AN HARMONIC ANALYSIS FOR OPERATORS
OPERATORS ON HilBERT SPACE AND
ANALYTIC OPERATORS'
II:
BY
KAREL DELEEUW
1. Introduction
We continue in this paper investigating the harmonic analysis for operators
introduced in [1J. In that paper, we associated to each bounded linear operator
T on a homogeneous Banach space B on the circle group T a formal Fourier
series,
+0Cl
(l.l)
T '"
t
Each
(n) is a bounded
equation
fen)
linear operator
T(n)Rr
where Rt is the translation
L
-00
=
eintRt
on B which satisfies the functional
l' (n),
operator defined on B by
(RJ)(s)
= f(s -
t),
Equivalently, T(n) is an operator of the form Mn
of multiplication by eill'
(1.2)
t E T,
(Mnf)(t)
= einy(t),
sET.
0
U, where Mn is the operation
t E T,
and U is an invariant operator, that is, one which commutes with translation,
so that URt = RtU, t E T.
1n [1] we established formal properties of the series (1.1) and showed that it is
C-l summable to T in the strong operator topology. In addition, we obtained
an extension to operators of the theorem of F. and M. Riesz which asserts that
a measure whose Fourier-Stieltjes transform vanishes on the negative integers
must be absolutely continuous.
In this paper we restrict ourselves to the case where B is L2(T), the space of
square integrable functions on the circle group. In Section 2 we state the formal
properties of the Fourier series relative to the adjoint operation and establish
an analogue for operators of the Parseval formula for functions. In Sections
3 and 4 we extend to operators the notions of support and analyticity. Finally,
in Section 5, we establish an analogue for operators of the second theorem of
F. and M. Riesz. This classical theorem asserts that a function having all its
Received October 3, 1975.
1 This research was supported in part by a National Science Foundation grant.
164
OPERATORS
ON HILBERT
165
SPACE
negative Fourier coefficients zero cannot vanish on a subset ofT having positive
Lebesgue measure unless it is zero almost everywhere.
2. Operators on Hilbert space-Parseval
formula
Throughout this paper we shall be concerned with the algebra ,51! of bounded
linear operators on the Hilbert space L2(T) of square integrable functions on the
unit group T. We shall denote by ( , ) the usual inner product on L2(T),
C/, g)
and by II
=
f
dt,
7! ~r+It/(t)g(t)
-rc
,(, g E
L~(T),
II the usual norm on VeT),
IIfll = U;/)1!2,
IE L2(T).
In this section we first summarize the main definitions and results from [1]
which are needed in what follows. We then show how the adjoint operation of
2? is related to our Fourier series for operators in ,51! and finally establish a
formula for operators in !l! which is an analogue of the Parseval formula for
functions.
Using the notation of [lJ, !f! 0 is the subspace of 2! consisting of operators T
which are invariant, Le., which commute with translation. It is well known (see
Cpapter ] 6 of [2J) that the operators in 2:0 are precisely those \vhich have the
exponentials {ein': 11 E Z} as eigenfunctions. They can also be characterized in
terms of the Fourier transform as follows. If the Fourier transform on L2(T)
is defined by J\ so that
fen)
,
=-
2n
-It
1 f+Jt
e-IIl'l'(t)
dt,
nEZ,
.
then an operator T in !1! is in 2:0 if and only if there is a bounded function ¢
/'-.
~
on Z so that Tf(n) = ¢(n)f{n), n E Z. (For this reason the operators in 2:0
are usually called multipliers.) In particular, !1! 0 is a commutative subalgebra
of 2:.
For each positive integer fl, 2?n is defined in [1] to be the closed linear sub~
space of ,ft' consisting of all operators T which satisfy the functional equation
TRt = eintRtT, t E T.
For each positive integer fl, the operator n" in 2! is defined in [1J in terms of
vector valued integrals:
[n,,(T)](/)
f
2n
=~
-It e-illfR_tTRJ
f+Jt
dt,
/E
L2(T)
If T E 2:, its Fourier transform is defined in [1] to be the Sf-valued function
defined on Z by
(2.] )
T(n)
=
n,,(T),
nEZ,
166
KAREL DELEEUW
and the formal series
+w
L
(2.2)
-00
T(n)
is called the Fourier series of T.
The following summarizes the main facts concerning the Fourier transform
which we will need in what follows.
