On The Solution Of A General Transform

On the Solution of a General Transform I. S. Reed PNAS 1944;30;169-172 doi:10.1073/pnas.30.7.169 This information is current as of March 2007. This article has been cited by other articles: www.pnas.org#otherarticles E-mail Alerts

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Notes:

MA THEMA TICS: I. S. REEDI

Voi..,. .0. 1944

i1,

This relation defines q for the rest of the paper, also. 7A more detailed proof will be included in the paper referred to in footnote 2. 8 Banach, S., Th6orie des operations lin&aires, Warsaw, 1932, p. 57; "Lemme." 9 Riesz, F., "Untersuchungen uber Systeme integrierbarer Funktionen," Math. Ann., ' 69,449-497 (1910); p. 475. o See Titchmarsh, E. C., Theory of Fourier Integrals, Oxford (1937), p. 105. "I Wiener, N., The Fourier Integral and Certain of Its Applications, Cambridge, England (1933), pp. 49-50. 1 The spectrum of a function of bounded variation consists of all points such that the total variation of the function is positive over every neighborhood of the point. 18 Salemn, R., "On Singular Monotonic Functions of the Cantor Type," Jour. Math, and Phys., 21, 69-82 (1942).14 Weil, A., "L'Integration dans les groupes topologiques et ses applications," Act. Sc. et Ind., No. 869, Paris, p. 55 (1940). 15 We are pleased to thank J. D. Tamarkin and A. Zygmund for suggesting a modification in the statement of Lemma 2 which results in a shortening of this part of our proof. 16 It seems noteworthy that this argument holds for p = 2 also, and together with the proof for Case -1, provides a new and somewhat simpler proof of Wiener's theorem for the space L2. 6

ON THE SOLUTION OF A GENERAL TRANSFORM BY I. S. REED DEPARTMENT OF MATHEMATICS, CALIFORNIA INSTITUTE OF TECHNOLOGY

Communicated May 15, 1944

The purpose of this paper is to give a brief extension to the solution of a Watson transform with an unsymmetric kernel. The solutions with symmetrical kernels are already well known from the work that has been done by Hardy, Watson, Titchmarsh, Goodspeed and others. THEOREM. If, (i) ac(u) and j3(u) are reciprocals of integral functions which take real values on the real axis. (ii) r(u) is an integralfunction such that if u - v + iw r(u) a +u) c(i

<

CeAv+Bjwf, r(- u (+ U)

<

C2eCv + Dlwl

0
(iii) k(s) = a(s)/((l -s), h(s) = P(s)/a(l - s) satisfy k(s)h(1- s) = 1 and are the Mellin transforms of K(x), H(x), respectively. k(1/2+ it), (iv) belongtoL2(-O, CO). .and he1/2 (iv) 1/ + it) beogt

MA THEMA TICS: I. S. REED

170

Then nsO-

a(n + 1)

PROC.- N. A-.- S.-

converges for O < x < e-A and is represented by f(x)

of L2(0, X ) for all positive x, andJoff(x)K(xy)dx = g(y) where g(y) is of L2(0,

cX) and is the analytic continuationfor all positive y of #(

)((n-

n =O

which converges for 0 < y < eC. Also fo`g(x)H(xy)dx = f(y) for y > 0. Hence f(x) and g(x) are the general transforms of the unsymmetrical kernels K(x) and H(x). Proof. Now by Cauchy's theorem,

f(x)

(1) =

r(n)xn n a(n+l)

-

.

