My question is related to this question Conditions for the differentiation of Fourier Transform of a function. Suppose my function $f(x)$ is from a Schwarz function https://en.wikipedia.org/wiki/Schwartz_space , since Fourier transform is an automorphism on the space of Schwarz space, we have that $\varphi(t)$ is also in Schwarz space.
This means $\varphi(t)$ is smooth and I can naturally differentiate under the integral sign without invoking a Leibnitz rule in measure space given in the first link I have provided?
In particular, my $f(x)$ is bounded smooth function with compact support.
Yes.
Since $\varphi$ is a Schwartz function, this means that all $x^n \varphi^{(m)}$ are in $L^1(\Bbb R)$ and we can apply the Dominated convergence theorem just fine.
Edit: To add more details, let's consider $$ \frac d{dx}[x^2\varphi(x)] = 2x\varphi(x) + x^2\varphi'(x) $$ which is bounded on $\Bbb R$ by the fact that $\varphi$ is Schwartz, says $\frac d{dx}[x^2\varphi(x)]\le M$. Thus $$ |(x+h)^2\varphi(x+h) - x^2\varphi(x)| \le hM $$ by the mean value theorem (wlog $h>0$), i.e. $$ \left|\left(1+\frac hx\right)^2\varphi(x+h) - \varphi(x)\right| \le \frac {hM}{x^2} $$ for large $|x|$. Thus $$\begin{align} |\varphi(x+h) - \varphi(x)| &\le \left|\varphi(x+h) - \left(1+\frac hx\right)^2\varphi(x+h)\right| + \left|\left(1+\frac hx\right)^2\varphi(x+h) - \varphi(x)\right| \\ &\le \frac {hC}{x^2} + \frac {hM}{x^2}. \end{align}$$ where $C$ can be obtained from the fact that $\varphi$ is Schwartz. Provide that $h<1$, we have $$\begin{align} \frac {|\varphi(x+h) - \varphi(x)|}h &\le \frac {C}{x^2} + \frac {M}{x^2} \end{align}$$ This shows that the difference quotient $\frac {|\varphi(x+h) - \varphi(x)|}h$ is bounded above (for small $h$) by an $L^1(\Bbb R)$ function. Hence we may apply the Dominated convergence theorem.