AND THEIR CONNECTION WITH THE THEORY OF INTEGRAL EQUATIONS. 417 



type in accordance with the above definition. Conversely, we may prove that, for 

 every function of positive type, the above integral equation has only positive singular 

 values. For, if we multiply along the homogeneous equation 



by / (s), and integrate with respect to s between the limits a and b, we obtain 



Since the double integral on the left cannot be negative, and X n is a finite number, 



it appears that 



A_>0. 



Thus the necessary and sufficient condition that a continuous symmetric function 

 should be of positive type is that the integral equation of the second kind of which it is 

 the characteristic function should have all its singular values positive.* 



In a similar manner it may be proved that this statement remains true when we 

 replace the word positive by negative, in both places where it occurs.* Moreover, 

 since a function must be of ambiguous type when it is of neither the positive nor the 

 negative type, we conclude that the necessary and sufficient condition for a continuous 

 symmetric function to be of ambiguous type is tlie existence of botfi, positive and 

 negative singular values of the integral equation of the second kind of which it is the 

 characteristic function. 



3. It is easy to see that, corresponding to a function K (s, t) whose type is 

 ambiguous, there exists a function 9 (s) which is not zero in the whole interval (a, b), 

 and satisfies the relation 



K(s,t)0(s)6(t)dsdt = ........ (A) 



J a J a 



For, if we employ the notation of (i) above, and suppose that k is any real constant, 

 we shall have 



f f K (s, t) [0, (s) + k0, (s)] [0! (t) + k6, (0] ds dt = P I* K (s, I) O l (s} 9, (t) ds dt 



Ja Ja Ja J a 



+ k f | ' K (s, t) [0, (s) 2 (t) + 2 (s) 6, (t)] ds dt + F f f K (s, t) 2 (s) 0. (t) ds dt. 



J a J a Ja J a 



* It follows from these results that, unless K (s, t) is identically zero, we cannot have 



I " K (s, t) 6 (s) 6 (f) ds dt = 0, 



for all members of 6. We shall prove this result in a different manner further on ( 12), but it is useful to 

 make the remark at this stage, since it shows conclusively that a function which is not identically zero 

 cannot be both of positive and negative type. 



VOI,. OCJX. A, 3 H 