2.1. (i) For each 11, nn is a projection of 2 onto 2n"
Oi)
An operator S in :e is in 2n if and only if it is of the form S = Mn 0 U,
where U is in an invariant operator and M" is multiplication by the function ein>,
Equivalently, S is in 2 n if and only if, for each 171, S(eim') is a constant multiple of
PROPOSITION
ei(n+m»,
(Hi) If T
E
2', the Fourier series of T is C-l summable to T in the strong
f
operator topology of .sf; equivalently, for each f E L!(T), the series L~~ (n)f
is C-l summable to f in the norm topology of L2(T).
then the series L:::::':~ S(n - m)f(m)
is C-l
(iv) If S, T E 2, nEZ,
....•.•
summable to ST(n) in the strong operator topology of .sf.
Proof
0), (ii), (iii), and (iv) follow immediately from Proposition 3.2,
Corollary 2.4, Proposition 3.8, and Proposition 4.2, respectively, of [1].
We next establish the simplest formal properties of Fourier series of operators
in .sf with respect to the adjoint operation. We first show that the adjoint
operation preserves the graded structure we have introduced in .sf, in particular,
= -n0
that
2: 2
LEMMA
Proof
2.2.
If T
2n> then T*
E
E
2' -n-
Let T E 2'n, t E T. Then
R_tT*Rt
Thus T* E 2'
= R:T*R~t =
(R_tTRt)*
= (eintT)* = e-intT*.
-n0
We next establish an analogue for operators of the fact that the nth Fourier
coetlicient of the complex conjugate of a function is the complex conjugate of
f
the ( - n )th coefficient
PROPOSITION
2.3.
off
Let T E
2, n E Z.
-"
T*(n) =
Proof
Then 7rn(T*)
=
7r_n(T)*,
[f( -n)]*.
Let}; g E L2(n). Then
([nn(T*)](f),
g)
,2n f+~
-1t
= (i.
=::)
(If
•.•7r
e-ins[R_sT*RsJfds,
e-III''[RsTR_sJ*f
-it
+1t
.
g)
ds, g .
)
i.e.,
OPERATORS
167
ON HILBERT SPACE
Thus,
([n,,(T*)J(f),
g)
=
f -;t
2n
(_1
2n
= ~
=
2n
_1
eim[R_tT
f
_IT
+1!
(f,
g)
dt,
g) dt
eillt(f, [R-tTRtJg)
_IT
1 f+1t
(
Rt]~f
_IT ei"t([R_tTRt]*f,
f+rr
f,-2n
=
=
+1!
etnTR_tTRtJgdt
.
dt
)
[lL,,(T)Jg).
Sincefand 9 were arbitrary in L2(1T), this shows that 1Tn(T*) = n-n(T)*.
As a corollary we obtain the fact that an operator in ::e is self-adjoint if and
only if its Fourier transform is "hermitian."
COROLLARY
2.4.
(2.3)
Let T E::e.
T( - n)
Then T is self-adjoint
=
[1'(n)]*
for all n E
if and only if
Z.
Proof
The necessity of (2.3) for se1f-adjointness follows from Proposition
2.3. Suppose now that (2.3) holds for an operator T in::e. Then, as a consequence of Proposition 2.3, each C-l sum of the Fourier series of T will be selfadjoint. That T is self-adjoint now follows from Oii) of Proposition 2.L
We next show how Proposition 2.3 leads to an analogue for operators of the
Parseval formula for functions.
THEOREM
2.5.
Let T E::e.
Then
+00
=
no(T*T)
with convergence
Proof
L
[nr(T)]*nr(T),
-if)
in the strong operator topology of ::e.
By (iv) of Proposition 2.1, the series
+w
(2.4)
L
1T_r(T*)1rr(T)
-OC)
is C-l summable to 1To(T*T) in the strong operator topology.
Proposition 2.3, the series (2.4) is the same as
+00
(2.5)
L
-w
[nr(T)]*1rr(T).
If for each positive integer N we define
SN
to be the Nth partial sum
+N
L
-N
[nr(T)]*TCr(T)
Because of
168
KAREL DELEEUW
of the series (2.5), we have
(2.6)
Sl
S S2 S S9 S ... ,
s
where
is the usual partial ordering of Y:' (see Chapter VII of [4]). The sequence (2.6) is bounded from above in this ordering by TJ:o(T*T)
since the
series (2.5) is C-l summable to TJ:(}(T*T) in the strong operator topology. The
theorem on p. 263 on [4J asserts that a monotonic sequence of self-adjoint
operators bounded from above must converge in the strong operator topology.
