1. ## 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.

2. Originally Posted by surjective
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.

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...

3. ## 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?

4. Originally Posted by surjective
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"

5. ## 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.

6. Originally Posted by surjective
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.