In this section we characterize a wide set of functions for which strong summability holds at each Lebesgue point. For the convergence of f∈W(Lp,lq)(R) (1<p<∞, 1≤q<∞) at pLebesgue points we proved the following result in [21]. Note that W(Lp,lp)(R)=Lp(R).
Obviously, the convergence holds almost everywhere. Corollary 4 does not hold for p=1 (see Hardy and Littlewood [5]). However, we [21] extended it for p=1, but for much more specialized points than the Lebesgue points, for the socalled Gabisoniya points, which were introduced in [9]. In the next theorem we generalize Theorem 3 and Corollary 4 for p=1 and for a subspace of W(L1,lq)(R).
Proof.
It is easy to see that
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1
T
∫
0
∞
θ
′
t
T
∏
j
=
1
d
s
t
f
x
j

f
x
j
d
t
=

1
T
·
∫
0
∞
θ
′
t
T
∏
j
=
1
d
1
2
π
∫
R
f
x
j

s
j
D
t
s
j
d
s
j

f
x
j
d
t
=

1
T
∫
0
∞
θ
′
t
T
∏
j
=
1
d
lim
n
→
∞
1
2
π
·
∫

n
n
f
x
j

s
j

f
x
j
D
t
s
j
d
s
j
d
t
.
Since(32)∫nnfxjsjDtsjdsj≤CtfWL1,lq,∫nnfxjDtsjdsj=fxj∫nnsintsjsjdsj=fxj∫ntntsinuudu≤Cfxj,we obtain
(33)

1
T
∫
0
∞
θ
′
t
T
∏
j
=
1
d
s
t
f
x
j

f
x
j
d
t
=
lim
n
→
∞

1
T
·
∫
0
∞
θ
′
t
T
·
1
2
π
d
∫

n
,
n
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=
1
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f
x
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lim
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∫

n
,
n
d

1
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∫
0
∞
θ
′
t
T
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=
1
d
f
x
j