This shows that the series (2.5) converges to some element V of Y:' in the strong
operator topology. But we know that it is C-l summable in that topology to
TJ:o(T*T)
and thus TJ:o(T*T) = V. This completes the proof of Theorem 2.5.
3. Support and cosupport
In this section we define the two notions of support and cosupport for operators in Sf. These reduce to the usual notion of support when the operator is
multiplication by a function. We show that a set is a support for an operator T
if and only if it is a co support for its adjoint T*.
Let.M be a measurable subset ofT. We will denote by MC its complement in
T and by L2(lvl) the subs pace of .e(T) consisting of those functions "supported"
by.!vI:
L2(Af)
=
{f:f
E L2(T),f
= 0 a.e. on
MC}.
Let T E 2. We would like to define the support of T to be the smallest
measurable subset K of T satisfying:
f = 0 a.e. off K implies Tf
= 0 a.e.
for all f E L2(T);
define the cosupport of T to be the smallest measurable subset K of T satisfying:
Tf
= 0 a.e. off
K
for all f
E
L2(T).
Unfortunately, because of sets of measure 0, these definitions do not make sense,
so we must proceed indirectly.
We define a measurable subset K of T to be a supporting set for T if
T(L2(KC»
S; {O}. Let Ii be Lebesgue measure on T. We define the constant ST
by
ST
3.1.
PROPOSlTION
=
inf {f'(K): K a supporting set for T}.
Let T
E
fE.
Then there is a measurable subset K of T
sat isfying the following:
(i)
(ii)
(iii)
K is a supporting set for T.
fl(K)
=
If
is
j1,f
ST'
any supporting set for T, f'{x: x E K, x
eft
A!}
=
O.
OPERATORS
ON HILBERT
169
SPACE
Proof.
Let Kj, K2' K3' ... be a sequence of supporting sets for T with
Sp A routine exhaustion argument shows that the set K defined by
K =: nl~~"
1 Km satisfies 0), Oi), and (iii).
Proposition
3.1 shows that T has a minimal supporting
set which is unique
up to a set of measure O. Such a set will be called a support for T.
We proceed similarly to define cosupport. A measurable subset K of T is
called a cosupporting set for T if T(L2(T» £: L2(K). We define the constant
p(Kn)
CST
-,
by
CST
=
inf
{fI(
K): K a cosupporting
3.2. Let T E .::t.
satisfying the following:
PROPOSITION
(i)
set for T}.
Then there is a rneasurable subset K of T
K is a co-supporting set for T.
=
(ii)
/l(I()
(iii)
If)\1 is any co-supporting set for T then
CST'
f1 {x:
X E K,
X if:
M}
= O.
Proof.
Similar to the proof of Proposition 3,1. Proposition 3.2 shows that
set which is unique up to a set of measure O.
Such a set will be called a cosupport for T.
Simple examples show that a support for an operator need not be a cosupport
for that operator. For example, if T E :J? is defined by
T has a minimal cosupporting
Tf
= [Jo" f(t)
df] g,
f E L2(T),
where g(t) = 1, all t E T, then [0, n] is a support for T while T is a cosupport.
However, we have the following.
PROPOSITION
3.3.
Let T
E
/i',
Ka measurable subset ofT.
Then thefollowing
are equivalent:
K is a support for T.
(ii) K is a cosupport for T *.
This proposition is an immediate consequence of the following.
(i)
LEMMA
3.4.
Let T
E :J?,
K a measurable subset ofT.
Then the following are
equiualent:
(i)
(ii)
K is a supporting set for T.
K is a cosupponing set for T*.
Proof (i) implies (ii). Letg E L2(T). We must show that T*g = 0 a.e. offK.
oft O.
But
If this were not so there would be some fE L2(KC) with (T*g,f)
= (g, Tf) = 0 since Tf = 0 for fE L2(KC) because of (1).
