Spectral Theories, II, p. 301
Bourbaki marks this one as harder than the rest.
Let k⩾2 and ℓ>k be natural numbers. For every integer N⩾1, let rk,ℓ(N) denote the number of families (n1,...,nℓ)∈Nℓ such that
n1k+⋯+nℓk=N
The object of this exercise is to prove that, if ℓ is sufficiently large as a function of k, then rk,ℓ(N)⩾1 ("Waring's problem") for every N⩾1.
Let e:T→C∗ denote the character defined by passing to the quotient from the character x↦e2iπx of R.
Let P⩾1 be an integer. Let SP denote the function on T such that
SP(x)=n=1∑Pe(xnk)
Let dx denote the normalized Haar measure on T.
a) If P>N1/k, then one has
rk,ℓ(N)=∫TSP(x)ℓe(−Nx)dx
b) Let δ be a real number such that 0<δ<1/3 and let P=[N1/k]. Suppose that N is sufficiently large, so that P⩾1. Let a and q be relatively prime integers such that 1⩽q⩽Pδ. Let Ma,q denote the image in T of the open interval
]qa−Pk−δ1,qa+Pk−δ1[.
For x∈Ma,q, one has
n=1∑Pe2iπxnk=q1S(a;q)I(x−qa)+O(P2δ)
as N→+∞, where
S(a;q)=x=1∑qe(qaxk),I(y)=∫0Pe(ytk)dt.
c) Let M denote the union of the sets Ma,q when q runs through the set of integers such that 1⩽q⩽Pδ and a the set of integers prime to q such that 1⩽a<q. If δ<1/5, then there exists a real number δ′>0 such that
∫MSP(x)ℓe(−Nx)dx=Pℓ−kS(Pδ,N)J(Pδ)+O(Pℓ−k−δ′)
as N→+∞, where one puts, for Q⩾1
q
1 ℓaN S(Q,N)=∑∑S(a;q)e−
1⩽q⩽Qa=1
∫Q∫1ℓ
J(Q) = e(ytk)dte(−y)dy.
−Q 0
d) One has
Γ(1+1/k)ℓ−(ℓ/k−1)
J(Q) = + O(Q )
Γ(ℓ/k)
as Q→+∞. (First prove that
1∫−1+1/k−(ℓ/k−1)
J(Q) = (y1⋯yℓ−1(1−y1−⋯−yℓ−1))dy+O(Q )
kXℓℓ
where
Xℓ={(y1,...,yℓ−1)∈(R+∗)ℓ−1∣y1+⋯+yℓ−1<1}
then evaluate this integral; cf. FVR VII, p. 8, Prop. 4.)
e) Suppose that ℓ>δ−1k2k. There exists δ′>0 such that
∫T−M∣SP(x)∣ℓe(−Nx)dx=O(Pℓ−k−δ′)
when N→+∞. (If N is sufficiently large and if x∈R has image in T− M, there exist integers a and q relatively prime such that
Pδ⩽q⩽Pk−δ,∣x−a∣⩽12
Then use Exercise 66.)
f) For every prime number p and every integer m⩾1, let us put
A(pm)=∑pm(pm1S(a;pm))ℓe(−pmaN)
a=1 (a,q)=1
and
N(pm)= Card({(x1,...,xℓ)∈(Z/pmZ)ℓ∣x1k+⋯+xℓk=N mod.pm}). When ℓ>2k, one has
+∞m
1+∑A(pm)= limm→+∞pm(ℓ−1)N(p).
m=1
g) When ℓ>2k, one has S(Q,N)=S(N)+o(1) when Q→+∞, where
S(N)=p∏(1+m⩾1∑A(pm))
the product being taken over the prime numbers is absolutely convergent. h) Let p be a prime number. Let ν⩾0 be the integer such that k=pνr where p does not divide r. Let us denote by γ the integer
{ν+1 if p⩾3
ν+2 otherwise.
If there exist integers (n1,...,nℓ) such that p does not divide all the ni and such that
x1k+⋯+xℓk≡0 mod.pγ,
then
limm→+∞pm(ℓ−1)N(pm)>0.
i) There exists ℓ>k such that, if N is sufficiently large, then S(N)>0. (Use the preceding question and question h) of Exercise 33.)
j) There exists an integer ℓ>k such that rk,ℓ(N)⩾1 for every integer N⩾1.
Nothing.
Nothing.