|=\|a_{n}\|_{\infty}<\infty\}$ and $l^{2}=\{(a_{n}):a_{n}\in\mathbb{C},(\sum|a_{n}|^{2})^{1/2}=\|a_{n}\|_{2}\}$. Define a map $T:l^{\infty}\rightarrow l^{2}$ as $$T(a_{n})=\{a_{1},\frac{a_{2}}{2},\frac{a_{3}}{3},\cdot\cdot\cdot\}$$. Which of the following is true?

$A.$ $T$ is a continuous map.

$B.$ $T$ is an onto map.

$C.$ $T^{-1}$ exist and is continuous.

$D.$ $T$ is uniformly continuous.

According to me we have $\|T(a_{n})\|\leq (\pi/\sqrt{2})\|a_{n}\|.$ So $A$ and $D$ are true. What about option $B$ and $C$? Please help me. Thanks.

neelkanth
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2 Answers2

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What about $b_n=\dfrac{1}{n^{\frac{3}{4}}}$ ($(b_n)\in \ell^2$). Is there a sequence $(a_n)\in \ell^{\infty}$ such that $T((a_n))=(b_n)$?

The image is $\Im(T)=\{(b_n)_n\in \ell^2\ \ /\ (nb_n)_n\ \ \text{ is bounded }\}$

$T^{-1}:\Im(T)\to \ell^{\infty}$ is not continiouse. Let $B_n=(0,0,\ldots,0,\frac{1}{n},\frac{1}{n+1},\ldots )$ i.e $B_{n,k}=0$ if $k<n$ and $B_{n,k}=\frac{1}{k}$ if $k\geq n$. ($(B_n)_n$ is a sequence of elements of $\ell^2$).

We have $\|B_n\|_2^2=\displaystyle\sum_{k=n}^{\infty}\dfrac{1}{k^2}\to 0$ (as $n\to +\infty$).

But $T^{-1}(B_n)=(0,\ldots,0,1,1,1,1\ldots)$ and $\|T^{-1}(B_n)\|_{\infty}=1$ do not converge to $0$. Hence $T^{-1}$ is not continuous.

Hamou
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  • okkk what about existence of inverse and its continuity??? – neelkanth Dec 19 '16 at 11:22
  • if $T$ is not onto, how do you define its inverse? – m-agag2016 Dec 19 '16 at 11:27
  • i mean on its image....... – neelkanth Dec 19 '16 at 11:28
  • Fir its image $\Im(T)={(b_n)_n\in \ell^2\ \ /\ (nb_n)_n\ \ \text{ is bounded }}$. – Hamou Dec 19 '16 at 11:34
  • To neelkanth : Any linear map $f$ from one normed linear space to another is continuous iff it is bounded, that is,iff $\sup {|f(x)|/|x| : x\ne 0}<\infty.$ Another example to show that $T^{-1}$ is not bounded on its image is to let $x_{k,n}=0$ if $k\ne n$ and $x_{k,k}=1$, and let $v_k=(x_{k,n})n.$ Then $|v_k|_2=1$ but $ |T^{-1}v_k|{\infty}=k.$ – DanielWainfleet Dec 19 '16 at 13:37
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I think the easier way to exclude the option B) is ,as we know in nls the continuity of a linear operator and boundedness are same and it is clear the inverse map is not bounded.Now if it map T is onto then it is a continuous surjective map between Banach spaces and so by open mapping theorem it is an open map i.e T inverse is continuous but T inverse is not bounded hence can't be continuous so (B) can't be true

user3333
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