(T*g,f)
170
KAREL DELEEUW
(ii) implies (i). Letfe I3(KC). We must show that Tf = 0 a.e. If this were
not as so there would be some g E L2(T) with (Tj, g) ¥= O. But (Tj, g) =
(f, T*g) = 0 since T*g E I3(K) because of (it).
Finally, let us point out that the notion of support and cosupport reduce to
the usual notion of support in the case of the operation of multiplication by a
function.
element of
!f!
3.5.
Let
defined by
PROPOSITION
<p
(Tf)(x)
Then {x: x E
T,
¢(x)
be a bounded measurable function
=
¢(x)f(x)
on T and T the
a.e. for x E T.
=1=
O} is both a support and cosupport for T.
Straightforward.
Proof
4. Analytic operators
An operator T in !f! will be caUed analytic if all of its negative Fourier coefficients vanish, i.e., T(n) = 0, all n < O. We shall denote by .sd the set of
analytic operators in Sf.
In this section we establish several basic properties of analytic operators. In
particular, we show that s1 is the strongly closed subalgebra of 5£ generated by
!f! 0' the space of invariant operators, and the multiplication operator (lvJ1f)(t) =
eij'(t),
tE
T.
Suppose that TEfl! is the operation of multiplication by a bounded measurable function 4>:
(4.1)
(Tf)(x)
=
¢(x)f(x)
a.e. for x E T.
Our first result shows that analyticity of T is equivalent to cf> giving the boundary
values of a function bounded and analytic in the unit disk. (Or what is equivalent, ¢(n) = 0 for n < 0; see p. 39 of [3J.)
PROPOSITION
4.1.
measurable function
(i)
(ii)
Suppose that T E Sf is defined by (4.1) for
on T. Then the following are equivalent:
T is analytic.
¢(n)
if n
=°
<
a bounded
O.
Proof
This is immediate from Proposition 3.3 of [1], which states that
T(n) ::::;:¢(n)Jvln for aU nEZ, where Mn is multiplication by the function eino,
i.e., (AI"f)(t)
= ei,,~(t), t E T.
The next few results establish that s4 is a subalgebra of
operator topology.
PROPOSITION 4.2.
d is the subalgebra
of .9?
51!
closed in the strong
OPERATORS
ON HILBERT
171
SPACE
Proof
That sl is a Enear subspace of .f£' [oHows from the linearity of the
We will show that ST E .~. Let n < O. It
Fourier transform. Let S, Ted.
/'.
suffices to show that ST (n) = O. By (iv) of Proposition 2.1, the series
Sen - m)T(m) is C-l summable in the strong operator topology to
I:::~~
..•....
ST (n)./'..But each term of this series must be 0 since Sand T are in s:t and n < O.
d.
Thus ST(n) = 0, so ST E
A lemma is needed before we are able to show that
operator topology.
LEMMA
4.3.
Let S
.:t!. Then, for each rand m in
E
=
(Seimo, ei(r+m}o)
(4.2)
<r#
(S(r)eim.,
is closed in the strong
Z,
ei(r+m)o).
Proof
By (iii) of Proposition 2.1, the series 2:::;;;::::~S(n)eim• is C-l summabIe to Seirno in the norm topology of L2(T). Taking inner products \vith
ei(r+m)., we see that the series
n= + iX]
L
(4.3)
n= -(J)
(S(n)eimo,
ei(r+rn).)
is C-l sum mabIe to (Seimo, ei(Hm)o). By (ii) of Proposition 2.1, each term of the
series (4.3) with n =;6 r must be zero, which shows that equality (4.2) holds.
PROPOSITION
4.4.
d is closed in the strong
operator topology of
:£.
Proof
Let Tj be a net in ~c1converging to T E :£ in the strong operator
topology. Let I' < O. We must show that T(r) = O. Let m be an arbitrary
integer. Since linear combinations of exponentials are dense in L\T), it suffices
to show that T(r)eim• = O. And since, by (ii) of Proposition 2.1, T(r)ei/rh is a
multiple of ei(d m)., it suffices to show that
(4.4)
(T(r)eimo,
ei(r+m).)
=
O.
By Lemma 4.3 and the fact that Tj -j. T in the strong operator topology we have
(T(r)eim.,
(4.5)
ei(r+m).)
=
=
(Teim.,
=
11m
ei(r+m)')
Jim (reim•
,J
' i(r+m).)
j
j
(1'.(r)eim•
J .,
ei(r+m)o)
,
Since r < 0 and each T;. Ed, each of the terms (~(r)eim.,
ei(r+m)o) is zero.
