This is the question I am trying to answer, I am having difficulties understnading what is going on. My first question is there a typo in the hint, i.e should it be a new function $g=f\ast h_{1/R}$ or is it actually $f=f\ast h_{1/R}$? I assume it is not but this makes no sense in the standard sense of a function. Therefore I assume Grafakos means we treat f as a tempered distribution and take the convolution in the sense of distributions so $$\langle f, \phi \rangle = \langle (f\ast h_{1/R}), \phi \rangle?$$
Now my main issue really is that I dont understand what the infinity norm of $\partial^{\alpha}f$ is. Is it the infinity norm of the function associated to the $\alpha$ distributional derivative of $f$? How do we necessarily know that this distribution can be identified with a function? I think it must be to do with the fact given about the Fourier transform but I don't understand why.
Any help would be appreciated
The author indeed means $f=f*h_{1/R}$. To achieve this, on the Fourier transform side we want $\hat f = \hat f \widehat{h_{1/R}}$. And this is possible if $\widehat{h_{1/R}}$ is equal to $1$ on the support of $\hat f$. Choosing $h$ in the way described in the hint achieves that.
When differentiating a convolution, we decide where the derivative goes. So, $$ |\partial^\alpha (f*h_{1/R})| =| f*\partial^\alpha (h_{1/R}) | \le \|f\|_{L^\infty} \|\partial^\alpha (h_{1/R})\|_{L^1} $$ The last step is to use the chain rule and change of variable to show that $$ \|\partial^\alpha (h_{1/R})\|_{L^1} = R^{|\alpha|}\|\partial^\alpha (h)\|_{L^1} $$ and since $h$ is chosen once and for all (for a given dimension), the quantity $\|\partial^\alpha (h)\|_{L^1}$ can be denoted $C_{\alpha_n}$.