s
j

f
x
j
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t
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d
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=
1
2
π
d
/
2
∫
R
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1
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.
For simplicity we will prove the rest of the theorem for d=3, only. It can be proved for higher dimensions similarly. As we mentioned earlier, we may suppose that s1>s2>s3>0 and s1s2s3>0. Let us fix a small r>0 and denote the square [0,r/2]3 by Sr/2. We can prove in the same way as we did in Theorem 4 of [21] that(34)∫Sr/2∏j=13fxjsjfxjKTθsds<ϵif r is small enough and T is large enough. The estimation of this integral on the set Sr/2c of that proof does not work now, because there we used a modified maximal function which is not necessarily bounded in our case. So we have to show here that(35)limT→∞∫Sr/2c∏j=13fxjsjfxjKTθsds=0.
To this end let us introduce the sets(36)A1≔s:s1>r2, 0<s3<s2<1T,A2≔s:s1>r2, 0<s3<1T<s2<δ,A3≔s:s1>r2, 1T<s3<s2<δ,A4≔s:s1>r2, 0<s3<1T<δ<s2,A5≔s:s1>r2, 1T<s3<δ<s2,A6≔s:s1>r2, 1T<δ<s3<s2,B1≔s:0<s1s2s3<1T,B2≔s:1T<s1s2s3<δ,B3≔s:δ<s1s2s3<s1s22,B4≔s:s1s22∨δ<s1s2s3<s1s2δ,B5≔s:s1s2δ∨δ<s1s2s3<s1s2,C1≔s:0<s1s2<δ,C2≔s:δ<s1s2<s12,C3≔s:s12<s1s2<s1δ,C4≔s:s1δ<s1s2<s1for a given small δ>0 and large T. Then we have to estimate the integral in (35) for Ai, i=1,…,6. First of all observe that Ai∩Bj=∅ for i=1,2,3 and j=1,2. On the set A1 we have s2,s3<1/T and so s1s2s3>s1/2. Hence(37)KTθs≤CT2αs11s1s2s3α≤CT2αs11α.Since f is locally bounded at xj, we get by (37) that(38)∫A1∩B3∪B4∪B5∏j=13fxjsjfxjKTθsds≤CT2α∫r/2∞∫01/T∫01/Ts11α∏j=13fxjsjds≤CTα∑i=0∞i∨11α∫ii+1fx1s1ds1≤CTαfWL1,l∞⟶0,as T→∞. On A2 we have(39)KTθs≤CT1αs11s21s1s2s3α≤CT1αs11αs21and so(40)∫A2∩B3∪B4∪B5∏j=13fxjsjfxjKTθsds≤CT1α∫r/2∞∫1/Tδ∫01/Ts11αs21∏j=13fxjsjds≤CTαlnT∑i=0∞i∨11α∫ii+1fx1s1ds1≤CTαlnTfWL1,l∞⟶0.Similarly,(41)∫A3∩B3∪B4∪B5∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫1/Tδ∫1/Tδs11αs21s31∏j=13fxjsjds≤CTαln2TfWL1,l∞⟶0.
On the set A4∩B1 we have s1s2<2/T<δ if T is large enough and so s2>s1/2. Then(42)KTθs≤CTs11s21≤CTs12,(43)∫A4∩B1∏j=13fxjsjfxjKTθsds≤CT∫r/2∞∫s1δs1∫01/Ts12∏j=13fxjsjds≤C∑i=0N01i∨12·∫ii+1∫s1δs1fx1s1fx2s2ds1 ds2+C∑i=N0∞i2·∫ii+1∫s1δs1fx1s1fx2s2ds1 ds2≤Cf×f1s1,s2:0<s1<N0,s1δ<s2≤s1WL1,l∞+CN01f×fWL1,l∞.The second term is less than ϵ if N0 is large enough and the first term is less than ϵ if δ is small enough. The set A4∩B2 can be handled in the same way.
On A4∩(B3∪B4∪B5) we have s1s2>δ. Thus (B3∪B4∪B5)∩C1=∅ and s1s2s3>s1s21/T>(s1s2)/2. Then(44)KTθs≤CT1αs11s21s1s2s3α≤CT1αs11s21s1s2α.On C2 we have(45)KTθs≤CT1αs12s1s2α≤CT1αs11ηs1s21α+η,where η is chosen such that 0<η<1∧α. Hence(46)∫A4∩B3∪B4∪B5∩C2∏j=13fxjsjfxjKTθsds≤CT1α∫r/2∞∫s1/2s1δ∫01/Ts11ηs1s21α+η·∏j=13fxjsjds≤CTα∑i=0∞i∨11η·∑0≤j<i+1/2j∨11α+η·∫ii+1∫s1j1s1jfx1s1·fx2s2ds1 ds2≤CTαfWL1,l∞2⟶0.On C3, (44) implies that(47)KTθs≤CT1αs11αs21≤CT1αs11α+ηs21η,∫A4∩B3∪B4∪B5∩C3∏j=13fxjsjfxjKTθsds≤CT1α∫r/2∞∫δs1/2∫01/Ts11α+ηs21η∏j=13fxjsjds≤CTα∑i=0∞i∨11α+η∑0≤j<i+1/2j∨11η·∫ii+1∫jj+1fx1s1fx2s2ds1 ds2≤CTαfWL1,l∞2⟶0.Observe that s2>δ contradicts C4.
Consider the set A5∩B1. If s1s2>2/T, then s3>s1s21/T>(s1s2)/2 and if s1s2<2/T, then s3>1/T>(s1s2)/2. Observe that s1s2<1/T+δ<2δ. Hence(48)KTθs≤Cs11s21s31≤Cs12s1s21/2s31/2.Since s3<δ, we can integrate in s3 to obtain(49)∫A5∩B1∏j=13fxjsjfxjKTθsds≤C∫r/2∞∫s12δs1∫s1s21/Ts1s2s12s1s21/2·s31/2∏j=13fxjsjds≤C∫r/2∞∫s12δs1s12fx1s1fx2s2ds1 ds2≤Cf×f1s1,s2:0<s1<N0,s12δ<s2≤s1WL1,l∞+CN01f×fWL1,l∞as in (43). Similarly, s1s2<2δ holds as well on A5∩B2. If s1s2s3>s3, then(50)KTθs≤CTαs11s21s31s1s2s3α≤CTαs12s31αand so(51)∫A5∩B2∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫s12δs1∫1/Tδs12s31α∏j=13fxjsjds≤C∫r/2∞∫s12δs1s12fx1s1fx2s2ds1 ds2<ϵas before. If s3>s1s2s3, then(52)KTθs≤CTαs12s1s2s31α,∫A5∩B2∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫s12δs1∫s1s2δs1s21/Ts12s1s2s31α·∏j=13fxjsjds≤C∫r/2∞∫s12δs1s12fx1s1fx2s2ds1 ds2<ϵ.If s1s2<2δ, then the integral on the set A5∩(B3∪B4∪B5) can be estimated in the same way. If s1s2>2δ, then A5∩B3=∅ because s3>(s1s2)/2 and s3<δ on this set. Writing 1/T instead of δ in the sets Bi (i=4,5), we obtain the definition of the sets Bi′. On B4′ we have by (50) that(53)KTθs≤CTαs11s21s31s1s2α,∫A5∩B4′∩C2∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫s1/2s1δ∫1/Tδs12s31s1s2α·∏j=13fxjsjds≤CTαlnT∑i=0∞i∨11η·∑0≤j<i+1/2j∨11α+η∫ii+1∫s1j1s1jfx1s1·fx2s2ds1 ds2≤CTαlnTfWL1,l∞2⟶0,∫A5∩B4′∩C3∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫δs1/2∫1/Tδs11αs21s31∏j=13fxjsjds≤CTαlnT∑i=0∞i∨11α+η∑0≤j<i+1/2j∨11η·∫ii+1∫jj+1fx1s1fx2s2ds1 ds2≤CTα·lnTfWL1,l∞2⟶0.Moreover, B5′ contradicts A5, more exactly, s3>1/T.
On A6 s3>δ and s1s2s3>0, thus s1s2>δ. Similar to (48),(54)KTθs≤Cs11s21s31≤Cs12s1s21on A6∩B1∩C2. Then
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∫
A
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B
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=
1
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which is small enough if N0 is large and δ is small enough. On A6∩B1∩C3 we use the estimation(56)KTθs≤Cs11s21s31≤Cs12s21to obtain
(57)
∫
A
6
∩
B
1
∩
C
3
∏
j
=
1
3
f
x
j