Thus (4.4) is a consequence of (4.5). This concludes the proof of Proposition
4.4. (Note that the above actually proves that .sI is closed in the weak operator
topology of :£.)
The two results which follow show the extent to which
is generated by
easily described operators.
d
172
KAREL DELEEUW
Recall that the multiplication operator !vfl is defined on I!(T) by (Mjf)(t)
:::::
eil(t)~ t E T. We shall denote by .sf 0 the (algebraic) subalgebra of :E generated
by .M 1 and the operators in .!£ 0, the invariant operators.
4.5.
PROPOSITION
do s; d.
Proof. Because ~4 is an algebra, it suffices to show that M1 E sd and fI!0 S;
d. Proposition 3.3 of [1] shows that Mj(n) = 0 unless n = 1, so M1 Ed.
Let T E 5f 0' Let n be an arbitrary nonzero integer,f E L2(T). Then
[r(n)J(f)
=
[n:n(T)](f)
=
2n:
J..
fh
-T/
=-
2n:
-11:
I f+1!
2n:
= (1-
=
=
° for
PROPOSITION
4.6.
Thus fen)
all n
::f::.
f
e-inTR_tTRtJf
e-intTf
-.-11:
+11:
e-int
dt
dt
dt) Tf
O.
° and
as a consequence, TEd.
~r#0 is dense in
d
in the strong operator topology.
Proof. Let TEd.
By (iii) of Proposition 2.1, the Cl sums of the Fourier
series of any operator in fI! converge to that operator in the strong operator
topology. Thus, it suffices to show that each term f(n) of the Fourier series of
Andf(O)E.Po,
Tiscontained in do. Ifn < O,f(n) = O,andthusf(n)Ed.
which is contained in do. Suppose now that n > O. By (ii) of Proposition 2.1,
fen) is of the form Mn 0 U, where U E 2!o and Mn is multiplication by the
function ein-. Thus, T(n) = Mn 0 U is in do, since Mn is the composite of Ml
with itself n times.
Summarizing the main results established thus far, we have:
THEOREM
4.7.
the multiplication
d
is the strongly closed subalgebra of !l' generated by
operator M[ defined by (M1f)(t)
= eij(t), t E T.
fI! 0
and
As the final result in this section, we give a characterization of analytic
operators in terms of invariant subspaces. Part of this result will be used in the
following section.
For each integer m we denote by L,;,(T) the subspace of L2(T) consisting of all
L2(T) which satisfy j(n) = 0 for n < m.
functionsfin
PROPOSITION
(i)
(ii)
4.8. Let T
E
2.
Then the following
T is analytic.
T(L~(T) <;; L;,(T) for all mE Z.
are equivalent:
OPERATORS
Proof (i) implies Oi).
Let m
(4.6)
E
173
ON HILBERT SPACE
Z.
Then the inclusion
T(L,;,(T))
£;
L,~(T)
do
is clear if T = j\;[j or if T E :£'0' Thus, (4.6) will hold for all T E do since
is generated by Mj and the operators in 2! o. That (4.6) holds for an arbitrary
T in d now follows from Proposition 4.6.
(ii) implies (i). Let T E :£' and assume that T satisfies (4.6) for all In E Z.
Let r < O. We must show that fer) = O. Let m be an arbitrary integer. Since
linear combinations
of exponentials are dense in U(T), it suffices to show that
(r )eim•
O. Since, by (ii) and (iii) of Proposition 2.1,
(r)eim• is a multiple
of ei(dm)., it suffices to show that
t
t
=
(4.7)
(T(r)eim>,
By Lemma
=
o.
4.3,
(Teim.,
(4.8)
ei(r+m»)
ei(r+m»)
=
Since (ii) holds, Tei/fl> E Lt~(T). Since
so the left side of equality (4.8) is zero.
(T(r)eim>,
r
e,(r+m»)
< 0, ei(r+m). is orthogonal to L;tCT),
Thus (4.7) follows from (4.8).
5. The second F, and M. Riesz Theorem
In this section, we prove the following theorem, which is our extension
operators in Sf of a classical theorem of F. and M. Riesz.
5.1. Let T be an analytic operator in::e. If T
a support and cosupport for T.
