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 Post subject: Functional analysis - ExercisesPosted: Thu Jul 14, 2016 7:20 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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,

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

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 Post subject: Re: Functional analysis - ExercisesPosted: Thu Jul 14, 2016 7:20 pm

Joined: Mon Jan 18, 2016 4:33 pm
Posts: 4
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.

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 Post subject: Re: Functional analysis - ExercisesPosted: Thu Jul 14, 2016 7:21 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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

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

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