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Calculating the norm of linear, continuous operator

  1. Nov 22, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##X=\{f \in C[0,1]: f(1)=0\}## with the ##\|x\|_{\infty}## norm. Let ##\phi \in X## and let ##T_{\phi}:X \to X## given by

    ##T_{\phi}f(x)=f(x)\phi(x)##.
    Prove that ##T## is a linear continuous operator and calculate its norm.

    The attempt at a solution.

    To check for linearity is pretty simple, for continuity I am not so sure if my proof is correct so I write it just in case:

    ##T## is continuous if and only if ##T## is bounded. So, I want to prove that there exists ##c>0: \forall f \in X##, ##\|T(f)\|_{\infty}\leq c\|f\|_{\infty}##. ##\|T(f)\|_{\infty}=sup_{x \in [0,1]}|f(x)\phi(x)|\leq sup_{x \in [0,1]}|f(x)|sup_{x \in [0,1]}|\phi(x)|##. As ##[0,1]## is compact, ##sup_{x \in [0,1]}|\phi(x)|=max_{x \in [0,1]}|\phi(x)|##, if I call ##c=max_{x \in [0,1]}|\phi(x)|##, then ##sup_{x \in [0,1]}|f(x)\phi(x)|\leq csup_{x \in [0,1]}|f(x)|## for every ##f \in X##. It follows that ##T## is bounded ##\implies## ##T## is continuous.

    Now, I had problems to calculate the norm of this operator. One of the definition for the norm in the space of linear, continuous operators is ##||T||=inf\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}## (among other equivalent definitions). Sorry if I say something trivially false but I think that ##c=max_{x \in [0,1]}|\phi(x)|## is the norm. I have to prove that ##c## is a lower bound of the set ##\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}## and that it is the upper lower bound.

    Here's a tentative solutions after the suggestions:
    Lets show that ##c=max_{x \in [0,1]}|\phi(x)|## is a lower bound of the set ##A=\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}##. Let ##M \in A##, then ##\|T(\phi)\|_{\infty}\leq M \|\phi\|_{\infty}##. So ##c^2=sup_{x \in [0,1]}|\phi(x)\phi(x)| \leq Mc \implies c\leq M##. This means that ##c## is a lower bound of ##A##, but ##sup_{x \in [0,1]}|f(x)\phi(x)|\leq csup_{x \in [0,1]}|f(x)|## for every ##f \in X##, so ##c \in A##. It follows that ##||T||=inf\{M: \|T(f)\|_{\infty}\leq M\|f\|_{\infty}\}=c##.
     
    Last edited: Nov 22, 2013
  2. jcsd
  3. Nov 22, 2013 #2

    LCKurtz

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    Look at the norm of ##T_\phi (\phi)##.
     
  4. Nov 22, 2013 #3

    Dick

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    I think your proof has a lot of the right ideas in it. But it's really unclear. And it does contain false statements. ##sup |f \phi| \le sup |f| sup |\phi|##. They aren't equal. And X isn't compact. [0,1] is compact. And LCKurtz has a great hint. And I don't think f(1)=0 has much to do with this part of the exercise. Can you rearrange the whole thing and try again?
     
  5. Nov 22, 2013 #4
    It really helped that hint, thanks!
     
  6. Nov 22, 2013 #5
    I meant ##[0,1]##, no idea why I wrote ##X##. If the elements of two sets ##A## and ##B## consist of non-negative elements, are you sure that ##sup AB \leq sup A sup B## and not always equal? I'll think about a counterexample to convince myself. In the last part of the exercise I've used that if the elements of ##A## are non-negative numbers, then ##supAA=supAsupA##.
     
  7. Nov 23, 2013 #6

    Dick

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    I'm absolutely sure. Try f(x)=1-x and g(x)=(1-x)*x. And you are dealing with sup|f|, it doesn't much matter whether anything is nonnegative or not.
     
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