THEOREM
#-
to
0, then T is both
We need two lemmas before proceeding to the proof of Theorem 5.1. The
first is an immediate consequence of the classical theorem af F. and M. Riesz
(this is the second corollary on p. 52 of [3]).
LEMMA 5.2.
Let f be an integrable function on T, In E Z. Assume that either
J(n) = ifn < mol' J(n) = 0 (in> m. Thenf cannot vanish on a subset ofT
having positive Lebesgue measure unless it vanishes a.e. on T.
°
Proof The F. and M. Riesz Theorem asserts that the conclusion of Lemma
5.2 holds if J(n) = 0 for n < O. The other cases may be reduced to this case by
consideration
of the functions e - im) and eim~r.
Recall that for K a measurable subset af T, e(K) is
{f:
f E e(T), f
= 0 a.e. off K},
and that far all m E Z, L~(T) is
{f:fE
L\T),J(n) =
° ifn < m}.
174
KAREL DELEEUW
5,3, Let K be a measurable subset ofT of positive Lebesgue measure
and mE Z, Then:
LEMMA
(i) L/~(T)
(ii) L,;(T)
n L2(KC) =
+
{O},
L2(K) is dense in
I!(T).
Proof
(i) is immediate from Lemma 5.2, since K has positive Lebesgue
Assume that L,;(T) + L2(K) is not
measure. We prove (ii) by contradiction.
dense in L2(T). Then there is a nonzero elementf of L2(T) orthogonal to both
L,;(T) and L\K). Since is orthogonal to L,;(T),
f
j(n)
(5.1)
Since
f is orthogonal
==
0
if
11 :2;
m.
to I3(K),
f(x)
(5.2)
= 0 a.e.
on K.
f
Lemma 5.2 shows that (5.1) and (5.2) cannot both hold unless
is the zero
element of L2(T).
We can now proceed to the proof of Theorem 5.1. We show first that T is a
support for T. We argue by contradiction, 1fT were not a support for T, there
would be a set K of positive Lebesgue measure so that T(L2(K» S; {O}. Then,
because of the analyticity of T and Proposition 4.8,
(5.3)
T(L,;(T)
+ L2(K»
S;
L,~(T) for all m E Z.
By (5.3) and (ii) of Proposition
5.2,
(5.4)
s; L,;,(T)
T(L2(T»
for all mE Z.
Since nmEZ L~(T) = {O}, (5.4) shows that T must have been the zero operator.
This completes the proof that T is a support for T.
We show next that T is a cosupport for T. If T were not a cosupport, there
would be a subset K of T whose complement KC has positive Lebesgue measure
and which satisfies
(5.5)
T(Ll(T»
s; L2(K),
By (5.5), the analyticity of T and Proposition (4.8),
(5.6)
T(L,;(T»
c: L2(K) n L,;(T)
Part (i) of Lemma 5.3 shows that L2(K)
Lebesgue measure, so (5,6) becomes
(5.?)
n L,;(T) =
E
Z.
{O}, since
KC
has positive
T(L,~(T» s; {O} for all m E Z.
Let P be the set of trigonometric
we have
(5.8)
for all m
polynomials on T. Since P s; U ~::;;L,~(T),
T(P) S;
{OJ
OPERATORS
ON HILBERT
SPACE
175
as a consequence of (5.7). Since P is dense in L2(T), (5.8) shows that T must
have been the zero operator. This completes the proof that T is a co support
for T.
Finally, we point out that if T is the operation of multiplication by a bounded
measurable function q), Theorem 5.1 reduces to the classical F. and M. Riesz
Theorem for cp. This is immediate from Proposition 3.5 and Proposition 4.1.
BIBUOGRAPHY
1. K.
An harmonic analysis for operators I: Formal properties, Illinois J. Math.,
vo!. 19 (1975), pp. 593-606.
2. R. E. Em.vARDs, Fourier series-A modern inrroduction, Rinehart and Winston, New York,
1967.
3. K. HOFFMAN, Banach spaces of analytic functions, Prentice-Hall, Englewood Cliffs, N.J.,
1962.
4. F.
DELEEUW,
RIESZ
STANFORD
AND
B. SZNAGY, Functional analysis, F. Ungar, New York, 1955.
UNIVERSITY
STANFORD,
CALIFORNIA