Proof of Theorem 5.
Necessity. Suppose that operator (5) is bounded from lp,v to lq,u that equivalently means the validity of the following inequality:
(44)∥Sn-1*f∥q,u≤C∥f∥p,v, f≥0.
Then due to (32) inequality (14) holds for any r=0,1,…,n-1. Therefore, by Lemma 7 we have Br(n)≪C, 0≤r≤n-1, that means B(n)≪C, where C is the best constant in (44); that is, C= ∥Sn-1*∥.
Sufficiency. will be proved by the induction method.
For n=1 the operator Sn-1*=S0* is the Hardy operator. Thus by Lemma 3 if B0(1)<∞, then the operator S0* is bounded from lp,v to lq,u with the estimate ∥S0*∥≪B0(1).
Next we assume that for n=1,2,…,l, 1≤l<n-1, if B(n)<∞, then the operator Sn-1*, n=1,2,…,l, is bounded from lp,v to lq,u with the estimate ∥Sn-1*∥ ≪B(n).
Now we need to prove that for n=l+1 if B(l+1)=max0≤r≤lBr(l+1)<∞, then the operator Sl* is bounded from lp,v to lq,u with the estimate ∥Sl*∥ ≪B(l+1).
Let f≥0 and Tj={k∈ℤ:2-k≤(Sl*f)j}, j∈ℕ. Suppose that kj=infTj for Tj≠∅ and kj=∞ for Tj=∅. When kj<∞, then 2-kj≤(Sl*f)j<2-kj+1. Further, for exception of the trivial case we suppose that (Sl*f)1>0.
Let m1=1 M1={j∈ℕ:kj=km1}. Assume that supM1+1=m2. If supM1<∞, then m2<∞ and supM1=maxM1=m2-1. It is obvious that m1<m2. Suppose that we have found m1<m2<⋯<ms<∞, s>1. Then we determine ms+1 as ms+1=supMs+1, where Ms={j∈ℕ:kj=kms}. By the definition 2-kms≤(Sl*f)j<2-kms+1, ms≤j≤ms+1-1, and kms<kms+1, s≥1.
Further, for convenience let kms=ns and ms+1-1=ms+1′; then
(45)2-ns≤(Sl*f)j<2-ns+1, ms≤j≤ms+1′,
and ns<ns+1, s≥1.
Let ℕ0={s∈ℕ:ms<∞}. Then from (45) it follows
(46)ℕ=⋃s∈ℕ0[ms,ms+1).
There are two possible cases: (1) ℕ0=ℕ and (2) ℕ0={1,2,…,s0}, s0>1.
Case (1) ℕ0=ℕ. Since ns<ns+1, s≥1, then -ns-1≤-ns+2+1. Therefore, using (45) and (6), we have
(47)2-ns-1=2-ns-2-ns-1≤2-ns-2-ns+2+1≤(Sl*f)ms+1′-(Sl*f)ms+2=∑i=ms+1′ms+2′Al,1(i,ms+1′)fi +∑i=ms+2∞[Al,1(i,ms+1′)-Al,1(i,ms+2)]fi≤∑i=ms+1′ms+2′Al,1(i,ms+1′)fi+∑i=0l-1Al,r+1(ms+2,ms+1′) ×∑i=ms+2∞Ar,1(i,ms+2)fi.
Using (45), (46), and (47), we get
(48)∥Sl*f∥q,uq=∑s≥1 ∑i=msms+1′uiq(Sl*f)iq<∑s(2-ns+1)q∑i=msms+1′uiq=22q∑s(2-ns-1)q∑i=msms+1′ui≪∑s(∑i=ms+1′ms+2′Al,1(i,ms+1′)fi)q∑i=msms+1′uiq +∑r=0l-1∑sAl,r+1q(ms+2,ms+1′) ×(∑i=ms+2∞Ar,1(i,ms+2)fi)q∑i=msms+1′uiq=Il+∑i=0l-1Ir.
