1302 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Lemma 38.1.3 The Schwartz class, S, is dense in W m,p (Rn) .

Proof: The set, Rn satisfies the segment condition, so C∞c (Rn) is dense in W m,p (Rn) .

However, C∞c (Rn)⊆S. This proves the lemma.

Recall now Plancherel’s theorem which states that || f ||0,2,Rn = ||F f ||0,2,Rn wheneverf ∈ L2 (Rn) . Also it is routine to verify from the definition of the Fourier transform that foru ∈S,

F∂ku = ixkFu.

From this it follows that||Dα u||0,2,Rn = ||xα Fu||0,2,Rn .

Here xα denotes the function x→ xα . Therefore,

||u||m,2,Rn =

(∫Rn

∑|α|≤m

x2α11 · · ·x2αn

n |Fu(x)|2 dx

)1/2

.

Also, it is not hard to verify that

∑|α|≤m

x2α11 · · ·x2αn

n ≤

(1+

n

∑j=1

x2j

)m

≤C (n,m) ∑|α|≤m

x2α11 · · ·x2αn

n

where C (n,m) is the largest of the multinomial coefficients obtained in the expansion,(1+

n

∑j=1

x2j

)m

.

Therefore, for all u ∈S,

||u||m,2,Rn ≤(∫

Rn

(1+ |x|2

)m|Fu(x)|2 dx

)1/2

≤C (n,m) ||u||m,2,Rn . (38.1.2)

This motivates the following definition.

Definition 38.1.4 Let Hm (Rn)≡{u ∈ L2 (Rn) : ||u||Hm(Rn) ≡

(∫Rn

(1+ |x|2

)m|Fu(x)|2 dx

)1/2

< ∞

}. (38.1.3)

Lemma 38.1.5 S is dense in Hm (Rn) and Hm (Rn) =W 2,m (Rn). Furthermore, the normsare equivalent.

Proof: First it is shown that S is dense in Hm (Rn) . Let u ∈ Hm (Rn) . Let µ (E) ≡∫E

(1+ |x|2

)mdx. Thus µ is a regular measure and u ∈ Hm (Rn) just means that Fu ∈

L2 (µ) , the space of functions which are in L2 (Rn) with respect to this measure, µ. There-fore, from the regularity of the measure, µ, there exists uk ∈Cc (Rn) such that

||uk−Fu||L2(µ)→ 0.