Justification of interchanging the limit and integral

Question

I have a question regarding interchanging the derivative and the integral, namely

$$\frac{d}{d\zeta}f^{1}(\zeta)=\frac{d}{d\zeta}\int_{-\infty}^{+\infty}f(x)e^{-2x\pi \zeta i}dx.$$ Where $f^{1}(\zeta)$ is fourier transform of $f$ , It is given that $xf(x)\in L^{1}(R)$.

I know that for DMT we need the integrand to be measurable, which I believe $f(x)e^{-2\pi x\zeta i}$ is and we need it to be bounded independently of the limit variable, that is $\zeta$, therefore $|f(x)e^{-2\pi i x \zeta}|\leq |f(x)|$ which is integrable on $R$ therefore the interchange is permitted, right? I have only seen DMT being used as $\lim_{n \to \pm \infty}$, I have not seen it being used for derivatives, is it possible?

Another theorem I believe that can be used is from multivariable calculus and requires uniform convergence with respect to the variable we are differentiatiing, that is $\zeta$ but that is easily given that $f\in L^{1}(R)$, the integrand is continuous and partial derivative would be uniformly continuous as well since $xf(x)$ is integrable on $L^{1}(R)$ and the partial derivative $xf(x)e^{-2\pi i x \zeta}$ is continuous, therefore it is permitted to interchange the derivative and integral. Is any of these two clarifications correct? I am not sure how to justify that $f(x)e^{-2\pi x \zeta}$ is continuous, given that $f$ is $L^{1}$ and not continuous entirely.

Answer 1

Define $\Phi(x,\zeta)=f(x)e^{-2\pi i x\zeta}$.

• $\Phi:\Bbb{R}\times\Bbb{R}\to\Bbb{C}$ is measurable
• $|\Phi(x,\zeta)|=|f(x)|$, so each $\Phi(\cdot,\zeta)\in L^1(\Bbb{R})$; this is why the Fourier transform is well-defined in the first place.
• $\frac{\partial\Phi}{\partial\zeta}(x,\zeta)=-2\pi i x\cdot f(x)e^{-2\pi i x\zeta}$ is also measurable, and we have $\left| \frac{\partial\Phi}{\partial\zeta}(x,\zeta)\right|=2\pi |xf(x)|$, and the RHS is integrable by hypothesis. The $2\pi$ isn’t the important thing (and neither is the equal sign). What is important here is that you have a $\zeta$-independent upper-bound on the derivative $\frac{\partial\Phi}{\partial\zeta}$, which is $x$-integrable.

This is enough to apply Leibniz’s integral rule (see the measure-theory statement); the proof here uses Lebesgue’s DCT (applied to the difference quotient) and the mean-value inequality. Note that the third bullet point is important, but you didn’t mention it, so no, your explanation is not correct.

The third bullet point is also why you were given the hypothesis of $x\mapsto xf(x)$ being in $L^1(\Bbb{R})$.