∫ ( )n∕2∫
≡ lim e−(εxn)2 -1- e− i(x′⋅y′+xnyn)u(y′,yn)dy′dyndxn
ε→0 ℝ 2π ℝn
( 1)n ∕2∫ ′ −ix′⋅y′ ∫ − (εx )2 −ix y ′
= lεim→0 2π- n u(y ,yn)e e n e nndxndy dyn.
ℝ ℝ
Now −
(εxn)
^{2}− ix_{n}y_{n} = −ε^{2}
( iyn)
xn + 2
^{2}− ε^{2}
yn2
4
and so the above reduces to an
expression of the form
∫ 1 −ε2y2n ∫ ′ − ix′⋅y′ ′ ∫ ′ −ix′⋅y′ ′
εli→m0 Kn ε e 4 n−1 u(y ,yn)e dy dyn = Kn n u(y,0)e dy
ℝ ℝ ℝ ′
= KnF γu (x)
and this proves the lemma with C_{n}≡ K_{n}^{−1}.
Earlier H^{t}
(ℝn)
was defined and then for U an open subset of ℝ^{n}, H^{t}
(U)
was defined
to be the space of restrictions of functions of H^{t}
(ℝn )
to U and a norm was given which
made H^{t}
(U)
into a Banach space. The next task is to consider ℝ^{n−1}×
{0}
, a smaller
dimensional subspace of ℝ^{n} and examine the functions defined on this set, denoted by
ℝ^{n−1} for short which are restrictions of functions in H^{t}
(ℝn )
. You note this is somewhat
different because heuristically, the dimension of the domain of the function is
changing. An open set in ℝ^{n} is considered an n dimensional thing but ℝ^{n−1} is
only n − 1 dimensional. I realize this is vague because the standard definition
of dimension requires a vector space and an open set is not a vector space.
However, think in terms of fatness. An open set is fat in n directions whereas
ℝ^{n−1} is only fat in n − 1 directions. Therefore, something interesting is likely to
happen.
Let S denote the Schwartz class of functions on ℝ^{n} and S^{′} the Schwartz class of
functions on ℝ^{n−1}. Also, y^{′}∈ ℝ^{n−1} while y ∈ ℝ^{n}. Let u ∈S. Then from Lemma 38.5.3
and s > 0,
∫ ( 2)s 2
1 +|y′| |Fγu (y′)| dy′
ℝn−∫1 ( ) |∫ |2
= C 1+ |y′|2 s|| F u(y′,y )dy || dy′
n ℝn−1 | ℝ n n|
∫ (1+ |y′|2)t−1∕2
C eiy⋅x--(------)---F u(y′)dy (38.5.19)
ℝn 1+ |y|2 t
(38.5.19)
Here the inside Fourier transform is taken with respect to ℝ^{n−1} because u is only defined
on ℝ^{n−1} and C will be chosen in such a way that γ ∘ζ = id. First the existence of C such
that γ ∘ ζ = id will be shown. Since u ∈S^{′} it follows
(1 + |y′|2)t−1∕2
y → --(-------)---Fu (y ′)
1 + |y|2 t
is in S. Hence the inverse Fourier transform of this function is also in S and so for
u ∈S^{′}, it follows ζu ∈S. Therefore, to check γ ∘ ζ = id it suffices to plug in x_{n} = 0.
From Lemma 38.5.2 this yields
This proves the theorem because S is dense in ℝ^{n}.
Actually, the assertion that γu
′
(x)
= u
′
(x ,0)
holds for more functions, u than just
u ∈S. I will make no effort to obtain the most general description of such functions but
the following is a useful lemma which will be needed when the trace on the boundary of
an open set is considered.
Lemma 38.5.5Suppose u is continuous and u ∈ H^{1}
(ℝn)
. Then there exists a set ofm_{1}measure zero, N such that if x_{n}
∕∈
N, then for every ϕ ∈ L^{2}
( )
ℝn−1
∫ xn
(γu,ϕ)H + (u,n(⋅,t),ϕ)H dt = (u (⋅,xn),ϕ)H
0
where here
∫
(f,g)H ≡ fgdx′,
ℝn−1
just the inner product in L^{2}
( n−1)
ℝ
. Furthermore,
u(⋅,0) = γu a.e. x ′.
Proof: Let
{uk}
be a sequence of functions from S which converges to u in H^{1}
(ℝn)
and let
{ϕk}
denote a countable dense subset of L^{2}
( )
ℝn−1
. Then
∫ xn
(γuk,ϕj) + (uk,n (⋅,t),ϕj) dt = (uk(⋅,xn),ϕj) . (38.5.20)
H 0 H H
and this converges to zero as k →∞. Therefore, using a diagonal sequence argument,
there exists a subsequence, still denoted by k and a set of measure zero, N ≡∪_{j=1}^{∞}N_{j}
such that for x^{′}
∕∈
N, you can pass to the limit in 38.5.20 and obtain that for all
ϕ_{j},