Calculus of Real and Complex Variables

10.5.3 Fourier Transforms of Functions In L²

n (ℝ )

Consider Ff and F⁻¹f for f ∈ L²(ℝⁿ). First note that the formula given for Ff and F⁻¹f when f ∈ L¹

will not work for f ∈ L²(ℝⁿ) unless f is also in L¹(ℝⁿ). Recall that a + ib = a − ib.

Proof: First note that for ψ ∈G,

-- ------- −1 -- ----- F (ψ ) = F −1(ψ ),F (ψ) = F (ψ ). (10.10)

(10.10)

This follows from the definition. For example,

Let ϕ,ψ ∈G. It was shown above that

∫ ∫ n(Fϕ)ψ(t)dt = n ϕ(F ψ)dx. ℝ ℝ

Similarly,

∫ ∫ −1 −1 ℝn ϕ(F ψ)dx = ℝn(F ϕ)ψdt. (10.11)

(10.11)

Now, 10.10 - 10.11 imply

Similarly

−1 ||ϕ||2 = ||F ϕ||2.■

Proof: Let ψ ∈G be given. Then

Also by Theorem 10.5.13 _k=1^∞ is Cauchy in L² and so it converges to some h ∈ L². Therefore, from the above,

∫ Ff (ψ ) = h (x)ψ(x) ℝn

which shows that F

∈ L² and h = F. The case of F⁻¹ is entirely similar. ■

Since Ff and F⁻¹f are in L²

, this also proves the following theorem.

Proof: Use the density of G in L²

to obtain a sequence, converging to f in L². Then by Lemma 10.5.14

||F f||2 = lim ||F ϕk||2 = lim ||ϕk||2 = ||f||2 . k→ ∞ k→ ∞

Similarly,

−1 ||f||2 = ||F f||2.■

The following corollary is a simple generalization of this. To prove this corollary, use the following simple lemma which comes as a consequence of the Cauchy Schwarz inequality.

Proof: 

|∫ ∫ | |∫ ∫ | || f g dx− f gdx||≤ || f g dx− f gdx||+ | ℝn k k ℝn | | ℝn k k ℝn k |

|∫ ∫ | || fkgdx − fgdx|| | ℝn ℝn |

≤ ||fk||2||g− gk||2 + ||g||2||fk − f||2.

Now

₂ is a Cauchy sequence and so it is bounded independent of k. Therefore, the above expression is smaller than ε whenever k is large enough.  ■

Proof: First note the above formula is obvious if f,g ∈G. To see this, note

--------------------- ∫ --- ∫ ---1---∫ −ix⋅t ℝn Ff Fgdx = ℝn Ff (x)(2π)n∕2 ℝn e g(t)dtdx

The formula with F⁻¹ is exactly similar.

Now to verify the corollary, let ϕ_k → f in L²

and let ψ_k → g in L². Then by Lemma 10.5.14

∫ ∫ ∫ ∫ --- ---- --- - ℝn F fFgdx = kli→m∞ ℝn F ϕkF ψkdx = kli→m∞ ℝn ϕkψkdx = ℝn fgdx

A similar argument holds for F⁻¹. ■

How does one compute Ff and F⁻¹f?

Proof: ||f −f_r||₂ → 0 and so ||Ff −Ff_r||₂ → 0 and ||F⁻¹f −F⁻¹f_r||₂ → 0 by Plancherel’s Theorem.  ■

What are Ff_r and F⁻¹f_r? Let ϕ ∈G

Since this holds for all ϕ ∈G, a dense subset of L²(ℝⁿ), it follows that

n ∫ Ffr(y) = (2π)−2 fr(x)e−ix⋅ydx. ℝn

Similarly

∫ F −1f (y) = (2π)− n2 f (x )eix⋅ydx. r ℝn r

This shows that to take the Fourier transform of a function in L²

, it suffices to take the limit as r →∞ in L² of (2π)−

n2

∫ _ℝⁿf_r(x)e^−ix⋅ydx. A similar procedure works for the inverse Fourier transform.

Note this reduces to the earlier definition in case f ∈ L¹

. Now consider the convolution of a function in L² with one in L¹.

Proof: An application of Minkowski’s inequality yields

( ) ∫ (∫ )2 1∕2 n n |h(x − y)||f (y)|dy dx ≤ ||f||1||h ||2. (10.16) ℝ ℝ

(10.16)

Hence ∫

dy < ∞ a.e. x and

∫ x → h(x − y)f (y)dy

is in L²

. Let E_r ↑ ℝⁿ,m < ∞. Thus,

hr ≡ XE h ∈ L2(ℝn)∩ L1 (ℝn ), r

and letting ϕ ∈G,

∫ F (hr ∗ f)(ϕ)dx

Since ϕ is arbitrary and G is dense in L²,

F (hr ∗ f) = (2π)n∕2F hrFf.

Now by Minkowski’s Inequality, h_r ∗f → h∗f in L²

and also it is clear that h_r → h in L²; so, by Plancherel’s theorem, you may take the limit in the above and conclude

F (h∗ f) = (2π)n∕2F hF f.

The assertion for F⁻¹ is similar and 10.15 follows from 10.16. ■