414 PROF. A. C. DIXON ON STURM-LIOUVILLK HARMONIC EXPANSIONS. 



Let e r , m be the part of S r>at which is also in \\. 

 Put 



= i in $i, m (m = 1, 2, ..., />). 

 except where the value Oo has been already assigned 



= ... = r in 5 r , ,(? = 1, 2, ...,p), 



except at such points as belong to <$ ,m, fV -, ot-j(*B = 1, >/') where the value 

 has been already assigned, and, lastly, 



= L in the rest of the domain. 

 Then \f(x) tf> (:*)) < (U L)//i in all parts of S r , m that are not in 



l\(r = 0, 1, ..., n; m = 1, 2, ..., jt>), 

 that is in intervals E whose sum is 



Now 



/< 



r = HI = I 



II II 



2 2 



r = ii i = 



/ ii it 



2 X <V<, > 2 A,-(t+l)e> (6-a)- 



and, therefore, 



2 2 e <(+l)e, 



/ = o w = i 



E > (b-a)-2(n+l)e. 



Hence the points where \f(x} <j>(x)\ > (U L)//t form a set whose measure 

 < 2(n+l)e, and 



-a) (U-L) 2 / 2 -f (U-L) 2 2 (n+\) *, 



In this expression we may suppose e = yt" 2 and make ?i as great as we wish so that 

 the whole is arbitrarily small. 



The most advantageous value for (x) in each interval is the average of f(x) over 

 the interval, for when c is to be a constant in the interval (x , JL\) the least value of 



is given by putting 



f^i 

 /(.<) dx -r- (xt- 

 - r 



G. The proof that has been given indicates a particular method of subdivision, say 

 the method A, but any other method, say B, may also be used. To prove this, let 



