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Math Help - Prove Lemma on the norms of a function.

  1. #1
    Member Mollier's Avatar
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    Prove Lemma on the norms of a function.

    Hi.

    Problem:

    (i) Suppose that the real-valued weight function w is defined, continuous, positive and integrable on the interval (a,b). Then, for any function f\in C[a,b],

    ||f||_2 \leq W||f||_{\infty},

    where

    W=\bigg[\int^b_aw(x)dx\bigg]^{1/2}.

    (ii) Given any two positive numbers \epsilon (however small) and M (however large), there exists a function f\in C[a,b] such that

    ||f||_2 < \epsilon,

    ||f||_{\infty}>M.

    Attempt:

    (i) By using integration by parts I get,

    \begin{aligned}<br />
\int^b_a|f(x)|^2w(x)dx=&\bigg[|f(x)|^2\int w(x)dx\bigg]^b_a-2\int^b_a|f(x)|f'(x)\bigg(\int w(x)dx\bigg)dx \\<br />
\leq& \bigg[|f(x)|^2\int w(x)dx\bigg]^b_a \\<br />
\leq& \bigg(\max_{x\in [a,b]}|f(x)|\bigg)^2\int^b_a w(x)dx \\<br />
\Rightarrow \bigg(\int^b_a|f(x)|^2w(x)dx\bigg)^{1/2}\leq& \max_{x\in [a,b]}|f(x)| \bigg(\int^b_a w(x)dx\bigg)^{1/2} \\<br />
\Rightarrow ||f||_2 \leq& W||f||_{\infty}<br />
\end{aligned}

    To me that looks good, but I've been mistaken before

    (ii) I do not see how I can prove existence of the function. This somehow reminds me of the dirac delta function, as I can make my function as pointy as I want...

    The book continues by saying that the Lemma indicates that the norms are not equivalent as they would be if this was a finite-dimensional linear space and we were dealing with vectors. That is if ||*||_{\infty} and ||*||_2 are vector norms on \mathbb{R}^n then

    n^{-1/2}||v||_{\infty}\leq ||v||_2 \leq n^{1/2}||v||_{\infty}, \forall v\in\mathbb{R}^n.

    Does this mean that if we had W^{-1}||f||_{\infty} \leq ||f||_2, then we could use the two norms interchangeably? I do not immediately see how the Lemma indicates that this is not so.

    Thanks!
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  2. #2
    A Plied Mathematician
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    No, I don't think your approach to (i) works. You have not assumed that f is differentiable. However, f is a continuous function on a closed interval. What do you know about such functions? Incidentally, I think there's an error in the problem statement. It should have

    \displaystyle W=\left(\int_{a}^{b}w^{2}(x)\,dx\right)^{1/2}.

    As for (ii), for the first inequality, I would experiment with (constant) functions like

    f(x)=\dfrac{\epsilon}{2(b-a)}.

    You may have to take square roots, or squares, or something to get that to work with the 2-norm.

    For (ii) on the second inequality, again, you're dealing with continuous functions on closed intervals. Just think about a constant function. Is there a constant function that might satisfy that inequality?
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  3. #3
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    Quote Originally Posted by Mollier View Post
    (i) Suppose that the real-valued weight function w is defined, continuous, positive and integrable on the interval (a,b). Then, for any function f\in C[a,b],

    ||f||_2 \leq W||f||_{\infty},

    where

    W=\bigg[\int^b_aw(x)dx\bigg]^{1/2}.


    Attempt:

    (i) By using integration by parts I get,

    \begin{aligned}<br />
\int^b_a|f(x)|^2w(x)dx=&\bigg[|f(x)|^2\int w(x)dx\bigg]^b_a-2\int^b_a|f(x)|f'(x)\bigg(\int w(x)dx\bigg)dx \\<br />
\leq& \bigg[|f(x)|^2\int w(x)dx\bigg]^b_a \\<br />
\leq& \bigg(\max_{x\in [a,b]}|f(x)|\bigg)^2\int^b_a w(x)dx \\<br />
\Rightarrow \bigg(\int^b_a|f(x)|^2w(x)dx\bigg)^{1/2}\leq& \max_{x\in [a,b]}|f(x)| \bigg(\int^b_a w(x)dx\bigg)^{1/2} \\<br />
\Rightarrow ||f||_2 \leq& W||f||_{\infty}<br />
\end{aligned}

    To me that looks good, but I've been mistaken before
    You are making this too complicated. All you need is to use the fact that |f(x)|\leqslant\|f\|_\infty for all x in [a,b]. The weighted L_2-norm is then given by

    \displaystyle \|f\|_2^2 = \int^b_a|f(x)|^2w(x)\,dx \leqslant \int^b_a\|f(x)\|_\infty^2w(x)\,dx=\|f(x)\|_\infty^  2W^2.

    Now take square roots.

    Quote Originally Posted by Mollier View Post
    (ii) Given any two positive numbers \epsilon (however small) and M (however large), there exists a function f\in C[a,b] such that

    ||f||_2 < \epsilon,

    ||f||_{\infty}>M.

    ...

    (ii) I do not see how I can prove existence of the function. This somehow reminds me of the dirac delta function, as I can make my function as pointy as I want...
    That is correct. It is easiest to construct such a function on the unit interval, for example f(x) = n^kx^n, where n is large, and n^k is some suitable power of n, probably n^{1/2} or something like that. Once you have found a function that works on the unit interval, you can translate it to the interval [a,b], to get f(x) = n^k\bigl(\frac{x-a}{b-a}\bigr)^n
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  4. #4
    Member Mollier's Avatar
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    I do tend to overcomplicate, which I guess is fine if the result is correct...
    Thanks.
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