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1. We define \(\displaystyle{T:\mathbb{l}_{2}(\mathbb{N})\longrightarrow \mathbb{l}_{2}(\mathbb{N})}\) by

\(\displaystyle{a=\left(a_{n}\right)_{n\in\mathbb{N}}\mapsto T(a):=\left(0,a_1,a_2,...,a_n,a_{n+1},...\right)}\) .

Prove that the function \(\displaystyle{T}\) is linear, bounded and \(\displaystyle{||T||=1}\) .

2. Let \(\displaystyle{\left(X,||\cdot||_{1}\right)\,,\left(Y,||\cdot||_{2}\right)}\) be \(\displaystyle{\mathbb{R}}\)- \(\displaystyle{\rm{Banach}}\) - spaces

and \(\displaystyle{T:X\longrightarrow Y}\) be a linear map. Prove that the function \(\displaystyle{T}\) is bounded if, and only if,

\(\displaystyle{f\circ T\in X^{\star}\,,\forall\,f\in Y^{\star}}\) .

Hello!Here is a solution for the second excercise:

Suppose \(T:X\rightarrow Y\) is bounded and \(f\in Y^{*}\).It follows that \(T\) is continuous on \(X\),so considering \(\{x_n,x\}\subset X\) such that \(x_n\rightarrow x \Rightarrow Tx_{n}\rightarrow Tx\).Since f is also continuous : \(f(Tx_n)\rightarrow f(Tx)\) so \(f\circ T\) is continuous on \(X\) which implies that it is bounded.

For the converse,suppose that \(f\circ T\in X^{*}\quad ,\forall f\in Y^{*}\) .This implies that the image of a bounded subset of \(X\) under \(f\circ T\) is a bounded subset of \(\mathbb R\quad ,\forall f\in Y^{*}\).Let \(\overline{B_{X}}\) denote the closed unit ball of \(X\).Then \(T(\overline{B_{X}})\) is a weakly bounded subset of \(Y\) therefore \(T\) is a weakly bounded linear operator (since \(\overline{B_{X}}\subset X\) is weakly bounded).We claim that the latter implies that \(T\) is bounded.
To prove this,suppose for contradiction that \(T\) is not bounded.Then \(T(\overline{B_{X}})\) is not bounded,so \(T(\overline{B_{X}})\subseteq Y\) cannot be weakly bounded.It follows that \(f(T(\overline{B_{X}}))\) is not bounded for some \(f\in Y^{*}\) which is a contradiction.

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

\(\displaystyle{(T(a))_{n}=\begin{cases}
\,\,\,0\,\,\,\,\,\,\,\,\,\,,n=1\\
a_{n-1}\,\,\,,n\geq 2
\end{cases}}\)

So,

\(\displaystyle{\sum_{n=1}^{\infty}(T(a))^2_{n}=\sum_{n=2}^{\infty}a_{n-1}^2=\sum_{n=1}^{\infty}a_{n}^2<\infty}\), 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 :

\(\displaystyle{\begin{aligned} ||T(a)||&=\sqrt{\sum_{n=1}^{\infty}(T(a))^2_{n}}\\&=\sqrt{\sum_{n=2}^{\infty}a_{n-1}^2}\\&=\sqrt{\sum_{n=1}^{\infty}a_{n}^2}\\&=||a||\end{aligned}}\)

which means that the function \(\displaystyle{T}\) is isometry. Therefore, the function \(\displaystyle{T}\) is linear, bounded (as isometry)

and

\(\displaystyle{||T||=\left\{||T(a)||\in\mathbb{R}: a\in\mathbb{l}_{2}(\mathbb{N}): ||a||\leq 1\right\}=\left\{||a||\in\mathbb{R}: a\in\mathbb{l}_{2}(\mathbb{N}): ||a||\leq 1\right\}=1}\) .