It is currently Mon Sep 24, 2018 10:48 am


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Integral operator
PostPosted: Fri Nov 13, 2015 5:53 pm 
Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Consider the real normed space \(\displaystyle{\left(C\,(\left[0,1\right]),||\cdot||_{\infty}\right)}\) and the map

\(\displaystyle{T:C\,(\left[0,1\right])\longrightarrow C\,(\left[0,1\right])\,,f\mapsto T(f):\left[0,1\right]\longrightarrow \mathbb{R}\,,T(f)(t)=\int_{0}^{t}f(s)\,\mathrm{d}s}\) .

A. Prove that the map \(\displaystyle{T}\) is \(\displaystyle{\mathbb{R}}\) - linear, bounded and one to one.

B. Find \(\displaystyle{T\,\left(C\,(\left[0,1\right]\right)\subseteq C(\left[0,1\right])}\) .

C. Is the operator \(\displaystyle{T^{-1}}\) bounded ?

D. Find \(\displaystyle{||T||=\sup\,\left\{||T(f)||_{\infty}\geq 0: f\in C\,(\left[0,1\right])\,,||f||_{\infty}\leq 1\right\}}\) .


Top
Offline Profile  
Reply with quote  

 Post subject: Re: Integral operator
PostPosted: Wed Jan 20, 2016 7:59 pm 
Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
We give a solution.

Α. If \(\displaystyle{f\,,g\in C(\left[0,1\right])}\) and \(\displaystyle{a\,,b\in\mathbb{R}}\) ,

then for each \(\displaystyle{t\in\left[0,1\right]}\) holds :

\(\displaystyle{\begin{aligned} T(a\,f+b\,g)(t)&=\int_{0}^{t}(a\,f+b\,g)(s)\,\mathrm{d}s\\&=\int_{0}^{t}\left(a\,f(s)+b\,g(s)\right)\,\mathrm{d}s\\&=a\,\int_{0}^{t}f(s)\,\mathrm{d}s+b\,\int_{0}^{t}g(s)\,\mathrm{d}s\\&=\left(a\,T(f)+b\,T(g)\right)(t)\end{aligned}}\)

so, \(\displaystyle{T(a\,f+b\,g)=a\,T(f)+b\,T(g)}\), that is \(\displaystyle{T}\) is \(\displaystyle{\mathbb{R}}\) - linear .

Also, for each \(\displaystyle{t\in\left[0,1\right]}\) holds :

\(\displaystyle{\begin{aligned} \left|T(f)(t)\right|&=\left|\int_{0}^{t}f(s)\,\mathrm{d}s\right|\\&\leq \int_{0}^{t}\left|f(s)\right|\,\mathrm{d}s\\&\leq \int_{0}^{t}\sup\,\left\{\left|f(s)\right|:0\leq s\leq 1\right\}\,\mathrm{d}s\\&=t\,||f||_{\infty}\\&\leq ||f||_{\infty}\end{aligned}}\) .

Therefore, \(\displaystyle{||T(f)||_{\infty}=\sup\,\left\{\left|T(f)(t)\right|: 0\leq t\leq 1\right\}\leq ||f||_{\infty}}\)

and \(\displaystyle{T}\) is bounded. Thus, \(\displaystyle{||T||\leq 1}\) .

Let now \(\displaystyle{f\,,g\in C(\left[0,1\right])}\) such that \(\displaystyle{T(f)=T(g)}\). Then,

\(\displaystyle{\forall\,t\in\left[0,1\right]: \int_{0}^{t}f(s)\,\mathrm{d}s=\int_{0}^{t}g(s)\,\mathrm{d}s}\)

and by differentiating the above relation with respect to \(\displaystyle{t}\), we get :

\(\displaystyle{f(t)=g(t)\,,\forall\,t\in\left[0,1\right]\implies f=g}\) .

So, the operator \(\displaystyle{T}\) is one to one.

B. If \(\displaystyle{g\in T(C(\left[0,1\right]))}\), then \(\displaystyle{g=T(f)}\) for some

\(\displaystyle{f\in C(\left[0,1\right])}\). Now, \(\displaystyle{g\in C^1(\left[0,1\right]}\) and

\(\displaystyle{g(0)=T(f)(0)=0}\) .

Conversely, if \(\displaystyle{g\in C^1(\left[0,1\right])}\) with \(\displaystyle{g(0)=0}\), then

\(\displaystyle{g^\prime\in C(\left[0,1\right]}\) and

\(\displaystyle{T(g^\prime)(t)=\int_{0}^{t}g^\prime(s)\,\mathrm{d}s=g(t)-g(0)=g(t)\implies T(g^\prime)=g\implies g\in T(C(\left[0,1\right])}\).

Therefore,

\(\displaystyle{T(C(\left[0,1\right]))=\left\{g\in C^1(\left[0,1\right]): g(0)=0\right\}\leq C(\left[0,1\right])}\) .

C. We have that

\(\displaystyle{T^{-1}: T(C(\left[0,1\right])\longrightarrow C(\left[0,1\right])\,,T(g)=g^\prime}\).

Let \(\displaystyle{g_{n}:\left[0,1\right]\longrightarrow \mathbb{R}\,,g_{n}(x)=x^n\,,n\in\mathbb{N}}\).

Then, \(\displaystyle{g_{n}\in T(C(\left[0,1\right]))}\) and

obviously, \(\displaystyle{||g_{n}||_{\infty}=1\,,\forall\,n\in\mathbb{N}}\) . But,

\(\displaystyle{T^{-1}(g_{n})(t)=g_{n}^\prime(t)=n\,t^{n-1}\,,0\leq t\leq 1\,,n\in\mathbb{N}}\)

and \(\displaystyle{||T^{-1}(g_{n})||_{\infty}=n}\).

So, if \(\displaystyle{T^{-1}}\) is bounded, then

\(\displaystyle{\left(\exists\,c>0\right)\,\left(\forall\,g\in T(C(\left[0,1\right]))\right),||T^{-1}(g)||_{\infty}\leq c\,||g||_{\infty}}\)

Especially,

\(\displaystyle{\forall\,n\in\mathbb{N}: n\leq c}\), a contradiction.

D. According to A. , we have that \(\displaystyle{||T||\leq 1}\) .

Let \(\displaystyle{f(t)=1\,,t\in\left[0,1\right]}\) with \(\displaystyle{||f||_{\infty}=1}\) . Then,

\(\displaystyle{\begin{aligned} ||T(f)||_{\infty}&=\sup\,\left\{\left|T(f)(t)\right|: 0\leq t\leq 1\right\}\\&=\sup\,\left\{\left|\int_{0}^{t}\,\mathrm{d}s\right|: 0\leq t\leq 1\right\}\\&=\sup\,\left\{t: 0\leq t\leq 1\right\}\\&=1\end{aligned}}\)

and therefore, \(\displaystyle{||T||\geq ||T(f)||_{\infty}=1}\).

Finally, \(\displaystyle{||T||=1}\) .


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 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 forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net