Continuous map from $l^{\infty}$ to $l^{2}.$ 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.
 A: 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.
A: 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
