1128 CHAPTER 33. FOURIER ANALYSIS IN Rn

Proof: By Plancherel’s theorem F−1ρ is in L2 (Rn). If f ∈ L1 (Rn), then by Minkow-ski’s inequality,

F−1ρ ∗ f ∈ L2 (Rn) .

Now let g ∈ L2 (Rn). By Holder’s inequality,

∫ ∣∣F−1ρ (x−y)

∣∣ |g(y)|dy≤(∫ ∣∣F−1

ρ (x−y)∣∣2 dy

)1/2(∫|g(y)|2 dy

)1/2

< ∞

and so the following is well defined a.e.

F−1ρ ∗g(x)≡

∫F−1

ρ (x−y)g(y)dy

also, ∣∣F−1ρ ∗g(x)−F−1

ρ ∗g(x′)∣∣ ≤ ∫ ∣∣F−1

ρ (x−y)−F−1ρ(x′−y

)∣∣ |g(y)|dy

≤∣∣∣∣F−1

ρ−F−1ρx′−x

∣∣∣∣ ||g||l2

and by continuity of translation in L2 (Rn), this shows x→ F−1ρ ∗ g(x) is continuous.Therefore, F−1ρ∗ maps L1 (Rn)+L2 (Rn) to the space of measurable functions. (Continu-ous functions are measurable.) It is clear that F−1ρ∗ is subadditive.

If φ ∈ G , Plancherel’s theorem implies as before,∣∣∣∣F−1ρ ∗φ

∣∣∣∣2 =

∣∣∣∣F (F−1ρ ∗φ

)∣∣∣∣2 =

(2π)n/2 ||ρFφ ||2 ≤ (2π)n/2 ||ρ||∞||φ ||2 . (33.3.24)

Now let f ∈ L2 (Rn) and let φ k ∈ G , with

||φ k− f ||2→ 0.

Then by Holder’s inequality,∫F−1

ρ (x−y) f (y)dy = limk→∞

∫F−1

ρ (x−y)φ k (y)dy

and so by Fatou’s lemma, Plancherel’s theorem, and 33.3.24,

∣∣∣∣F−1ρ ∗ f

∣∣∣∣2 =

(∫ ∣∣∣∣∫ F−1ρ (x−y) f (y)dy

∣∣∣∣2 dx

)1/2

≤

≤ lim infk→∞

(∫ ∣∣∣∣∫ F−1ρ (x−y)φ k (y)dy

∣∣∣∣2 dx

)1/2

= lim infk→∞

∣∣∣∣F−1ρ ∗φ k

∣∣∣∣2

≤ ||ρ||∞(2π)n/2 lim inf

k→∞

||φ k||2 = ||ρ||∞ (2π)n/2 || f ||2 .