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    I need to show that the operator norm of \ell_{\varphi}:C^0[a,b]\rightarrow\mathbb{R},

    \displaystyle \ell_{\varphi}(f) = \int_a^b \varphi(t)f(t)\;dt

    is

    \displaystyle \|\ell_{\varphi}\|_{\text{op}} = \int_a^b |\varphi(t)|\;dt

    where \varphi \in C^0[a,b].

    It's easy enough to see that the norm's bounded above by this, but I need to find a function f such that this norm is attained. The obvious choice is f(t) = \text{sgn}(\varphi(t)), but this isn't continuous. If \varphi has a finite number of zeros then it can just be approximated by joining the discontinuities in some neighbourhood (-\varepsilon, \varepsilon) of each zero. However, if \varphi is nowhere constant and its zeros are, for example, the Cantor set which is nowhere dense and contains no isolated points, I'm not sure how to construct an approximation.

    Any ideas?
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    The proof involves a trick:

    

\begin{array}{ccl}

\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t  )|^2}{1+n|\phi(t)|}}dt }\\

&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(  t)|}\Big) \\

&\le&\displaystyle{\frac{1}{n}}+  \|\ell_\phi\|.}

\end{array}
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    (Original post by mathprof)
    The proof involves a trick:

    

\begin{array}{ccl}

\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t  )|^2}{1+n|\phi(t)|}}dt }\\

&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(  t)|}\Big) \\

&\le&\displaystyle{\frac{1}{n}}+  \|\ell_\phi\|.}

\end{array}
    Presumably you mean \displaystyle{\ell_\phi \Big(\frac{n\phi(t)}{1+n|\phi(t)  |}\Big)} ?

    That's a really nice method! Is this alright as well?
    Spoiler:
    Show

    Fix  \epsilon>0 . Take  \phi_n simple and uniformly bounded by \sup|\phi| such that  ||\phi_n - \phi ||^{2}_2 \leq \epsilon (simple functions are dense in L^2)
    Take  f_n =sgn(\phi_n) . Let  f be continuous such that  ||f-f_n||^{2}_2 \leq \epsilon and  \sup |f| \leq 1 (continuous functions are dense in L^2)
    Then |(\phi,f)-\int |\phi| | \leq |(\phi_n,f)-(\phi,f)|+|(\phi_n,f)-(\phi_n,f_n)|+|(\phi_n,f_n)-(1,|\phi|)|
    By Cauchy Schwartz the first two terms of the RHS is less than or equal to  D \epsilon where D depends only on  \phi .
    Finally  |(|\phi_n|-|\phi|,1)| \leq ||1|| ||\phi_n|-|\phi|| \leq ||1|| ||\phi-\phi_n|| by the reverse triangle inequality.
    Thus  ||(phi,f)-\int |\phi|| \leq C \epsilon where C depends only on  \phi . The result follows by selecting f so that epsilon is arbitrarily small.
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    Yes, I meant it without the absoute value. I don't see a difference however since both are continuous and have the same value.

    By alright do you mean is it a valid proof? Can you see something wrong with it?
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    (Original post by mathprof)
    Yes, I meant it without the absoute value. I don't see a difference however since both are continuous and have the same value.
    Well if you remove the absolute value they don't have the same value, l_\phi(|\phi|) \neq \l_\phi(\phi) .

    By alright do you mean is it a valid proof? Can you see something wrong with it?
    Erm, I don't know? I am worried about assuming that the dense subsets I'm using can be bounded by the functions they are converging to but other than that I don't see anything wrong... I figured I'd ask? I mean obviously your proof is far simpler but I wanted to check I understood properly...
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    (Original post by mathprof)
    The proof involves a trick:

    

\begin{array}{ccl}

\int_a^b|\phi(t)|dt &=&\displaystyle{ \int_a^b\frac{|\phi(t)|+n|\phi(t  )|^2}{1+n|\phi(t)|}}dt }\\

&=&\displaystyle{ \int_a^b\frac{|\phi(t)|}{1+n| \phi (t)|}dt}+\displaystyle{\ell_\phi \Big(\frac{n|\phi(t)|}{1+n|\phi(  t)|}\Big) \\

&\le&\displaystyle{\frac{1}{n}}+  \|\ell_\phi\|.}

\end{array}
    Wow, I really like this proof, thank you!

    I think IrrationalNumber's right about the thing in the bracket though, as \phi(t)|\phi(t)| \neq \phi(t)^2 in general. I'm guessing the first term on the last line is meant to be (b-a)/n as well? (not that it affects the result in any way)

    (Original post by IrrationalNumber)
    Presumably you mean \displaystyle{\ell_\phi \Big(\frac{n\phi(t)}{1+n|\phi(t)  |}\Big)} ?

    That's a really nice method! Is this alright as well?
    Spoiler:
    Show

    Fix  \epsilon>0 . Take  \phi_n simple and uniformly bounded by \sup|\phi| such that  ||\phi_n - \phi ||^{2}_2 \leq \epsilon (simple functions are dense in L^2)
    Take  f_n =sgn(\phi_n) . Let  f be continuous such that  ||f-f_n||^{2}_2 \leq \epsilon and  \sup |f| \leq 1 (continuous functions are dense in L^2)
    Then |(\phi,f)-\int |\phi| | \leq |(\phi_n,f)-(\phi,f)|+|(\phi_n,f)-(\phi_n,f_n)|+|(\phi_n,f_n)-(1,|\phi|)|
    By Cauchy Schwartz the first two terms of the RHS is less than or equal to  D \epsilon where D depends only on  \phi .
    Finally  |(|\phi_n|-|\phi|,1)| \leq ||1|| ||\phi_n|-|\phi|| \leq ||1|| ||\phi-\phi_n|| by the reverse triangle inequality.
    Thus  ||(phi,f)-\int |\phi|| \leq C \epsilon where C depends only on  \phi . The result follows by selecting f so that epsilon is arbitrarily small.
    That seems to work too, and I'm guessing more along the lines of what they'd expect us to do in an exam. I don't see any problems with boundedness given that \phi is bounded? (continuous and defined on a closed interval)
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    (Original post by IrrationalNumber)
    Well if you remove the absolute value they don't have the same value, l_\phi(|\phi|) \neq \l_\phi(\phi) .
    I beg your pardon, I confused \phi |\phi| with \phi^2 , as has already been pointed out.


    Erm, I don't know? I am worried about assuming that the dense subsets I'm using can be bounded by the functions they are converging to but other than that I don't see anything wrong... I figured I'd ask? I mean obviously your proof is far simpler but I wanted to check I understood properly...
    What are these dense subsets you're referring to? I think the proof is OK. You need to take a limit but that's fine. Don't over-think this.
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    (Original post by .matt)
    I'm guessing the first term on the last line is meant to be (b-a)/n as well? (not that it affects the result in any way)
    Yes, sorry about that.
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    (Original post by mathprof)
    What are these dense subsets you're referring to? I think the proof is OK. You need to take a limit but that's fine. Don't over-think this.
    I was talking about continuous and simple functions being dense. And yeah, I am overthinking it, thanks
 
 
 
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