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I want to compute the Fourier transform of $\partial_{x} u(x)$

My definition is $\hat{u}(k)= \int_{\mathbb{R}}u(x)e^{-ik\cdot x}dx$. I am ignoring the constant from the definition.

I am told the answer is: $\widehat{\partial_{x}u} = \boxed{ik\hat{u}(k)}$

Why?

Writing out the integral we have: $\int_{-\infty}^{\infty} \partial_x u e^{-ik\cdot x}dx$.

Integration-by-parts: $\int fdg = fg - \int gdf$, I pick $f = e^{-ikx} \implies df = -ike^{-ikx}$ and $dg = \partial_x u \implies g = u$.

So we get:

$u(x) e^{-ikx} \mid_{-\infty}^{\infty} +ik \underbrace{\int_{-\infty}^{\infty}u e^{-ikx}dx}_{\hat{u}(k)}$

So that first summand is supposed to go away, but how?

KZ-Spectra
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1 Answers1

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You’re correct, the proof only works if $u$ goes to 0 at infinity.

However, if you take $v_n(x)=u(x)$ on $[-n,n]$ and 0 otherwise, then you can apply the proof to $v_n$. For reasonable assumptions on $u$ (Like it being L2), then $v_n$ converges to $u$ so the Fourier transforms and derivatives correspondingly converge and you get the desired result.

Eric
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