# Operator norm

1. Dec 9, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data

http://img252.imageshack.us/img252/4844/56494936eo0.png [Broken]

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

3. 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)

Last edited by a moderator: May 3, 2017
2. Dec 9, 2008

### Office_Shredder

Staff Emeritus
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

3. Dec 10, 2008

### dirk_mec1

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?

4. Dec 10, 2008

### Office_Shredder

Staff Emeritus
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.

5. Dec 10, 2008

### dirk_mec1

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.

6. Apr 3, 2011

### mastoras

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

Last edited by a moderator: May 5, 2017