SC-90-09.pdf - ZIB

Indeed (1.6) follows from (1.21) by taking z — y and u = x + s(y — x), while
(1.12) follows from (1.21) for z = x and u = x + s(y — x). Let u, x € D
and denote by L the line containing u and x, and by L\ the half line of origin
u, contained in L but not containing x. Assumption (1.6) says that (1.21) is
satisfied only for z € L\ and not necessarily for all z £ L. However, we will
prove that in case F is twice Frechet differentiate assumptions (1.6) and (1.21)
are equivalent. First note that if F is twice Frechet differentiate then, by
dividing both sides of (1.6) by s, and then setting s —* 0, we obtain
\\F'( y )- 1 F"(x)(y-xY\\(u) - F'(x))(u -x)\\< »J llZ {~* { ly 2 dt
which proves that (1.21) is satisfied.
l
= u 1 \\u - x\\ 2 dt = u\\u -x\\ 2 ,
o
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