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**surjective** Good evening to all,

For $\displaystyle c>0$, I am considering the mapping:

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

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

I have done the following: First I notice that:

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

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

Using rules of calculus I can argue that the function $\displaystyle \frac{1}{\sqrt{c}}f(\frac{x}{c})$ is continous. Now I just need to show that this function has compact support and my job is done. The support of a function $\displaystyle f$ is the smallest closed set outside which the function is equal to zero, i.e. $\displaystyle supp f$. 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 $\displaystyle \frac{1}{\sqrt{c}}f(\frac{x}{c})$ has compact support?

Thanks.