T d r(-u) a-so sinhru a(l-u)Xd

when 0 < a < 1. By condition (iv) the integral representation holds first for 0 < x < e A and by analytical continuation it converges uniformly in x X.1 Alsof(x) = O(xa) for all positivex. Thus any interval O< xo < < by moving the line of integration across the origin, changing the integral only by a constant, and then taking a < - 1/2 for the lower limit of integration and a > - 1/2 for the upper limit, we see that fJo {f(x) } 2dx < O or is integrable in the Lebesgue sense L2(0, co). It follows then by Theorem 129 of Titchmarsh's Fourier Integral and its extension to unsymmetrical formulae mentioned in Section 8.9 that from condition (iv) and from f(x) being a function of L2(0, co'), that the formula g(y) = fo0f(x)K((xy)dx defines almosteverywhere a function g(y) belonging to L2(0, co) and that also the reciprocal formula f(y) = fo0g(x)H(xy)dx holds. In particular, f f(x)K(xy)dx = ('/27r+)5+ idu r(-u) K(xu)xudx O a-*0 sin (7u) a(1 -u)o

12,ra+i.f

wdu r(-u) sin (ru) a(1 u) Ja-io ('/27ri), + rdu r(-u) yu_du sin (S9) V(u) du f + 7rdv r(-v 1) sin (wv)

Jn-ia = ~(~y)n

n-O

r-n

13~f(n

-

1)

+ 1)

j3(v + 1)y

-g(y).

171

MA THEMA TICS: 1. S. REED

VOL. 30, 1944

Thus the proof is complete. Example. When -1/2 < v < 1/2 it is well knowi that if K(x) - x"" Yv(x)

then

H(x)

=

x1/SHp(x)2

and in this case if a(s)

=

2 /u2r(1/2v + I/2s + 1/4)sec(1/2v

-

1/2s

+

/4)Ir

then

(3(s) = 2/2r(l/2p + 1/2S + 1/4)csc('/2s + 1/21 + '/4)ir hence if N

=

f(x)

1/2n + 1/21 + 3/4 = j

n-0

(-xl 4 2)'r(n) cos (1/2v- /2n + 1/4)T/r(N)

g(x) = n-0j (-x/42)1r(-n If, moreover, r(n) = 1/2n + 3/4)Tr and so

g(x)

=

sec(/21

-

1) sin (1/2P + l/2n +

-

1/,)T/r(N).

1/2n + 1/4), then r(-n - 1) = sec(Q/2v +

(-x/4 2)n tan (Nw)/1r(N) =

n-O

X(x) tan (1/2P + '/4)r +

I(X) cot (1/2V + '/4)T

where

X(X)

=

(1/2X2)M/r(1/2i + n + 3/4), ,(x)

=

E (1/2X2)f+l/2/r(I/2P +

n

+ /4)

n-O

Also

f(x) = E (-x/42)"/r(N) = X(x) -(x). n-0

Again, if r(-n - 1) = csc (Ni), then r(n) = csc (1/2- '/2n + '/4w)r and g(x) = X(x) -,(x),f(x) = X(x) cot ('/21 + '/4)lr + A (x) tan ('/21 + '/4)ir Also it is known that3 (y)-

A(y)

so we obtain the identity

=

f0ox(x) - (x)}(xy)12Jv(xy)dx

172

MA THEMA TICS: I. S. REED

NRoci N. A. S.

- i(x) } (xy)'12Jv(xy)dx [X(x) cot ('/2V + 1/4)w +.#(x) tan (1/2v + 1/4)X]

fo {IX(x)

X(y) -

S

fo

-

-

-

fO' [X(x) cot (1/2V +,1/4)w + #(x) tan (1/2P + 1/4)7r]

(xy)'1'Hv(xy)dx (xy) 1' Yi'(xy)dx

also

ff7' [iX(x) cot (1/2v ±

1/4)7r + ,A(x) tan (1/2P + 1/4)] (Xy)1[Hv(xy)

+

Yv(xy)]dx

=

0.

1 Hardy, G. H., "Ramanujan and the- Theory of Fourier Transforms," Quart. Jour. Math. (Oxford series), 8, 245-254 (1937). 2 Titchmarsh, E. C., Theory of Fourier Integrals, p. 215 (1937), Oxford. 3 Hardy, G. H., Ramanujan, p. 208.