Lecture 2: Continuous functions

Transcription

1 Lecture 2: Continuous functions Rafikul Alam Department of Mathematics IIT Guwahati

2 Continuous functions Task: Analyze continuity of the functions: Case I: f : A R n R Case II: f : A R R n Case III: f : A R n R m Question: What does it mean to say that f is continuous?

3 Example: Consider f : R 2 R given by { xy if (x, y) (0, 0), f (x, y) := x 2 +y 2 0 if (x, y) = (0, 0). Then R R, x f (x, y) is continuous for each fixed y R R, y f (x, y) is continuous for each fixed x Is f continuous at (0, 0)?

4 Continuity of f : R n R Definition 1: Let f : R n R and a R n. Then f continuous at a if for any ɛ > 0 there is δ > 0 such that x a < δ = f (x) f (a) < ɛ. f is continuous on R n if f is continuous at each x R n. Example: Consider f : R 2 R given by f (0, 0) := 0 and f (x, y) := xy/(x 2 + y 2 ) for (x, y) (0, 0). Then f is NOT continuous at (0, 0). Moral: Let f : R 2 R. Then continuity of x f (x, y) and y f (x, y) do not guarantee continuity of f.

5 Example: Consider f : R 2 R given by { x sin(1/y) + y sin(1/x) if xy 0, f (x, y) := 0 if xy = 0. Then f is continuous at (0, 0). But x f (x, y) is NOT continuous at 0 for y 0 y f (x, y) is NOT continuous at 0 for x 0 Remark: Let f : R 2 R. Then continuity of f at (a, b) does NOT imply continuity of t f (t, y) and s f (x, s) at a and b, respectively, for each (x, y).

6 Let S := {(x, y, z) R 3 : x 2 + y 2 + z 2 = 1}. Define f : S R by f (x, y, z) := x y + z. Is f continuous? Definition 2: Let f : A R n R and a A. Then f continuous at a if for any ɛ > 0 there is δ > 0 such that x A and x a < δ = f (x) f (a) < ɛ. f is continuous on A if f is continuous at each x A. Example: Let A := S {(0, 0, 0)}. Consider f : A R given by f (0, 0, 0) := 1 and f (x, y, z) := x + y + z for (x, y, z) S. Then f is continuous on A.

7 Sequential characterization Theorem: Let f : A R n R and a A. Then the following are equivalent: f is continuous at a If (x k ) A and x k a then f (x k ) f (a). Proof: Examples: Examine continuity of f : R 2 R given by 1. f (x, y) := sin(xy). 2. f (x, y) := e x2 +y 2 3. f (0, 0) := 0 and f (x, y) := x 2 y for (x, y) (0, 0). x 4 + y 2

8 Sum, product and composition Theorem: Let f, g : A R n R be continuous at a A. Then f + g : A R, x f (x) + g(x) is continuous at a, f g : A R, x f (x)g(x) is continuous at a, If h : R R is continuous at g(a) then h g : A R, x h(g(x)) is continuous at a. Proof: Use sequential characterization.

9 Examples: Examine continuity of f : R 2 R given by 1. f (x, y) := e x sin(x 2 y) + e xy+1 + x 2 + y 2, 2. f (0, 0) := 0 and f (x, y) := xy for (x, y) (0, 0), x 2 + y 2 3. f (0, 0) := 0 and f (x, y) := sin2 (x y) x + y for (x, y) (0, 0).

10 Continuity of f : R n R m Definition 3: Let f : A R n R m and a A. Then f continuous at a if for any ɛ > 0 there is δ > 0 such that x A and x a < δ = f (x) f (a) < ɛ. f is continuous on A if f is continuous at each x A. Examples: Examine continuity of f : R n R m given by 1. f (x) := (sin(x), cos(x), x) 2. f (x, y) := (e x sin(y), y cos(x), x 3 + y) sin(x y) 3. f (x, y, z) := ( 1 + x + y, y 2 ex2 z2 ).

11 Componentwise continuity characterization Let f : A R n R m. Then f (x) = (f 1 (x),, f m (x)) where f i : A R. Theorem: Let f : A R n R m be given by f (x) = (f 1 (x),, f m (x)). Then f is continuous at a A f i is continuous at a for i = 1, 2,..., m. Proof: Use f i (x) f (x) and f (x) m i=1 f i(x).

12 Sum, product and composition Theorem: Let f, g : A R n R m be continuous at a A. Then f + g : A R m, x f (x) + g(x) is continuous at a, f g : A R, x f (x), g(x) is continuous at a, If h : R m R p is continuous at g(a) then h g : A R p, x h(g(x)) is continuous at a. Proof: Use sequential characterization.

13 Uniform continuity of f : R n R Definition: Let f : A R n R. Then f is uniformly continuous on A if for any ɛ > 0 there is δ > 0 such that x, y A and x y < δ = f (x) f (y) < ɛ. Example: The function f : R n R given by f (x) := x is uniformly continuous. What about g(x) := x 2? Fact: (x k ) A Cauchy + f uniformly cont. = (f (x k )) Cauchy.

14 Sequential characterization Theorem: Let f : A R n R. Then the following are equivalent: f is uniformly continuous on A. If (x k ) A and (y k ) A such that x k y k 0 then f (x k ) f (y k ) 0. Proof: Examples: 1. f : R 2 R, (x, y) x 2 + y 2 is NOT uniformly continuous. 2. f : (0, 1) (0, 1) R, (x, y) 1/(x + y) is NOT uniformly continuous.

15 Lipschitz continuity: f : R n R m Definition: Let f : A R n R m. Then f is Lipschitz continuous on A if there is M > 0 such that x, y A = f (x) f (y) M x y. Lipschitz continuity = Uniform Continuity = Continuity Examples: 1. f : R n R, x x is Lipschitz continuous. 2. f : [0, 1] R, x x is uniformly continuous but NOT Lipschitz. 3. f : (0, 1) R, x 1/x is continuous but NOT uniformly continuous. *** End ***