First we estimate Il. Using Hölder’s inequality twice, we have
(49)Il≤∑s(∑i=ms+1′ms+2′Al,1p′(i,ms+1′)vi-p′)q/p′ ×∑i=msms+1′uiq(∑i=ms+1′ms+2′(vifi)p)q/p≤(∑s(∑i=ms+1′ms+2′Al,1p′(i,ms+1′)vi-p′)q(p-1)/(p-q)nnnnn×(∑i=msms+1′uiq)p/(p-q))(p-q)/p ×(∑s∑i=ms+1′ms+2′(vifi)p)q/p≪B~l(p-q)/p∥f∥p,vq,
where
(50)B~l=∑s(∑i=ms+1′ms+2′Al,1p′(i,ms+1′)vi-p′)q(p-1)/(p-q) ×(∑i=msms+1′uiq)p/(p-q).
Next we need the following obvious inequality:
(51)Δi+(∑j=i∞Al,1p′(j,i)vj-p′) =Al,1p′(i,i)+∑j=i+1∞Δi+Al,1p′(j,i)vj-p′ ≥Al,1p′(i,i)+∑j=i+1ms+2′Δi+Al,1p′(j,i)vj-p′ =Δi+(∑j=ims+2′Al,1p′(j,i)vj-p′), ms+1′≤i≤ms+2′.
Now, taking into account the above inequality (51) and using (10) and (7), we estimate B~l:
(52)B~l≪∑s∑i=ms+1′ms+2′(∑j=ims+2′Al,1p′(j,i)vj-p′)p(q-1)/(p-q) ×(∑k=msms+1′ukq)p/(p-q)Δi+(∑j=ims+2′Al,1p′(j,i)vj-p′)≤∑s∑i=ms+1′ms+2′(∑j=i∞Al,1p′(j,i)vj-p′)p(q-1)/(p-q) ×(∑k=1iukq)p/(p-q)Δi+(∑j=i∞Al,1p′(j,i)vj-p′)≪∑i=1∞(∑j=i∞Al,1p′(j,i)vj-p′)p(q-1)/(p-q) ×(∑k=1iukq)p/(p-q)Δi+(∑j=i∞Al,1p′(j,i)vj-p′)=(Bl(l+1))pq/(p-q).
From (49) and (52) we have
(53)Il≪Blq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
Now we estimate Ir, r=0,1,…,l-1. Let ℕ1={k=ms+2:s∈ℕ}.
Assume that Δkq=Al,r+1q(ms+2,ms+1′)∑i=msms+1′uiq if k=ms+2∈ℕ1 and Δkq=0 if k∉ℕ1. Then
(54)Ir=∑k=1∞Δkq(∑i=k∞Ar,1(i,k)fi)q=∥Sr*f∥q,Δq, r=0,1,…,l-1.
The operator Sr* is the operator Sn-1* for n=r+1 and 1≤r+1≤l. Therefore, by our assumption we have ∥Sr*∥ ≪B~(r+1)=max0≤t≤rB~t(r+1). Hence,
(55)Ir≪max0≤t≤r B~t(r+1),
where
(56)(B~t(r+1))pq/(p-q)=∑i=1∞(∑j=i∞At,1p′(j,i)vj-p′)p(q-1)/(p-q) ×(∑k=1iAr,t+1q(i,k)Δkq)p/(p-q) ×Δi+(∑j=i∞At,1p′(j,i)vj-p′).
We estimate the expression ∑k=1iAr,t+1q(i,k)Δkq:
(57)∑k=1iAr,t+1q(i,k)Δkq =∑ms+2≤iAr,t+1q(i,ms+2)Al,r+1q(ms+2,ms+1′)∑j=msms+1′ujq
(use the left side of (6) and nonincreasing of Al,t+1q(i,j) in the second argument)
(58)≤∑ms+2≤iAl,t+1q(i,ms+1′)∑j=msms+1′ujq≤∑ms+2≤i ∑j=msms+1′Al,t+1q(i,j)ujq≤∑j=1iAl,t+1q(i,j)ujq.
Substituting the obtained estimate in (56), we have B~t(r+1)≤Bt(l+1). Therefore, from (55) we get
(59)Ir≪max0≤t≤r Btq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq, lr=0,1,…,l-1.
Then from (48), (53), and (59) we obtain
(60)∥Sl*f∥q,u≪B(l+1)∥f∥p,v,(61)∥Sl*∥≪B(l+1).
Now we turn to case (2) ℕ0={1,2,…,s0}, 1≤s0<∞. In this case we have ms0<∞ and ms0+1=∞. Here there are two possible cases: ns0<∞ and ns0=∞.
Below we suppose that ∑s=kt=∑s=1t if k<0.
Let ns0<∞. Then
(62)∥Sl*f∥q,uq=∑s=1s0 ∑j=msms+1(Sl*f)jquj=∑s=1s0-2 ∑j=msms+1′(Sl*f)jqujq+∑j=ms0-1ms0′(Sl*f)jqujq +∑j=ms0∞(Sl*f)jqujq=J1+J2+J3.
If J1≠0, then s0>2, and as in the case ℕ0=ℕ using (47), we estimate J1 and as a result we get
(63)J1≪Bq(l+1)∥f∥p,vq.
Using (45) and Hölder’s inequality, we estimate the value J2:
(64)J2<2q(2-ns0-1)q∑j=ms0-1ms0′ujq ≪(∑i=ms0′∞Al,1(i,ms0′)fi)q∑j=ms0-1ms0′ujq ≤(∑i=ms0′∞Al,1p′(i,ms0′)vi-p′)q/p′ ×∑j=ms0-1ms0′ujq(∑i=ms0′∞|vifi|p)q/p ≤[(∑i=ms0′∞Al,1p′(i,ms0′)vi-p′)q(p-1)/(p-q)nnnnnn×(∑j=ms0-1ms0′ujq)p/(p-q)](p-q)/p∥f∥p,vq.
Using (7) and (10), we estimate the expression in the square brackets:
(65)(∑i=ms0′∞Al,1p′(i,ms0′)vi-p′)q(p-1)/(p-q)(∑j=ms0-1ms0′ujq)p/(p-q) ≪∑i=ms0′∞(∑j=i∞Al,1(j,i)vj-p′)p(q-1)/(p-q) ×Δi+(∑j=i∞Al,1(j,i)vj-p′)(∑j=ms0-1ms0′ujq)p/(p-q) ≤∑i=ms0′∞(∑j=i∞Al,1(j,i)vj-p′)p(q-1)/(p-q) ×(∑j=1iujq)p/(p-q)Δi+(∑j=i∞Al,1(j,i)vj-p′) ≤(Bl(l+1))pq/(p-q).
Therefore,
(66)J2≪(Bl(l+1))q∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
Now we estimate J3. Since 2-ns0≤(Sl*f)j<2-ns0+1 for all j≥ms0, then (Sl*f)j≤2(Sl*f)t for all j,t≥ms0. Hence,
(67)J3≤supt≥ms0∑j=ms0t(Sl*f)jqujq≤2qsupt≥ms0(∑i=t∞An-1,1(i,t)fi)q∑j=ms0tujq.
Further, as for the estimate of J2 by Hölder’s inequality, we have
(68)J3≪supt≥ms0[(∑i=t∞Al,1q(i,t)vi-p′)q(p-1)/(p-q)nnnnnnnnn×(∑j=ms0tujq)p/(p-q)](p-q)/p∥f∥p,vq≪Blq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
From (62), (63), (66), and (68) we have (60) and (61). If ns0=∞ that means kms=∞, then Tj=∅ for j≥ms0; that is, (Sl*f)j=0 for j≥ms0. By the assumption (Sl*f)1>0, therefore, s0>1. Then m2<∞ and s0≥2. Hence,
(69)∥Sl*f∥q,uq=∑s=1s0-2 ∑j=msms+1′(Sl*f)jqujq+∑j=ms0-1ms0′(Sn*f)jqujq=J1+J2.
This, together with estimates (63) and (66), gives (60) and (61). Thus, if B(l+1)<∞, then the operator Sl* is bounded from lp,v to lq,u and estimate (61) holds. Consequently, for any n≥1 from B(n)<∞ we have that the operator Sn-1* is bounded from lp,v to lq,u and the estimate ∥Sn-1*∥ ≪B(n) holds. The last estimate and the estimate ∥Sn-1*∥ ≫B(n) obtained in the necessity part yield ∥Sn-1*∥ ≈B(n).