Proving $\mathcal{F}\left(\frac{d^n}{dx^n} f(x)\right)=(ik)^n\mathcal{F}(f(x))$.

Question

I am trying to prove the $n^{th}$ transform of the Fourier transform: $$\mathcal{F}\left(\frac{d^n}{dx^n} f(x)\right)=(ik)^n\mathcal{F}(f(x)) \tag{1}.$$ I have solved the problem for the case $n=1$ by using integration by parts: $$\mathcal{F}(f'(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f'(x) e^{-ikx} \ dx=ik\mathcal{F}(f(x)).$$ I think that solving $(1)$ requires $n$ iterations of integration by parts, by i'm having trouble writing this formally. Some help would be really appreciated.

Answer 1

Using formal induction

We want to show that $\mathcal{F}\{f^{(n)}(x)\} = (ik)^n \, \mathcal{F}\{f(x)\}.$

Basecase: For $n=0$ the identity is trivial.

Induction step: Assume that the identity is valid for some integer $n=p \geq 0$. Then, $$\begin{align} \mathcal{F}\{f^{(p+1)}(x)\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f^{(p+1)}(x) \, e^{-ikx} \, dx \\ &= \frac{1}{\sqrt{2\pi}} \left( \left[f^{(p)}(x) \, e^{-ikx}\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f^{(p)}(x) \, (-ik)e^{-ikx} \, dx \right) \\ &= ik \, \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f^{(p)}(x) \, e^{-ikx} \, dx \\ &= ik \, \mathcal{F}\{f^{(p)}(x)\} \\ &= ik \, \left((ik)^p \, \mathcal{F}\{f(x)\} \right) \\ &= (ik)^{p+1} \, \mathcal{F}\{f(x)\}, \\ \end{align}$$ i.e. it's also valid for $n=p+1.$

Conclusion: By the induction principle, the identity is valid for all integer $n \geq 0.$