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§2.5 Properties of Differentiable Functions 47

2.5 Properties of Differentiable Functions

2.5.1 Continuity of Differentiable Functions

Theorem 2.40. Let U Ď Rn be open, and f : U Ñ Rm be differentiable at x0 P U . Then f
is continuous at x0.

Proof. Since f is differentiable at x0, there exists L P B(Rn, Rm) such that

D δ1 ą 0 Q ››f (x) ´ f (x0) ´ L(x ´ x0)››Rm ď }x ´ x0}Rn @ x P B(x0, δ1) .

As a consequence,

() (2.2)
››f (x) ´ f (x0)››Rm ď }L} + 1 }x ´ x0}Rn @ x P B(x0, δ1) .

For a given ε ą 0, let δ = ! ε ) Then δ ą 0, and if x P B(x0, δ),
min δ1, .
2(}L} + 1)

››f (x) ´ f (x0)››Rm ď ε ă ε. ˝
2

Remark 2.41. In fact, if f is differentiable at x0, then f satisfies the “local Lipschitz
property”; that is,

D M = M (x0) ą 0 and δ = δ(x0) ą 0 Q if }x´x0}X ă δ, then }f (x)´f (x0)}Y ď M }x´x0}X

since we can choose M = }L} + 1 and δ = δ1 (see (2.2)).

Example 2.42. Let f : R2 Ñ R be given in Example 2.27. We have shown that f is not
differentiable at (0, 0). In fact, f is not even continuous at (0, 0) since when approaching
the origin along the straight line x2 = mx1,

lim f (x1, mx1) = lim mx12 = m2 ‰ f (0, 0) if m ‰ 0.
(m2 + 1)x21 m2 + 1
(x1,mx1)Ñ(0,0) x1Ñ0

Example 2.43. Let f : R2 Ñ R be given in Example 2.28. Then f is not continuous at
(0, 0); thus not differentiable at (0, 0).

Example 2.44. Let f : R2 Ñ R be given by

$ x3 if (x, y) ‰ (0, 0) ,
&
f (x, y) = x2 + y2

% 0 if (x, y) = (0, 0) .