MHB Norm of a linear integral operator

[sarrah1]

i have a problem concerning the norm of linear integral operator. ii found the answer in a book called unbounded linear operators theory and applications by Dover books author seymour Goldberg. the proof runs as follows ||T|| is less than max over x in [a,b] of integral (|k(x,y)|dy) Then he proved that ||T|| is greater than it, thus ||T|| is equal to it. The proof was using sign function. Yet I couldn't understand it. can you kindly explain it to me how he showed that it is greater.

 

[Opalg]

i have a problem concerning the norm of linear integral operator. ii found the answer in a book called unbounded linear operators theory and applications by Dover books author seymour Goldberg. the proof runs as follows ||T|| is less than max over x in [a,b] of integral (|k(x,y)|dy) Then he proved that ||T|| is greater than it, thus ||T|| is equal to it. The proof was using sign function. Yet I couldn't understand it. can you kindly explain it to me how he showed that it is greater.

You need to supply some more information. For a start, you have not defined the operator $T$. I assume that it acts on some space of functions by a formula of the form $$(Tf)(x) = \int_a^b \!\!\!k(x,y)f(y)\,dy$$? Then you need to say what this space of functions consists of (continuous functions on the interval $[a,b]$? integrable functions? square-integrable functions? ...), and what the norm on that space is. Finally, what do you know about the kernel function $k$? Again, it could be continuous, integrable or square-integrable.

 

[sarrah1]

thank you oplag

the operator K is defined as follows: (Ky)(x)=\int k(x,s)y(s)ds. over [a,b). The problem I can't write in Latex or whatever that is. k(x,s) is continuous on the square a<x,s<b so one can easily see that K is bounded and ||K||< max over(a,b) of \int |k(x,s)|ds. To show that ||K|| is equal to the latter is to show that it is greater than the latter. So this book does so.
I don't want to be pushy but really by e-mail: <email addy removed by admin> can facilitate the matter for me as I can attach a word file. Life is hard for me here, I am ignorant of the script
thanks

 

[MarkFL]

thank you oplag

the operator K is defined as follows: (Ky)(x)=\int k(x,s)y(s)ds. over [a,b). The problem I can't write in Latex or whatever that is. k(x,s) is continuous on the square a<x,s<b so one can easily see that K is bounded and ||K||< max over(a,b) of \int |k(x,s)|ds. To show that ||K|| is equal to the latter is to show that it is greater than the latter. So this book does so.
I don't want to be pushy but really by e-mail: <email addy removed by admin> can facilitate the matter for me as I can attach a word file. Life is hard for me here, I am ignorant of the script
thanks

Hello Sarrah,

The vast majority of people helping on forums do not want to correspond by email (also it's a bad idea to make your addy public, and so I have removed it from your post for your safety), and download attachments meant for programs to which they may not have access, etc. Not to be blunt, but those asking for help should be the ones making the effort to make things easier for those giving help, wouldn't you agree?

There are lots of online tutorials for learning to use $\LaTeX$, and if I may say so myself, MHB has one of the best set of tools to aid in the creation of $\LaTeX$ expressions you will find anywhere. Commonly used symbols and commands can be found on our Quick $\LaTeX$ element and you can preview them live in our $\LaTeX$ Live Preview element.

 

[I like Serena]

The problem I can't write in Latex or whatever that is... I am ignorant of the script

Hi sarrah,

Just put (for instance) dollars around your formulas. That's all there is too it.
Then instead of (Ky)(x)=\int k(x,s)y(s)ds, you'll get $(Ky)(x)=\int k(x,s)y(s)ds$.

 

[sarrah1]

this is for rehearsal only. trying to write and see the result
(Kx)(t)=\int_{k(s,t}^{a,b} \,ds

 

[I like Serena]

But... you didn't put dollars (\$) around it.

