LECTURES ON THE EDGE-OF-THE-WEDGE THEOREM
(b) If
Rn
b
for some fixed b and all ae = ÷ + iy with y K, then the same inequality holds for
all y€L·
Proof. Fix 0. Define a measure ì on Rn by Üì = | cfc. Then y e is a
continuous map of Ê into L (ì) . The compactness of X implies that there are fin-
itely many points yx, · · · , y in Ê such that to every ã Å Ê corresponds at least
one y. with ||e - e || e, the norm being that of L1^)* Since each e is in Ll(y),
there is a compact ball  in Rn such that
J e
#
ø £ for 1 ß m,
where ä is the complement of ß. Hence
(1) J e ø 2e for every y K.
Â' ¾
Let Õ be the set of all y Rn for which the inequality (1) holds. Then Õ is convex.
To see this, suppose y', y " Y, 0 ó 1, y = ( l ó) y' + ay" . Since
Holder's inequality (with (1 ó) and ó~é as conjugate exponents) shows that
y 6 Y. Thus (1) holds for every ã G K.
If now y and y ' are in K, (1) implies that
e - e ,
I y y
J ^+ /l«
y
-v
1

Since ì(â ) « and e , —- e uniformly on ß as y ' —» y, it follows that |je
#
e |j 0
as y'—*y. This proves (a).
To prove (b), observe first that (a) implies that the function Uy defined on Rn + iK
by
(2) l/U) = J" Â(ß)â
é,-*þ
R"
is continuous and bounded. Let Õ be the set of all y X for which
(3) sup | l / ( * + i y ) | b.
X
Since Ê C Õ by assumption, it is enough to show that Õ is convex. So suppose
Previous Page Next Page