MA 102 (Multivariable Calculus) Rupam Barman and Shreemayee Bora - PowerPoint PPT Presentation

Continuous functions MA 102 (Multivariable Calculus) Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity 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? R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Example: Consider f : R 2 → R given by xy � if ( x , y ) � = (0 , 0) , x 2 + y 2 f ( x , y ) := 0 if ( x , y ) = (0 , 0) . Then ❼ R → R , x �→ f ( x , y ) is continuous for each fixed y ❼ R → R , y �→ f ( x , y ) is continuous for each fixed x Is f continuous at (0 , 0)? R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Continuity of f : R n → R Definition 1: Let f : R n → R and X 0 ∈ R n . Then ❼ f continuous at X 0 if for any ε > 0 there is δ > 0 such that � X − X 0 � < δ = ⇒ | f ( X ) − f ( X 0 ) | < ε. ❼ f is continuous on R n if f is continuous at each X ∈ R n . Example: Consider f : R 2 → R given by � xy √ if ( x , y ) � = (0 , 0) , x 2 + y 2 f ( x , y ) := 0 if ( x , y ) = (0 , 0) . Then f is continuous at (0 , 0). R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Continuity of f : R n → R 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) . However, we have seen that ❼ R → R , x �→ f ( x , y ) is continuous for each fixed y ❼ R → R , y �→ f ( x , y ) is continuous for each fixed x Moral: Let f : R 2 → R . Then continuity of x �→ f ( x , y ) and y �→ f ( x , y ) do not guarantee continuity of f . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity 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 ❼ x �→ f ( x , y ) is NOT continuous at 0 for y � = 0 ❼ 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 ) . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity 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 X 0 ∈ A . Then ❼ f continuous at X 0 if for any ε > 0 there is δ > 0 such that X ∈ A and � X − X 0 � < δ = ⇒ | f ( X ) − f ( X 0 ) | < ε. ❼ 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 . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Sequential characterization Theorem: Let f : A ⊂ R n → R and X 0 ∈ A . Then the following are equivalent: ❼ f is continuous at X 0 ❼ If ( X k ) ⊂ A and X k → X 0 then f ( X k ) → f ( X 0 ) . Proof: Examples: Examine continuity of f : R 2 → R given by 2. f ( x , y ) := e x 2 + y 2 1. f ( x , y ) := sin( xy ) . x 2 y 3. f (0 , 0) := 0 and f ( x , y ) := x 4 + y 2 for ( x , y ) � = (0 , 0) . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Sum, product and composition Theorem: Let f , g : A ⊂ R n → R be continuous at X 0 ∈ A . Then ❼ f + g : A → R , X �→ f ( X ) + g ( X ) is continuous at X 0 , ❼ f · g : A → R , X �→ f ( X ) g ( X ) is continuous at X 0 , ❼ If h : R → R is continuous at g ( X 0 ) then h ◦ g : A → R , X �→ h ( g ( X )) is continuous at X 0 . Proof: Use sequential characterization. R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Continuity of f : R n → R m Definition 3: Let f : A ⊂ R n → R m and X 0 ∈ A . Then ❼ f continuous at X 0 if for any ε > 0 there is δ > 0 such that X ∈ A and � X − X 0 � < δ = ⇒ � f ( X ) − f ( X 0 ) � < ε. ❼ 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 ) 3. f ( x , y , z ) := ( sin( x − y ) 1 + | x | + | y | , e x 2 − y 2 − z 2 ) . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity 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 X 0 ∈ A ⇐ ⇒ f i is continuous at X 0 for i = 1 , 2 , . . . , m . Proof: Use | f i ( X ) | ≤ � f ( X ) � and � f ( X ) � ≤ � m i =1 | f i ( X ) | . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Sum, product and composition Theorem: Let f , g : A ⊂ R n → R m be continuous at X 0 ∈ A . Then ❼ f + g : A → R m , X �→ f ( X ) + g ( X ) is continuous at X 0 , ❼ f • g : A → R , X �→ � f ( X ) , g ( X ) � is continuous at X 0 , ❼ If h : R m → R p is continuous at g ( X 0 ) then h ◦ g : A → R p , X �→ h ( g ( X )) is continuous at X 0 . Proof: Use sequential characterization. R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Uniform continuity of f : R n → R Definition: Let f : A ⊂ R n → R . Then f is uniformly continuous on A if for any ε > 0 there is δ > 0 such that X , Y ∈ A and � X − Y � < δ = ⇒ | f ( X ) − f ( Y ) | < ε. Example: The function f : R n → R given by f ( X ) := � X � is uniformly continuous. What about g ( X ) := � X � 2 ? R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Uniform Continuity Not Uniformly Continuous Uniformly Continuous The tube/pipe of diameter ǫ and of same length δ can move freely without touching the graph of f over the set D . R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity Sequential characterization Theorem: Let f : A ⊂ R n → R . Then the following are equivalent: ❼ f is uniformly continuous on A . ❼ If ( X k ) ⊂ A and ( Y k ) ⊂ A such that � X k − Y k � → 0 then | f ( X k ) − f ( Y k ) | → 0 . Proof: Fact: ( X k ) Cauchy in A + f uniformly cont. = ⇒ ( f ( X k )) Cauchy. 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. R. Barman & S. Bora MA-102 (2017)

Continuous functions Uniform continuity 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 , ∞ ) → R , x �→ √ x is uniformly continuous but NOT Lipschitz. 3. f : (0 , 1) → R , x �→ 1 / x is continuous but NOT uniformly continuous. R. Barman & S. Bora MA-102 (2017)