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Math Help - Mapping

  1. #1
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    Mapping

    Good evening to all,

    For c>0, I am considering the mapping:

    D_{c}:C_{c}(\mathbb{R}) \to C_{c}(\mathbb{R}), (D_{c}f)(x)=\frac{1}{\sqrt{c}}f(\frac{x}{c}), \hspace{0,5cm} x\in \mathbb{R}

    I would like to show that D_{c} actually maps C_{c}(\mathbb{R}) to C_{c}(\mathbb{R}).

    I have done the following: First I notice that:

    C_{c}(\mathbb{R})= \{f: \mathbb{R} \to \mathbb{C} \big| f \hspace{0,2cm}\text{is continuos and has compact support} \}

    Hence to show the intended I must show that when the operator D_{c} is applied on a function f\in C_{c}(\mathbb{R}) then the outcome is a function which is also in C_{c}(\mathbb{R}).

    Using rules of calculus I can argue that the function \frac{1}{\sqrt{c}}f(\frac{x}{c}) is continous. Now I just need to show that this function has compact support and my job is done. The support of a function f is the smallest closed set outside which the function is equal to zero, i.e. supp f. Because I don't have a specific function to work with I'm finding it a bit difficult to show the case of compact support. How do I show that the function \frac{1}{\sqrt{c}}f(\frac{x}{c}) has compact support?

    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Good evening to all,

    For c>0, I am considering the mapping:

    D_{c}:C_{c}(\mathbb{R}) \to C_{c}(\mathbb{R}), (D_{c}f)(x)=\frac{1}{\sqrt{c}}f(\frac{x}{c}), \hspace{0,5cm} x\in \mathbb{R}

    I would like to show that D_{c} actually maps C_{c}(\mathbb{R}) to C_{c}(\mathbb{R}).

    I have done the following: First I notice that:

    C_{c}(\mathbb{R})= \{f: \mathbb{R} \to \mathbb{C} \big| f \hspace{0,2cm}\text{is continuos and has compact support} \}

    Hence to show the intended I must show that when the operator D_{c} is applied on a function f\in C_{c}(\mathbb{R}) then the outcome is a function which is also in C_{c}(\mathbb{R}).

    Using rules of calculus I can argue that the function \frac{1}{\sqrt{c}}f(\frac{x}{c}) is continous. Now I just need to show that this function has compact support and my job is done. The support of a function f is the smallest closed set outside which the function is equal to zero, i.e. supp f. Because I don't have a specific function to work with I'm finding it a bit difficult to show the case of compact support. How do I show that the function \frac{1}{\sqrt{c}}f(\frac{x}{c}) has compact support?

    Thanks.
    I'm not sure if this answers your question....

    If f has compact support then so does \frac{1}{\sqrt{c}}f\left(\frac{x}{c}\right) to do this note that L_c:\mathbb{R}\to\mathbb{R} is a hoemorphism and so L_c\left(\text{supp }f\right) is compact and...
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  3. #3
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    Compact support

    Hello,

    Thanks for the reply. I must admit that I don't even knnow what a hoemorphism is. I there any other way of showing the case of compact support?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Hello,

    Thanks for the reply. I must admit that I don't even knnow what a hoemorphism is. I there any other way of showing the case of compact support?
    All you need to know is that if E is compact so is \frac{1}{c}E=\left\{\frac{e}{c}:e\in E\right\}.

    P.S. It was a typo, it should have been "homeomorphism"
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  5. #5
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    compact support

    Hello,

    So I could write: "since f(x) has compact support then so does \frac{1}{\sqrt{c}}f(\frac{x}{c})". Why is this the case?

    Furthermore, is there a difference between something being compact and something having compact support.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Hello,

    So I could write: "since f(x) has compact support then so does \frac{1}{\sqrt{c}}f(\frac{x}{c})". Why is this the case?

    Furthermore, is there a difference between something being compact and something having compact support.
    See my answer to your other post, but please do note the disclaimer.
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