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![Definition 5.1: Derivative • Let f be defined (and real-valued) on [a,b] • ∀x∈[a,b], let ϕ(t)=(f(t)−f(x))/(t−x) (a t b, t≠x) • Define f^′ (x)=(lim)_(t→x)ϕ(t), provided that this limit exists • f′ is called the derivative of f • If f′ is defined at point x, f is differentiable at x • If f′ is defined ∀x∈E⊂[a,b], then f is differentiable on E Theorem 5.2: Differentiability Implies Continuity • Statement ○ Let f be defined on [a,b] ○ If f is differentiable at x∈[a,b] then f is continuous at x • Proof ○ lim_(t→x)(f(t)−f(x))=lim_(t→x)((f(t)−f(x))/(t−x) (t−x))=lim_(t→x)(f^′ (x)(t−x))=0 ○ So lim_(t→x)f(t)=f(x) Theorem 5.5: Chain Rule • Statement ○ Given § f is continuous on [a,b], and f^′ (x) exists at x∈[a,b] § g is defined on I⊃im(f), and g is differentiable at f(x) ○ If h(t)=g(f(t)) (a≤t≤b), then ○ h is differentiable at x, and h^′ (x)=g^′ (f(x))⋅f^′ (x) • Proof ○ Let y=f(x) ○ By the definition of derivative § f(t)−f(x)=(t−x)(f^′ (x)+u(t)), where t∈[a,b], lim_(t→x)u(t)=0 § g(s)−g(y)=(s−y)(g^′ (y)+v(s)), where s∈I, lim_(s→y)v(s)=0 ○ Let s=f(t), then § h(t)−h(x)=g(f(t))−g(f(x)) § =(f(t)−f(x))(g^′ (y)+v(s)) § =(t−x)(f^′ (x)+u(t))(g^′ (y)+v(s)) ○ If t≠x, then § (ht)−hx))/(t−x)=(f^′ (x)+u(t))(g^′ (y)+v(s)) ○ As t→x § u(t)→0, and v(s)→s § So s=f(t)→f(x)=y by continuity § Therefore h′ (x)=lim_(t→x)〖(ht)−hx))/(t−x)〗=f^′ (x) g^′ (y)=g^′ (f(x)) f^′ (x) Definition 5.7: Local Maximum and Local Minimum • Let X be a metric space, f:X→R • f has a local maximum at p∈X if ∃δ 0 s.t. ○ f(q)≤f(p),∀q∈X s.t. d(p,q) δ • f has a local minimum at p∈X if ∃δ 0 s.t. ○ f(q)≥f(p),∀q∈X s.t. d(p,q) δ Theorem 5.8: Local Extrema and Derivative • Statement ○ Let f be defined on [a,b] ○ If f has a local maximum (or minimum) at x∈(a,b) ○ Then f^′ (x)=0 if it exists • Proof ○ By Definition 5.7, choose δ, then § a x−δ x x+δ b ○ Suppose x−δ t x § (f(t)−f(x))/(t−x)≥0 § Let t→x (with t x), then f^′ (x)≥0 ○ Suppose x t x+δ § (f(t)−f(x))/(t−x)≤0 § Let t→x (with t x), then f^′ (x)≤0 ○ Therefore f^′ (x)=0](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aea9e3142970.png)
![Definition 2.45: Connected Set • Let X be a metric space, and A,B⊂X • A and B are separated if ○ A∪B ̅=∅ and A ̅∪B=∅ ○ i.e. No point of A lies in the closure of B and vice versa • E⊂X is connected if ○ E is not a union of two nonempty separated sets Theorem 2.47: Connected Subset of R • Statement ○ E⊂R is connected if and only if E has the following property ○ If x,y∈E and x z y, then z∈E • Proof (⟹) ○ By way of contrapositive, suppose ∃x,y∈E, and z∈(x,y) s.t. z∉E ○ Let A_z=E∩(−∞,z) and B_z=E∩(z,+∞) ○ Then A_z and B_z are separated and E=A_z∪B_z ○ Therefore E is not connected • Proof (⟸) ○ By way of contrapositive, suppose E is not connected ○ Then there are nonempty separated sets A and B s.t. E=A∪B ○ Let x∈A,y∈B. Without loss of