430 ME. J. MERCER: FUNCTIONS OF POSITIVE AND NEGATIVE TYPE, 



continuity of the functions ; but it cannot be closed, unless it contains every point of 

 the interval. Moreover, it can be proved that a and ft, its lower and upper limits 

 respectively, do not belong to the set, unless they coincide with the end points of the 

 interval. 



At each point of the set 2 the quotient 



will have a definite value, because </> (.v) is never zero. We may therefore define a 

 single-valued function f(*), whose domain is 2, and whose value at any point is that 

 of this quotient. It will appear in the sequel that the properties of K(S, t) depend 

 very largely on the nature of f(s), and accordingly, in anticipation of this, we shall 

 speak of it as the discriminator of K (.v, t). T lie discriminator will evidently be 

 continuous in its domain, but it will never have the value zero. 



15. Let us now suppose that K (.s-, t) is of positive type, and is not zero everywhere 

 in the square Q. We have proved ( 11) that, under these circumstances, the function 

 *(*!< ' s 'i), which in the present case is simply (&i) (f> (sj, cannot be zero in the whole 

 of (>, />); also, at points where it does not vanish, we know that K(SI, s^ is positive 

 (6, 10). It follows that, for a function of positive type, the set 2 certainly exists, 

 and that in it the discriminator only takes positive values. 



Again, when N L and .s- 3 are any two points of ^, and ft., > .v,, we have 



hence, since /'(*,) is a positive number, it follows by the theorem of 10 that 



This result may be combined with the previous one in the statement that the 

 discriminator of /c(.v, /) is a non-decreasing function whose values are all positive. 



We have next to consider the points of (a, b) at which one or both of the functions 

 6 (.s-), </> (x) vanish. These fall naturally into three sets, according as they belong to 

 (1) the closed interval (a, a), (-2) the closed interval (ft, 1>), or (3) the open interval 

 (a, ft). As regards (l), it is not difficult to show that (s) vanishes in the whole 

 interval. For, if a, is any point of (K, a), one at least of the numbers (%), < (oj) 

 must tie zero ; and hence, since K (a ]; a^ is zero, the function K (s, a t ) is zero at each 

 point of (a, b) (11). 



Now when s > c^ we have 



K (s, ((l ) = (a,} 4> (s), 



and, at points of 2, $ (s) does not vanish ; we must therefore have (a^ = 0. It can 

 be proved in a similar manner that <f> (s) vanishes everywhere in the interval (ft, b). 



Finally, we can show that, at points of the open interval (a, ft) which do not belong 

 to 2, both (s) and <f> (s) vanish. In fact, if a t is any one of these points, there are 



