1110 CHAPTER 32. FOURIER TRANSFORMS

This shows that to take the Fourier transform of a function in L2 (Rn), it suffices to take thelimit as r→ ∞ in L2 (Rn) of (2π)−

n2∫Rn fr(x)e−ix·ydx. A similar procedure works for the

inverse Fourier transform.Note this reduces to the earlier definition in case f ∈ L1 (Rn). Now consider the convo-

lution of a function in L2 with one in L1.

Theorem 32.3.19 Let h ∈ L2 (Rn) and let f ∈ L1 (Rn). Then h∗ f ∈ L2 (Rn),

F−1 (h∗ f ) = (2π)n/2 F−1hF−1 f ,

F (h∗ f ) = (2π)n/2 FhF f ,

and||h∗ f ||2 ≤ ||h||2 || f ||1 . (32.3.11)

Proof: An application of Minkowski’s inequality yields(∫Rn

(∫Rn|h(x−y)| | f (y)|dy

)2

dx

)1/2

≤ || f ||1 ||h||2 . (32.3.12)

Hence∫|h(x−y)| | f (y)|dy < ∞ a.e. x and

x→∫

h(x−y) f (y)dy

is in L2 (Rn). Let Er ↑ Rn, m(Er)< ∞. Thus,

hr ≡XEr h ∈ L2 (Rn)∩L1 (Rn),

and letting φ ∈ G , ∫F (hr ∗ f )(φ)dx

≡∫

(hr ∗ f )(Fφ)dx

= (2π)−n/2∫ ∫ ∫

hr (x−y) f (y)e−ix·tφ (t)dtdydx

= (2π)−n/2∫ ∫ (∫

hr (x−y)e−i(x−y)·tdx)

f (y)e−iy·tdyφ (t)dt

=∫

(2π)n/2 Fhr (t)F f (t)φ (t)dt.

Since φ is arbitrary and G is dense in L2 (Rn),

F (hr ∗ f ) = (2π)n/2 FhrF f .

Now by Minkowski’s Inequality, hr ∗ f → h∗ f in L2 (Rn) and also it is clear that hr→ h inL2 (Rn) ; so, by Plancherel’s theorem, you may take the limit in the above and conclude

F (h∗ f ) = (2π)n/2 FhF f .

The assertion for F−1 is similar and 32.3.11 follows from 32.3.12.