s
j

f
x
j
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θ
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d
s
≤
C
∫
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2
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s
1
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2
∫
s
1

s
2

1
/
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s
1

s
2
s
1

2
s
2

1
∏
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=
1
3
f
x
j

s
j
d
s
≤
C
f
×
f
×
f
1
s
:
0
<
s
1
<
N
0
,
δ
<
s
2
≤
s
1
/
2
,
s
1

s
2

δ
<
s
3
<
s
1

s
2
W
L
1
,
l
∞
+
C
N
0

1
/
2
f
×
f
×
f
W
L
1
,
l
∞
<
ϵ
,
as just before.
On A6∩B2 s1s2>δ again. If s1s2>2δ, then s3>s1s2δ>(s1s2)/2 and if s1s2<2δ, then s3>δ>(s1s2)/2. This case can be handled in the same way as the set A6∩B1.
On A6∩B3 we get(58)KTθs≤CTαs11s21s31s1s2s3α≤CTαs11s21s1s21s1s2s3α,which implies that(59)∫A6∩B3∩C2∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫s1/2s1δ∫s1s2/2s1s2δs12s1s21·s1s2s3α∏j=13fxjsjds≤CTα∑i=0∞i∨11μ∑0≤j<i+1/2j∨11μ·∑0≤k<j+1/2k∨11α+2μ·∫ii+1∫s1j1s1j∫s1s2k1s1s2kfx1s1fx2s2·fx3s3ds1 ds2 ds3≤CTαfWL1,l∞3⟶0,where 0<μ<1/2∧α/2. Similarly,(60)∫A6∩B3∩C3∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫δs1/2∫s1s2/2s1s2δs12s21s1s2s3α·∏j=13fxjsjds≤CTα∑i=0∞i∨11μ·∑0≤j<i+1/2j∨11μ∑0≤k<j+1/2k∨11α+2μ·∫ii+1∫jj+1∫s1s2k1s1s2kfx1s1fx2s2·fx3s3ds1 ds2 ds3≤CTαfWL1,l∞3⟶0.
Moreover, on A6∩B4(61)KTθs≤CTαs11s21s31s1s2α,∫A6∩B4∩C2∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫s1/2s1δ∫δs1s2/2s12s1s2α·s31∏j=13fxjsjds≤CTα∑i=0∞i∨11μ·∑0≤j<i+1/2j∨11α+2μ∑0≤k<j+1/2k∨11μ·∫ii+1∫s1j1s1j∫kk+1fx1s1fx2s2·fx3s3ds1 ds2 ds3≤CTαfWL1,l∞3⟶0,∫A6∩B4∩C3∏j=13fxjsjfxjKTθsds≤CTα∫r/2∞∫δs1/2∫δs1s2/2s11αs21s31∏j=13fxjsjds≤CTα∑i=0∞i∨11α/2∑0≤j<i+1/2j∨11α/2·∑0≤k<j+1/2k∨11α/2∫ii+1∫jj+1∫kk+1fx1s1·fx2s2fx3s3ds1 ds2 ds3≤CTαfWL1,l∞3⟶0.Since A6∩B5=∅, the proof of the theorem is complete.