21-356 Lecture 5

# Theorem 16 let e rn let f e r let x0 e assume

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: for di"erentiability at a point x0 . Theorem 16 Let E ' RN , let f : E # R, let x0 & E " . Assume that there ex!f ists r > 0 such that B (x0 , r) ' E and the partial derivatives ! xj , j = 1, . . . , N , exist for every x & B (x0 , r) and are continuous at x0 . Then f is di!erentiable at x0 . Proof. Let x, y & B (x0 , r). Write x = (x1 , . . . , xN ) and y = (y1 , . . . , yN ). Then f (x) ! f (y) = (f (x) ! f (x1 , . . . , xN #1 , yN )) + · · · + (f (x1 , y2 , . . . , yN ) ! f (y)) . By the mean value theorem applied to the function 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 ! "...
View Full Document

## This document was uploaded on 03/31/2014.

Ask a homework question - tutors are online