4 ILHAM A. ALIEV, BORIS RUBIN, SINEM SEZER, AND SIMTEN B. UYHAN
then
(2.5)
100 100
Waf(x,
t)-
=
lim Waf(x,
t)-
=
cl-'f(x)
0
t
e--+0 e
t
where the limit exists in the Lp-norm and pointwise for almost all x. Iff
E
Co,
this limit is uniform on Rn.
PROOF.
Consider the truncated integral
(2.6)
1
00
dt
lef(x)
=
Waf(x, t)-,
e
t
c 0.
Our aim is to represent it in the form
(2.7) lef(x)
=
1
00
Qesf(x) e-aes k(s) ds
where
(2.8)
k
E L1(0, oo) and
Once (2.7)
is
established, all the rest follows from properties (a)-(c) in Definition
2.1 according to the standard machinery of approximation to the identity; see (St].
Equality (2.7) can be formally obtained by changing the order of integration,
namely,
roo
100
dt
lef(x)
=
Jo
dJJ.(TJ) e Qtf1f(x) e-atf1
t
roo dJJ.(TJ)
100
Qesf(x) e-aes ds
lo
11
s
=
1
00
Qesf(x) e-aesk(s) ds, k(s)
= s-
1
1 8 dJJ.(TJ).
Furthermore, since JJ.([O, oo))
=
0, then
1oolk(s)ids
1111s
dJJ.(TJ)~~s
+
1ool1oo
dJJ.(TJ)~~s
r1
diJJ.I(TJ)
11
ds
+
roo diJJ.I(TJ)
[11
ds
lo
11
s
J1 J1
s
1
00
llogTJidiJJ.I(TJ) oo.
Similarly we have
100 100
1
k(s)ds
=
log- dJJ.(TJ)
=
cl-',
0 0
11
which gives (2.8). Thus, to complete the proof, it remains to justify application of
Fubini's theorem leading to (2.7). To this end, it suffices to show that the repeated
integral
is finite for almost all x in Rn. We write it as A(x)
+
B(x), where
1
00
dt r1/t
A(x)
=
e
t
Jo
1Qtf1f(x)l diJJ.I(TJ),
1
00
dt
1
00
B(x)
= -
1Qt11 f(x)l diJJ.I(TJ).
e
t
1/t
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