Proof.
For given φ∈A0(R) we suppose that suppφ(x)⊆[c,d], without loss of generality. Then using the embedding rule and the substitution t=(y-x)/ε we have the representatives of the distribution x+-r-1/2 in Colombeau algebra:
(23)x+-r-1/2~(φε,x)=(-1)r2r(2r-1)!!1ε∂r∂x∫0∞y-1/2φ(y-xε)dy=2r(2r-1)!!1εr+1∫0∞y-1/2φ(r)(y-xε)dy=2r(2r-1)!!1εr∫-x/εd(x+εt)-1/2φ(r)(t)dt.
Similar, using the embedding rule and the substitution s=(y-x)/ε we have the representatives of the distribution x--r-1/2 in Colombeau algebra:
(24)x--r-1/2~(φε,x)=(-1)r2r(2r-1)!!1ε∂r∂x×∫-∞0(-y)-1/2φ(y-xε)dy=2r(2r-1)!!1εr+1×∫-∞0(-y)-1/2φ(r)(y-xε)dy=2r(2r-1)!!1εr∫c-x/ε(-x-εs)-1/2φ(r)(s)ds.
Then, for any ψ(x)∈𝒟(R) we have
(25)〈x+-r-1/2~(φε,x)·x--r-1/2~(φε,x),ψ(x)〉 =∫-∞∞x+-r-1/2~(φε,x)x--r-1/2~(φε,x)ψ(x)dx =22r((2r-1)!!)21ε2r∫-dεcεψ(x)∫-x/εdφ(r)(t) ×∫c-x/ε(x+εt)-1/2(-x-εs)-1/2φ(r)(s)ds dt dx =22r((2r-1)!!)21ε2r∫cdψ(-εω)∫ωdφ(r)(t) ×∫cω(t-ω)-1/2(ω-s)-1/2φ(r)(s)ds dt dω,
using the substitution ω=-x/ε.
By the Taylor theorem we have that
(26)ψ(-εω)=∑k=02rψ(k)(0)k!(-εω)k+ψ(2r+1)(ηω)(2r+1)!(-εη)2r+1,
for η∈(0,1). Using this for (25) we have
(27)〈x+-r-1/2~(φε,x)·x--r-1/2~(φε,x),ψ(x)〉 =22r((2r-1)!!)2∑k=02r(-1)kψ(k)(0)k!ε2r-kIk+O(ε),
where
(28)Ik=∫cdφ(r)(t)∫tdφ(r)(s)×∫st(t-ω)-1/2(ω-s)-1/2ωkdω ds dt,
for k=0,1,…,2r and we have changed the order of integration.
Putting ω-s=(t-s)v we have
(29)∫st(t-ω)-1/2(ω-s)-1/2ωkdω =∫01v-1/2(1-v)-1/2[sv+(1-v)t]kdv =∑p=0kk!p!(k-p)!∫01vp-1/2(1-v)k-p-1/2sptk-pdv =∑p=0kk!p!(k-p)!B(p+12,k-p+12)sptk-p.
Thus
(30)Ik=∑p=0kk!p!(k-p)!B(p+12,k-p+12)×∫cdtk-pφ(r)(t)∫tdspφ(r)(s)ds dt=∑p=0kk!p!(k-p)!B(p+12,k-p+12)Jk,p.
Next, suppose that k is even, less than 2r and that p is less than r. It follows from Lemma 2 that ∫tdspφ(r)(s)ds is an even or odd function accordingly as r+p is odd or even. We thus have that tk-pφ(r)(t)∫tdspφ(r)(s)ds is an odd function and Jk,p=0. If p≥r then k-p<r and by changing the order of integration we can prove that again Jk,p=0. If we suppose that k is odd and less than 2r we can prove in a similar manner that Jk,p=0.
For the case k=2r if p≠r again we have Jk,p=0. For k=2r and p=r using Lemma 3 and changing the order of integration we have
(31)J2r,r=∫cdtrφ(r)(t)∫tdsrφ(r)(s)ds dt=∑i=0r(-1)ir!(r-i)!∫cdt2r-iφ(r)(t)φ(r-i-1)(t)dt=(-1)rr!∫cdtrφ(r)(t)∫tdφ(s)ds dt=(-1)rr!∫cdφ(s)∫cstrφ(r)(t)dt ds=(-1)rr!∑i=0r(-1)ir!(r-i)!∫cdsr-iφ(r-i-1)(s)φ(s)ds=(r!)2∫cdφ(s)(∫csφ(t)dt)ds.
Further,
(32)∫cdφ(s)(∫csφ(t)dt)ds=∫cd(∫csφ(t)dt)d(∫csφ(t)dt)=12(∫csφ(t)dt)2|cd=12,
and J2r,r=(r!)2/2. So, Ik=0 for k=0,1,…,2r-1 and
(33)I2r=(2r)!2B(r+12,r+12)=((2r-1)!!)2π22r+1.
Finally we have
(34)〈x+-r-1/2~(φε,x)·x--r-1/2~(φε,x),ψ(x)〉 =π2ψ(2r)(0)(2r)!+O(ε) =π2(2r)!〈δ(2r)(x),ψ(x)〉+O(ε).
Therefore passing to the limit, as ε→0, we obtain (22) proving the theorem.