Thank you Gigaster.
1. The function \(\displaystyle{T}\) is well defined. Indeed, let \(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\in\mathbb{l}_{2}(\mathbb{N})}\) .
The image \(\displaystyle{T(a)}\) of \(\displaystyle{a}\), is a real sequence with
\(\displaystyle{(T(a))_{n}=\begin{cases} \,\,\,0\,\,\,\,\,\,\,\,\,\,,n=1\\ a_{n1}\,\,\,,n\geq 2 \end{cases}}\)
So,
\(\displaystyle{\sum_{n=1}^{\infty}(T(a))^2_{n}=\sum_{n=2}^{\infty}a_{n1}^2=\sum_{n=1}^{\infty}a_{n}^2<\infty}\), since
\(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\in\mathbb{l}_{2}(\mathbb{N})}\) and thus
\(\displaystyle{T(a)\in\mathbb{l}_{2}(\mathbb{N})}\) .
Let \(\displaystyle{x=\left(x_{n}\right)_{n\in\mathbb{N}}\,,y=\left(y_{n}\right)_{n\in\mathbb{N}}\in\mathbb{l}_{2}(\mathbb{N})}\)
and \(\displaystyle{a\,,b\in\mathbb{R}}\) . Then,
\(\displaystyle{\begin{aligned}\left(T(a\,x+b\,y)\right)_{n\in\mathbb{N}}&=\left(0,a\,x_1+b\,y_1,a\,x_2+b\,y_2,...,a\,x_{n}+b\,y_{n},...\right)\\&=\left(0,a\,x_1,a\,x_2,...,a\,x_n,...\right)+\left(0,b\,y_1,b\,y_2,...,b\,y_n,...\right)\\&=a\,\left(0,x_1,x_2,...,x_n,...\right)+b\,\left(0,y_1,y_2,...,y_{n},...\right)\\&=\left(a\,T(x)+b\,T(y)\right)_{n\in\mathbb{N}}\end{aligned}}\)
which means that the fucntion \(\displaystyle{T}\) is a linear map.
Now, for each sequence \(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\in\mathbb{l}_{2}(\mathbb{N})}\) holds :
\(\displaystyle{\begin{aligned} T(a)&=\sqrt{\sum_{n=1}^{\infty}(T(a))^2_{n}}\\&=\sqrt{\sum_{n=2}^{\infty}a_{n1}^2}\\&=\sqrt{\sum_{n=1}^{\infty}a_{n}^2}\\&=a\end{aligned}}\)
which means that the function \(\displaystyle{T}\) is isometry. Therefore, the function \(\displaystyle{T}\) is linear, bounded (as isometry)
and
\(\displaystyle{T=\left\{T(a)\in\mathbb{R}: a\in\mathbb{l}_{2}(\mathbb{N}): a\leq 1\right\}=\left\{a\in\mathbb{R}: a\in\mathbb{l}_{2}(\mathbb{N}): a\leq 1\right\}=1}\) .