 

[Opalg]

this is for rehearsal only. trying to write and see the result
(Kx)(t)=\int_{k(s,t}^{a,b} \,ds

To underline ILS's comment, if you type $(Kx)(t)=\int_{k(s,t}^{a,b} \,ds$, then it will appear as $(Kx)(t)=\int_{k(s,t}^{a,b} \,ds$ when you Preview or Submit it. If you just want to experiment with using LaTeX, you don't need to use the Submit button at all. Just use Preview until you have finished experimenting, then abandon it.

the operator K is defined as follows: (Ky)(x)=\int k(x,s)y(s)ds. over [a,b). The problem I can't write in Latex or whatever that is. k(x,s) is continuous on the square a<x,s<b so one can easily see that K is bounded and ||K||< max over(a,b) of \int |k(x,s)|ds. To show that ||K|| is equal to the latter is to show that it is greater than the latter.

There is still one essential piece of information missing. You want to find $\|K\|$, which is defined as $$\sup_{\|x\| \leqslant1}\|Kx\|,$$ the supremum of $\|Kx\|$ as $x$ runs through the unit ball of some normed space of functions. But you haven't said what that space is, and (most important of all) you haven't said what norm you are using on that space. Until we know that information, it's not possible to offer any help with the problem.

 

[MarkFL]

Another $\LaTeX$ tip:

To practice, you can enter your expression into our $\LaTeX$ Live Preview to see how it is going to render in your post once submitted. Click one of the tags buttons (the dollar signs or the $\Sigma$ button), then put your code in between them, and you will instantly see a preview:

View attachment 4204

Once you are satisfied, you can then copy the code to your post at the current cursor location by clicking the "Copy to Post" button. If you are just practicing, you can enter other expressions and then once done, just abandon the post. :D

 

[sarrah1]

thank you very much.
the integral operator defines a set of completely continuous linear operators in the Banach space C[a,b] if the kernel k(x,t), is bounded in the square a<x,t<b.
What I am looking for is the norm in L(infinity).
I know it is equal to max(a<x<b)\ of Int |k(x,t)|dt. the proof that it is less than the latter is direct, but why it is greater so that it's equal
thanks

 

[Opalg]

thank you very much.
the integral operator defines a set of completely continuous linear operators in the Banach space C[a,b] if the kernel k(x,t), is bounded in the square a<x,t<b.
What I am looking for is the norm in L(infinity).
I know it is equal to max(a<x<b)\ of Int |k(x,t)|dt. the proof that it is less than the latter is direct, but why it is greater so that it's equal
thanks

So you know that $$(Kf)(x) = \int_a^b\!\!\!k(x,t)f(t)\,dt$$, and that if $\|f\|_\infty = 1$ then $$\|K\| \geqslant \|Kf\|_\infty = \max_{a\leqslant x\leqslant b}|(Kf)(x)|$$ What happens if you take $f$ to be the constant function $1$? Having done that, can you see what will happen (for a fixed value of $x$) if you take $f(t)$ to be a continuous function that is equal to $1$ when $k(x,t)$ is positive and $-1$ when $k(x,t)$ is negative. Of course, such a function is unlikely to be continuous, so you may have to approximate it by a continuous function that switches rapidly between $-1$ and $+1$ when necessary.

 

[sarrah1]

very grateful and sorry for the enough trouble I caused you although I couldn't understand a word of what you wrote. I am a numerical analyst not versed in functional analysis. I was preparing a paper on the solution of integral equations, so I wrote a few lines on the norm of the kernel involved. The referee asked me to define it in a compact form based upon sup||Kx||/||x||. So I refereed to a book and changed a bit the few lines I wrote. I just need to know if what I wrote is compatible and suitable for the referee to accept it.
Unless I can use a word file which I can't upload here, any help although appreciated won't be of any use. can i upload a word file ?
thanks once again
sarrah

 

[MarkFL]

You can attach files with the extension "doc" as long as the file is not more than 20KB in size. :D