Quadrature Methods for Integral Equations

267

Proof of Lemma. (<^) Assume that (2 1) and (2.2) hold First, to show that K^ is a bounded operator, define

1= 1 then It is easily seen that

||K „ ||=supK „ ( t ) .

te [ a b]

But It follows easily from (2.2) that /c„ is a continuous function on the compact interval [fl,b], hence к;„ is bounded on la^b^, and so ||K„|| < oo

We recall the definition of collective compactness (see [1, p. 4]) the quence of bounded operators {KJ is collectively compact if the set

S = {K„g:geB,n^ll (24)

where В denotes the closed unit ball m С[а,Ь], has compact closure By the Arzelà-Ascoli theorem, it is necessary and sufficient that 5 be a bounded, equicontmuous subset of C[fl,fe]

To show that S is equicontmuous, let geB and r', t6[a,b], and consider

m

\K „ git' ) - K^g { t ) \= X \-W„,(t')-W^M8{sJ

1= 1 m

èl\W^ , ( t' ) - W „ , it ) \ 1= 1

^sup t\W„,it')-WJt)\.

n 1=1

The right-hand side is independent of m and g for geB, and converges to zero as t'-> t by virtue of (2.2), hence the set

{ K^g { t ) g6ß,m^l}

is equicontmuous at each point Гб[а,Ь]. But it is known (see for example [12, p 94]) that equicontinuity at each point of a compact interval implies equicon- tmuity over the interval, hence the equicontmuity of S is proved

Next , given geC[a,^], we seek to show that \\K^g — Kg\\^^^0 as n->oo. From (2.1) we have that K^g{t)-^Kg{t) for all ге[а,Ь]. But by the argument in the previous paragraph, the sequence {K^g} is an equicontmuous family; and It IS known (see for example [1, p. 7]) that pointwise convergence of an equicontmuous family on a compact interval is sufficient to ensure uniform convergence. Hence we conclude that

lim||K „ g - Kg|U=0 (2.5)

« - <• 00

Finally , we show that the set S defined by (2.4) is bounded. If g6C[fl,b] then It follows from (2.5) that

sup||K „ g||<oo .