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Math Help - Dual space isometry

  1. #1
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    Dual space isometry




    Is the above proof correct?


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  2. #2
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    Quote Originally Posted by forumbug View Post

    Is the above proof correct?
    As far as I can say, yes.
    Quote Originally Posted by forumbug View Post

    You have
    \left|\int^1_0 g(t)h(t) dt\right| \le \int^1_0 |g(t)| |h(t)| dt \le \int^1_0 |g(t)| dt
    therefore
    ||\varphi_g|| \le \int^1_0 |g(t)| dt .

    I think that ||\varphi_g|| = \int^1_0 |g(t)| dt holds. A naive approach to show this would be trying to approximate the step function h(t)=g(t)/|g(t)| by continuous functions; but I guess there is a simpler solution.

    BTW the same question was posted here Dual space isometry
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  3. #3
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    Is this correct?

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  4. #4
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    Quote Originally Posted by forumbug View Post
    Is this correct?
    It would be much easier to answer, if you posted everything in latex format instead of images. (You obviously latexed this.)

    Do you see that in your answer to question 1 you replaced supremum of some set with supremum of one-element set. Doesn't this seem suspicious to you?

    You did not write what W means, but I guess W=\{\varphi_g; g\in X\}.
    W.l.o.g. you can assume f_2=0. (Do you know why?)
    Now if a continuous function is non-zero, then there exists x such that f(x)\ne 0.
    But then there exists \varepsilon>0 such that f does not change sign on (x-\varepsilon,x+\varepsilon). Can you use this facts (or similar ideas) to find a function g such that \varphi_g(f)\ne0?

    I think that your answer to the 3rd problem is correct. Another possibility is evaluation map like f\mapsto f(1/2).

    You get \varphi_g(x_n)\to 0 whenever \int_0^1 x_n(t) dt converges to 0. So you can for instance choose
    <br />
x_n(t)=\begin{cases}<br />
1-nx,&x\le1/n,\\<br />
0,&\text{otherwise}<br />
\end{cases}<br />

    Again, the same question was posted here: Dual space isometry
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  5. #5
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    Quote Originally Posted by kompik View Post
    Do you see that in your answer to question 1 you replaced supremum of some set with supremum of one-element set. Doesn't this seem suspicious to you?
    Now, when I think about it, you perhaps meant \exists t\in[0,1] |h(t)|=1 instead of \forall t\in[0,1] h(t)=1
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  6. #6
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    Thank you for your replies. I have been trying to finish off the question.

    I tried latex the question in the forum but couldn't write multi-line equations using
    Code:
    \begin{align*}
    \end{align*}
    For part 1, I meant that the second last step is true because the function h(t) = 1 \forall<br />
t \in [0,1] is in X = C[0,1] and since all other h satisfying
    \displaystyle \sup_{t \in [0,1]} |h| = 1 are bounded above by
    h=1, \displaystyle \int^1_0 g(t) h(t) dt where \displaystyle<br />
\sup_{t \in [0,1]} \abs{h} = 1 and h(t) \ne 1 is smaller than
    \displaystyle \int^1_0 g(t) h(t) dt where \displaystyle \sup_{t<br />
\in [0,1]} |h| = 1 and h(t) = 1. Hence the supremum is
    unchanged by limiting the set to having just h(t) = 1.

    Is this correct?

    For part 2, thanks for your clue, I put down the following.

    Since f_1 and f_2 are both continuous, if f_1 \ne f_2 at
    some point t_0 \in [0,1], then there exists \delta > 0 such that
    f_1(t) - f_2(t) > 0 \forall t \in [t_0 - \delta, t_0 + \delta].
    Then any g \in X such that g(t) \ne 0 \forall t \in [t_0 -<br />
\delta, t_0 + \delta] will ensure \varphi_g(f_1) \ne<br />
\varphi_g(f_2).

    For part 4, I am not sure if the question meant ||\cdot||_\infty to be supremum norm or essential supremum norm. For essential supremum norm, I don't think such an example exists.

    Thanks a lot for helping me get through this question.
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  7. #7
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    Quote Originally Posted by forumbug View Post
    For part 1, I meant that the second last step is true because the function h(t) = 1 \forall<br />
t \in [0,1] is in X = C[0,1] and since all other h satisfying
    \sup_{t \in [0,1]} |h| = 1 are bounded above by
    h=1, \displaystyle \int^1_0 g(t) h(t) dt where \displaystyle<br />
\sup_{t \in [0,1]} \abs{h} = 1 and h(t) \ne 1 is smaller than
    \displaystyle \int^1_0 g(t) h(t) dt where \displaystyle \sup_{t<br />
\in [0,1]} |h| = 1 and h(t) = 1. Hence the supremum is
    unchanged by limiting the set to having just h(t) = 1.

    Is this correct?
    Well, you're right in claiming that \int^1_0 g(t) h(t) dt \ le \int^1_0 g(t) dt.
    But you need to estimate \left|\int^1_0 g(t) h(t) dt\right|, which might be completely different if g(t).h(t) is not non-negative.

    Quote Originally Posted by forumbug View Post
    For part 2, thanks for your clue, I put down the following.

    Since f_1 and f_2 are both continuous, if f_1 \ne f_2 at
    some point t_0 \in [0,1], then there exists \delta > 0 such that
    f_1(t) - f_2(t) > 0 \forall t \in [t_0 - \delta, t_0 + \delta].
    Then any g \in X such that g(t) \ne 0 \forall t \in [t_0 -<br />
\delta, t_0 + \delta] will ensure \varphi_g(f_1) \ne<br />
\varphi_g(f_2).
    Seems ok to me. You should also notice that you have used the fact that g does not change the sign on that interval.
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  8. #8
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    Quote Originally Posted by kompik View Post
    Well, you're right in claiming that \int^1_0 g(t) h(t) dt \ le \int^1_0 g(t) dt.
    But you need to estimate \left|\int^1_0 g(t) h(t) dt\right|, which might be completely different if g(t).h(t) is not non-negative.
    Should I consider h(t) = \pm 1 and hence get \sup \left\{|\int^1_0 g(t) dt|, |\int^1_0 - g(t) dt|<br />
\right\}?

    Quote Originally Posted by kompik View Post
    Seems ok to me. You should also notice that you have used the fact that g does not change the sign on that interval.
    Thanks, got it.
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  9. #9
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    Quote Originally Posted by forumbug View Post
    Should I consider h(t) = \pm 1 and hence get \sup \left\{|\int^1_0 g(t) dt|, |\int^1_0 - g(t) dt|<br />
\right\}?
    I do not think this is correct.
    Notice that for g(t)=\sin2\pi t then \int_0^1 g(t)h(t) dt=0 for h(t)=\pm1, but there exist function such that \int_0^1 g(t)h(t) dt \ne 0.
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  10. #10
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    Quote Originally Posted by kompik View Post
    I do not think this is correct.
    Notice that for g(t)=\sin2\pi t then \int_0^1 g(t)h(t) dt=0 for h(t)=\pm1, but there exist function such that \int_0^1 g(t)h(t) dt \ne 0.
    I don't quite understand this. I will come back to this post when I find out the solution.
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