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Unformatted text preview: positive homogeneity of the
norm. This shows that the segment joining x and y is contained in B (x0 , r). 3 Directional Derivatives and Di!erentiability Let E $ RN , let f : E ' R and let x0 ! E . Given direction v ! RN , let L be
the line through x0 in the direction v, that is,
L := x ! RN : x = x0 + tv, t ! R ,
4 2-bis Continuous functions in RN
Let E ⊂ Rn , let f : E → R and let x0 ∈ E . We say that f is continuous at x0
if for every > 0 there exists a real number δ = δ (, x0 ) > 0 such that for all
x ∈ E with x ∈ B (x0 , δ ) we have |f (x) − f (x0 )| < .
Remark 0.1. Even if f is continuous in each variable, this DOES NOT imply
that f is continuous.
Example 0.2. Let N = 2, and consider the function f : R2 → R deﬁned as x2 if y = 0 f (x, y ) := −1 if y = 0 and y = x 0
otherwise. Let us prove that f is not continuous in the origin. We need to show that there
exists 0 > 0 such that for all δ > 0 there exists (xδ , yδ ) satisfying
(xδ , yδ ) − (0, 0) < δ
Let us consider 0 = 1
2 and |f (xδ , yδ ) − f (0, 0)| > 0 . and for every δ > 0 let us take
x δ = yδ = 2 . Then,
(xδ , yδ ) − (0, 0...
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This document was uploaded on 03/31/2014.
- Spring '14