Deriving Fourier transform of differentiation into multiplication

Question

I am trying to understand the details of this answer.

I am told that, if $F(\omega)$ is the transform of $f(t)$, then the Fourier transform changes differentiation into multiplication as follows:

$$\mathcal{F}(D_tf)(\omega)=\int_{-\infty}^{\infty}D_tf(t)e^{-j\omega t}\mathrm{d}t = f(t)e^{-j\omega t}|_{-\infty}^{\infty}+j\omega\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\mathrm{d}t = 0 + j\omega F(\omega),$$

where $j$ is imaginary.

I'm wondering what the steps of the derivation for this is? In addition to this, I'm particularly curious about the following:

1. How we treat the term $f(t)e^{-j\omega t}|_{-\infty}^{\infty}$, so that it doesn't diverge, since we have that $f(t)e^{-j\omega t}|_{-\infty}^{\infty} = \dfrac{f(t)}{e^{j \omega t}} - f(t)e^{j\omega \infty}$. Is $f(t)$ restricted so that we have that $\dfrac{f(t)}{e^{j \omega t}} - f(t)e^{j\omega \infty} = 0 - f(t)e^{j\omega \infty}$? But, in that case, what conditions do we need to ensure that $f(t)e^{j\omega \infty}$ doesn't diverge?

2. How we treat $j\omega\int_{-\infty}^{\infty} f(t)e^{-j\omega t}\mathrm{d}t$. Is this just a matter of iteratively applying integration by parts?

I would greatly appreciate it if someone would please take the time to show how this Fourier transform is derived, clarifying my points of interest in the process.

Answer 1

Regarding $(1)$: Since $\omega$ is real, $e^{\pm j\omega t}$ has modulus $1$, so all that we really need is for $f$ to vanish at $\pm\infty$. A natural class to define the Fourier transform on (at least initially) is the space of Schwartz functions (https://en.wikipedia.org/wiki/Schwartz_space), which will guarantee this (and more).

Regarding $(2)$: Since $$\mathcal{F}(f)(\omega)=\int\limits_{-\infty}^\infty f(t)e^{-jt\omega}\, dt,$$ it follows immediately that

$$j\omega \int\limits_{-\infty}^\infty f(t)e^{-jt\omega}\, dt=j\omega \mathcal{F}(f)(\omega).$$ Assuming sufficient smoothness, repeated integration by parts will just give you higher powers of $j\omega$ in front of $\mathcal{F}(f)(\omega)$, which is an illustration of how the Fourier transform exchanges smoothness and decay.

In the case that I misunderstood your second question, and you were unsure where that term came from, the first equation you wrote came directly from integration by parts. The term you asked about in $(2)$ is the term that you get when you move the derivative from one term to the other. The price that we have to pay for this luxury comes in the form of the boundary terms, which I talked about in the first portion of my answer.

EDIT:

When I said that we can define the Fourier transform on Schwartz functions, I mean that we can take the Fourier transform of any Schwartz function. It is readily seen that Schwartz functions are integrable (see How to prove that a Schwartz function belongs to $L^p$?), in which case we will be able to define its Fourier transform (as the Fourier transform sends $L^1$ to $L^\infty$). Since Schwartz functions are smooth and decay faster than any polynomial (as do its derivatives), we are justified in the integration by parts. In fact, it can be shown using integration by parts that the Fourier transform of a Schwartz function is itself a Schwartz function (and it's an isomorphism).

The general gist is this: the Schwartz space is a very nice class of functions because they're closed under differentiation and multiplication by polynomials, which means that many formal calculations (not just regarding Fourier transforms) are justified when using them. One can often then extend by density, as they're dense in a lot of important spaces. In particular, working with Schwartz functions when taking Fourier transforms makes it easier to derive lots of properties, since we're not worried about issues like regularity or convergence.