5)50 KELLOGG. 



n(x) is the total mass on the segment (0, x), and is defined as follows: 



12 1 



/i(0) = 0, n(l) = 1, on - ^ x ^ -, yu(.r) = -; in general, fi(x) being 



defined on the gaps in S n phis their end points (i.e. on S — S n plus 

 its derivative), it is defined on the new gaps of <S„ and their end points 

 (i.e. on Sn-i — Sn plus its derivative) by assigning to it the arith- 

 metic mean of its values at the two nearest points to right and left 

 at which it has been defined. In this way it is determined at points 

 everywhere dense, and the definition is completed by demanding that 

 it be continuous, a thing evidently possible. 



The function L(P) is seen to be harmonic in the entire plane, apart 

 from the points of <S. Moreover, it is continuous at the points of S. 

 To show this, we first determine a simple function dominating A/i = 



H(X 2 ) — /i(.tj), (.r 2 > X}). 



If Xi = 0, then A/i = n(xz) = —when x% — — . If we eliminate 



n between the equations x = — and y = —, we obtain the function 



y = x p (p = log 2/log 3). Geometric intuition suggests that x p 

 dominates fx(.v), and further, that (A/z) ^ (A.r) p , since // = fi(x) never 

 increases more rapidly than at the origin. These inferences are cor- 

 rect. We first note that y = /x(.r) never rises above the polygonal 



line through the points ( — — ] of y — x p . For as x increases from — 



2.1 

 to—, fx(x) remains constant at—, and fx(x) does not increase beyond 



o — 



1 2 



halt way to r^j until x has progressed - of the remaining distance irom 

 — o 



2 1 12 2 



3S to g^I- Hence *<>* — ^ x ^ 3^ + 3~^I> V = /*(*) hes under the 



chord joining ^— , ^J and ^— , —J. But from — + — to ^+ 



2 2 



— - +^7, by the same reasoning, /z(.r) lies under its steeper chord 



joining (— + — , — + —J and (— lf —J. A continuation of the 



argument, and the use of the continuity of n(x) completes the proof 

 that n(x) lies under the polygonal line in question, and hence under 

 the concave downward curve y = x p . 



Next, suppose that X\ is the abscissa of the left end of one of the 



