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I am trying to understand the following argument given in a text book:

Suppose $f \in L^1(\mathbb R^n)$, consider the function $\hat{f}(\zeta)= \int_{\mathbb R^n} \exp(-2\pi i X.\zeta)f(X)dX$.

Suppose $x_j f(X) \in L^1(\mathbb R^n)$ for $1 \leq j \leq n$ then $$\delta_j (\hat{f}(\zeta))=-2\pi i \int_{\mathbb R^n} \exp(-2\pi i X.\zeta)x_j f(X)dX $$ where $\delta_j$ denote partial derivative w.r.t. $\zeta_j$

Why can we interchange differentiation and integration as is done in the above argument?

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Dontknowanything
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1 Answers1

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The integral you are dealing is a so called Fourier transform - an important type of integral transform. The transform is defined, as in the link, for functions $g\in L^1(\mathbb R^n)$, hence the assumption for $f \in L^1(\mathbb R^n)$.

The interchange of derivative and integral can be justified - at least in the simple $1$-dimensional case - with the Leibniz rule, which can also be extended to the $\mathbb R^d$ case. Another reference would be Real Analysis - Modern Techniques and Their Applications by Gerald B. Folland who states it (don't recall the exact wording):

Let $(X,M, \mu)$ be a measure space and $f ∶ X \times [a, b] \mapsto \mathbb C, (−\infty < a < b < \infty)$, and $f (\cdot,t) ∶ X \mapsto \mathbb C$ is integrable for all $t\in [a, b]$. Set further $$F(t) = \int_X f (x,t) d\mu(x). $$ If we now assume that $\frac{\partial f}{\partial t}$

  • exists
  • and $\exists g\in L^1(\mu)$ such that $\forall t,x:\big\lvert\frac{\partial f}{\partial t}(x,t)\big\rvert\leq g(x)$

then $F$ is differentiable and we can interchange the order of integration and differentiation and write $$ F^{'}(t)=\int_X\frac{\partial f}{\partial t}(x,t)d\mu(x)$$ In your case you need also $\frac{\partial f}{\partial t}\in L^1$ to justify the existence of the Fourier transform.

EDIT: to clarify why the conditions are satisfied in the 1-dimensional case, you can generalize this with a little effort to higher dimensions and differentials

We have for $f(x)\in L^1$ $$ g(t,x)=\exp(-2\pi i xt)f(x) \text{ with } \partial_t g(t,x)=(-2\pi i)\exp(-2\pi i xt)xf(x) $$ we get $$ \lvert \partial_t g(t,x) \rvert=\lvert (-2\pi i)\exp(-2\pi i xt)xf(x) \rvert= 2\pi \lvert xf(x) \rvert $$ Since we assumed that $xf(x)\in L^1$ we know that the derivative is bounded by a $L^1$ function and therefore we can interchange the order of integration and differentiation.

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