45.2. A RIGHT INVERSE FOR THE TRACE FOR A HALF SPACE 1485

and (∫∞

0|| f (t)||p1,p,Rn−1 dt

)1/p

+

(∫∞

0

∣∣∣∣ f ′ (t)∣∣∣∣p0,p dt)1/p

< ∞.

Then taking a measurable representative, we see f ∈W 1,p(Rn+

)and f,xn = f ′. Also, as an

equation in Lp(Rn−1

), the following holds for all t > 0.

f (·, t) = f (0)+∫ t

0f,xn (·,s)ds

But also, for a.e. x′,the following equation holds for a.e. t > 0.

f(x′, t)= γ f

(x′)+∫ t

0f,xn

(x′,s)

ds, (45.1.1)

showing that

γ f = f (0) ∈W 1− 1p ,p(Rn−1)≡ T

(W 1,p (Ω) ,Lp (Ω) , p,

1p

).

To see that 45.1.1 holds, approximate f with a sequence from C∞(Rn+

)and finally

obtain an equation of the form∫Rn−1

∫∞

0

[f(x′, t)− γ f

(x′)−∫ t

0f,xn

(x′,s)

ds]

ψ(x′, t)

dtdx′ = 0,

which holds for all ψ ∈C∞c(Rn+

). This proves the lemma.

Thus taking the trace on the boundary loses exactly 1p derivatives.

45.2 A Right Inverse For The Trace For A Half SpaceIt is also important to show there is a continuous linear function,

R : W 1− 1p ,p(Rn−1)→W 1,p (Rn

+

)which has the property that γ (Rg) = g. Define this function as follows.

Rg(x′,xn

)≡∫Rn−1

g(y′)

φ

(x′−y′

xn

)1

xn−1n

dy′ (45.2.2)

where φ is a mollifier having support in B(0,1) .

Lemma 45.2.1 Let R be defined in 45.2.2. Then Rg∈W 1,p(Rn+

)and is a continuous linear

map from W 1− 1p ,p(Rn−1

)to W 1,p

(Rn+

)with the property that γRg = g.

Proof: Let f ∈W 1,p(Rn+

)be such that γ f = g. Let ψ (xn) ≡ (1− xn)+ and assume f

is Borel measurable by taking a Borel measurable representative. Then for a.e. x′ we havethe following formula holding for a.e. xn.

Rg(x′,xn

)=

∫Rn−1

[ψ (xn) f

(y′,ψ (xn)

)−∫

ψ(xn)

0(ψ f ),n

(y′, t)

dt]

φ

(x′−y′

xn

)x1−n

n dy′.