1316 CHAPTER 38. SOBOLEV SPACES BASED ON L2

Lemma 38.5.3 Let u ∈S. Then there exists a constant, Cn, depending on n but indepen-dent of u ∈S such that

Fγu(x′)=Cn

∫R

Fu(x′,xn

)dxn.

Proof: Using the dominated convergence theorem,∫R

Fu(x′,xn

)dxn ≡ lim

ε→0

∫R

e−(εxn)2Fu(x′,xn

)dxn

≡ limε→0

∫R

e−(εxn)2(

12π

)n/2 ∫Rn

e−i(x′·y′+xnyn)u(y′,yn

)dy′dyndxn

= limε→0

(1

)n/2 ∫Rn

u(y′,yn

)e−ix′·y′

∫R

e−(εxn)2e−ixnyndxndy′dyn.

Now −(εxn)2− ixnyn =−ε2

(xn +

iyn2

)2− ε2 y2

n4 and so the above reduces to

limε→0

(1

)n/2 ∫Rn

u(y′,yn

)e−ix′·y′

∫R

e−ε2(

xn+iyn2

)2−ε2 y2

n4 dxndy′dyn

= limε→0

Kn

∫Rn

u(y′,yn

)e−ix′·y′e−ε2 y2

n4

∫R

e−ε2(

xn+iyn2

)2

dxndy′dyn

= limε→0

Kn

∫Rn

u(y′,yn

)e−ix′·y′e−ε2 y2

n4

dy′dyn

which is an expression of the form

limε→0

Kn

∫R

e−ε2 y2n4

∫Rn−1

u(y′,yn

)e−ix′·y′dy′dyn = Kn

∫Rn

u(y′,0

)e−ix′·y′dy′

= KnFγu(x′)

and this proves the lemma with Cn ≡ K−1n .

Earlier Ht (Rn) was defined and then for U an open subset of Rn, Ht (U) was defined tobe the space of restrictions of functions of Ht (Rn) to U and a norm was given which madeHt (U) into a Banach space. The next task is to considerRn−1×{0} , a smaller dimensionalsubspace of Rn and examine the functions defined on this set, denoted by Rn−1 for shortwhich are restrictions of functions in Ht (Rn) . You note this is somewhat different becauseheuristically, the dimension of the domain of the function is changing. An open set in Rn

is considered an n dimensional thing but Rn−1 is only n− 1 dimensional. I realize this isvague because the standard definition of dimension requires a vector space and an open setis not a vector space. However, think in terms of fatness. An open set is fat in n directionswhereas Rn−1 is only fat in n− 1 directions. Therefore, something interesting is likely tohappen.

Let S denote the Schwartz class of functions on Rn and S′ the Schwartz class offunctions on Rn−1. Also, y′ ∈ Rn−1 while y ∈ Rn. Let u ∈ S. Then from Lemma 38.5.3

1316 CHAPTER 38. SOBOLEV SPACES BASED ON L?Lemma 38.5.3 Let u € G. Then there exists a constant, C,, depending on n but indepen-dent of u € © such thatFyu (x’) = Cn | Fu (x’, xn) dXp.RProof: Using the dominated convergence theorem,2[Fu x’ Xn) ) dx, = lim | e~@")" Fu (x’, Xn) dXpe-0/R1 n/2 > “(I wllim fe (ew (=) / ely) u (y!,yn) dy’dyndy. 1 n/2 _ yl! _ 2 _.= lim (=) [wlvime “y fe (€%n)" @Pindn dxndy! dyn.: 2 2Now — (€xn)* — iXpVn = —€7 (sn + i) — 7h and so the above reduces to1 n/2 lt —¢2 yn 2 2hca (23) [owed eS) 8 assayE> R” R2 _ 2 tyn= limK, | u(y',yn)e “xy ge [ e° (m+ )’ dxndy' dyne>0 R" JR2]= limK, win'y! 9e'F <dyldlim Kn f., u(ysynje™ Fe © 4 cdy'dynwhich is an expression of the formeeelim Kn [ge fu (y/.yn) eV dy'dyn€-0K, | u(y’,0) e® Y' gy!R”= K,Fyu(x’)and this proves the lemma with C, = K,!.Earlier H’ (IR") was defined and then for U an open subset of R”, H‘ (U) was defined tobe the space of restrictions of functions of H‘ (R”) to U and a norm was given which madeH' (U) into a Banach space. The next task is to consider R’~! x {0} , a smaller dimensionalsubspace of IR” and examine the functions defined on this set, denoted by R”~! for shortwhich are restrictions of functions in H’ (IR). You note this is somewhat different becauseheuristically, the dimension of the domain of the function is changing. An open set in R”is considered an n dimensional thing but R”~! is only n—1 dimensional. I realize this isvague because the standard definition of dimension requires a vector space and an open setis not a vector space. However, think in terms of fatness. An open set is fat in n directionswhereas R”~! is only fat in n — 1 directions. Therefore, something interesting is likely tohappen.Let G denote the Schwartz class of functions on R” and G’ the Schwartz class offunctions on R’~!. Also, y’ € R’~! while y € R”. Let uw € G. Then from Lemma 38.5.3