How Do Banach Spaces and Bounded Linear Operators Interact in Homework Problems?

[dirk_mec1]

Homework Statement

http://img252.imageshack.us/img252/4844/56494936eo0.png

2. relevant equations
BL = bounded linear space (or all operators which are bounded).

The Attempt at a Solution

I got for the first part:

||A||_{BL} =||tf(t)||_{\infty} \leq ||f||_{\infty} so

||A||_{BL} \leq 1

In second part I have to proof that: ||A||_{BL} \geq 1 and thus the conclusion can be drawn.

But I'm stuck at how to use the fact that E is banach. (At least that's the one thing I haven't used yet so I presume I need it for the second part)

 

[Office_Shredder]

The general strategy for this is to prove that ||A|| is bounded below the proposed value, then find a single example where the norm is greater than or equal to the proposed value.

For example, if the map was f(t) -> 2f(t) we could show ||A||>=2 by citing the example f=1... |A(1)| = |2| = |2|*|1| and by definition this gives (assuming we've shown A is bounded) ||A||>=2.

Usually it's just guesswork to find this and doesn't require extra information. In this case try f = t

 

[dirk_mec1]

The general strategy for this is to prove that ||A|| is bounded below the proposed value, then find a single example where the norm is greater than or equal to the proposed value.

For example, if the map was f(t) -> 2f(t) we could show ||A||>=2 by citing the example f=1... |A(1)| = |2| = |2|*|1| and by definition this gives (assuming we've shown A is bounded) ||A||>=2.

Usually it's just guesswork to find this and doesn't require extra information. In this case try f = t

Yes that is clear but I'm worried that I might have to use the fact that E is banach. By the way you choose f(t)=t but why can't I choose for example f(t)=1 it is continuous and the supremum norm is one. Aren't there a lot possibilities for choosing this function?

 

[Office_Shredder]

There are a lot of possibilities for the function you pick...

The fact that E is Banach is unnecessary, but keep in mind though that you DID in fact use it when you chose not to prove that you're working over a normed vector space.

 

[dirk_mec1]

There are a lot of possibilities for the function you pick...

The fact that E is Banach is unnecessary, but keep in mind though that you DID in fact use it when you chose not to prove that you're working over a normed vector space.

If that's the case it would be unneccessary to put the exclamation sign there but my instructor DID so I'm probably missing something.

 

[mastoras]

Homework Statement

http://img252.imageshack.us/img252/4844/56494936eo0.png

2. relevant equations
BL = bounded linear space (or all operators which are bounded).

The Attempt at a Solution

I got for the first part:

||A||_{BL} =||tf(t)||_{\infty} \leq ||f||_{\infty} so

||A||_{BL} \leq 1

In second part I have to proof that: ||A||_{BL} \geq 1 and thus the conclusion can be drawn.

But I'm stuck at how to use the fact that E is banach. (At least that's the one thing I haven't used yet so I presume I need it for the second part)

what u write is wrong

So,

||A||_{BL} =sup{||A(f)||_{\infty} ,for ||f||_{\infty} \leq 1

=sup{||tf(t)||_{\infty} ,for ||f||_{\infty} \leq 1

=sup{|t|||f(t)||_{\infty} ,for ||f||_{\infty} \leq 1

\leq sup{|t|||f|_{\infty}|t| ,for ||f||_{\infty} \leq 1

\leq 1

Now consider f^*(t)=-t+1 for all t in [0,1] => ||f^*||_{\infty}=1

notice that f^*(t) is continuous in [0,1] and f(1)=0

also

1=||A(f^*)||_{\infty} \leq ||A||_{BL}||f^*||_{\infty}=||A||_{BL}


enjoy