It is currently Tue Sep 25, 2018 8:17 am

 All times are UTC [ DST ]

 Page 1 of 1 [ 3 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: Functional analysis - ExercisesPosted: Thu Jul 14, 2016 7:20 pm
 Team Member

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,

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

Top

 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.

Top

 Post subject: Re: Functional analysis - ExercisesPosted: Thu Jul 14, 2016 7:21 pm
 Team Member

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

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

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 3 posts ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta