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