This is a problem from the book from Stephane Mallat "A wavelet Tour of signal processing: a sparse way".
A function $f$ is bounded and $p$ times continuously differentiable with bounded derivatives if $\int_\mathbb{R}|\hat{f(\omega)}|(1+|\omega|^p) d\omega<+\infty$.
If $\hat{f}$ has a compact support then above inequality implies $f\in C^\infty.$ I am having a doubt that how $f$ lies in $C^\infty$ by using this inequality.
I started to proceed with the fact that the space $C^p$ with compact support is dense in $C^\infty$ i.e., $C^p\subset C^\infty$ but how can we take function $f\in C^p$ to be with compact support when we knew that only $\hat{f}$ is with compact support given. Also I know the facy that $f$ and $\hat{f}$ both cannot be with compact support(by Heisenberg uncertainity principle).
You already know that $\int_\mathbb{R}|\widehat{f}(\omega) |(1+|\omega|^p) d\omega<+\infty$ implies $f$ is $p$ times continuously differentiable.
Now suppose $\hat f$ has compact support. Under mild assumptions on $f$ (namely, $f\in L^1+L^2$) we know that $\hat f$ is locally integrable. The integral $\int_\mathbb{R}|\widehat{f}(\omega) |(1+|\omega|^p) d\omega$ is finite for every $p$, because the function we integrate has compact support as is locally integrable. By the result quoted above, $f$ is $p$ times continuously differentiable, and this is true for every $p$. This means precisely that $f\in C^\infty$.