21-356 Lecture 4

xn 1 yn 10 f zn xn yn xn where zn

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n of one variable f (x1 , . . . , xN #1 , ·), f (x) # f (x1 , . . . , xN #1 , yN ) = 10 !f (zN ) (xN # yN ) , ! xN where zN := (x1 , . . . , xN #1 , "N xN + (1 # "N ) yN ) for some "N ! (0, 1). Note that \$zN # y\$ & \$x # y\$ . Similarly, f (x1 , . . . , xi , yi+1 , . . . , yN )#f (x1 , . . . , xi#1 , yi , yi+1 , . . . , yN ) = !f (zi ) (xi # yi ) , ! xi where zi := (x1 , . . . , xi#1 , "i xi + (1 # "i ) yi , yi+1 , . . . , yN ) for some "i ! (0, 1) and \$zi # y\$ & \$x # y\$ . Write f (x) # f (y) # "f (x0 ) · (x # y) " N # ! !f !f = (zi ) # (x0 ) (xi # yi ) . ! xi ! xi i=1 Then * N* * |xi # yi | |f (x) # f (y) # "f (x0 ) · (x # y)| # * ! f !f * * & * ! xi (zi ) # ! xi (x0 )* \$x # y\$ . \$x # y\$ i=1 Since |xi #yi | \$x#y\$ & 1, we have that * N* * |f (x) # f (y) # "f (x0 ) · (x # y)| # * ! f !f * * 0& & * ! xi (zi ) # ! xi (x0 )* . \$x # y\$ i=1 (4) Taking y = x0 and using the fact that \$zi # x0 \$ & \$x # x0 \$ ' 0 as x ' x0 , !f together with the continuity of ! xi at x0 , gives...
View Full Document

This document was uploaded on 03/31/2014.

Ask a homework question - tutors are online