generality, assume x y ○ Let z≔sup(A∩[x,y]). Then by Theorem 2.28, z∈A ̅ ○ By definition of E, z∉B. So, x≤z y ○ If z∉A § x∈A and z∉A § ⇒x z y § ⇒z∉E ○ If z∈A § Since A and B are separated, z∉B ̅ § So ∃z_1 s.t. z z_1 y and z_1∉B § Then x z_1 y, so z_1∉E Theorem 4.22: Continuous Mapping of Connected Set • Statement ○ Let X,Y be metric spaces ○ Let f:X→Y be a continuous mapping ○ If E⊂X is connected then f(E)⊂Y is also connected • Proof ○ Suppose, by way of contradiction, that f(E) is not connected ○ "i.e." f(E)=A∪B, where A,B⊂Y are nonempty and separated ○ Let G≔E∩f^(−1) (A) and H≔E∩f^(−1) (B) ○ Then E=G∪H, where G,H≠∅ ○ Since A⊂A ̅, we have G⊂f^(−1) (A ̅ ) ○ Since f is continuous and A ̅ is closed, f^(−1) (A ̅ ) is also closed ○ Therefore G ̅⊂f^(−1) (A ̅ ), and hence f(G ̅ )⊂A ̅ ○ Since f(H)=B and A ̅∩B=∅, we have G ̅∩H=∅ ○ Similarly, G∩H ̅=∅ ○ So, G and H are separated ○ This is a contradiction, therefore f(E) is connected Theorem 4.23: Intermediate Value Theorem • Statement ○ Let f:R→R be continuous on [a,b] ○ If f(a) f(b) and if c statifies f(a) c f(b) ○ Then ∃x∈(a,b) s.t. f(x)=c • Proof ○ By Theorem 2.47, [a,b] is connected ○ By Theorem 4.22, f([a,b]) is a connected subset of R ○ By Theorem 2.47, the result follows](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aea9aa22812b.png)
![Continuous • Definition ○ Suppose X,Y are metric spaces, E⊂X, p∈E, and f:E→Y ○ f is continuous at p if for every ε 0, there exists δ 0 s.t. ○ If x∈E,d_X (x,p) δ, then d_Y (f(x),f(p)) ε ○ If f is continuous at every point p∈E, then f is continuous on E • Note ○ f must be defined at p to be continous at p (as opposed to limit) ○ If p is an isolated point of E ○ Then every function f whose domain is E is continous at p Theorem 4.6 • In the context of Definition 4.5, if p is also a limit point of E, then • f is continious at p if and only if (lim)_(x→p)f(x)=f(p) Theorem 4.7 • Statement ○ Suppose X,Y,Z are metric spaces, E⊂X,f:E→Y, g:f(E)→Z, and ○ h:E→Z defined by h(x)=g(f(x)),∀x∈E ○ If f is continuous at p∈E, and g is continuous at f(p) ○ Then h is continuous at p • Note: h is called the composition of f and g and is called g∘f • Proof ○ Let ε 0 be given ○ Since g is continuous at f(p),∃η 0 s.t. § If y∈f(E) and d_Y (y,f(p)) η, then d_Z (g(y),g(f(p))) ε ○ Since f is continuous at p, ∃δ 0 s.t. § If x∈E and d_X (x,p) δ, then d_Y (f(x),f(p)) η ○ Consequently, if d_X (x,p) δ, and x∈E, then § d_Z (g(f(x)),g(f(p)))=d_Z (hx),hp)) ε ○ So, h is continuous at p by definition Theorem 4.8 • Statement ○ Given metric spaces X,Y ○ f:X→Y is continuous if and only if ○ f^(−1) (V) is open in X for every open set V⊂Y • Proof (⟹) ○ Suppose f is continuous on X, and V⊂Y is open ○ We want to show all points of f^(−1) (V) are interior points ○ Suppose p∈X, and f(p)∈V, then p∈f^(−1) (V) ○ V is open, so ∃ε 0 s.t. y∈V if d_Y (f(p),y) ε ○ Since f is continuous at p, ∃δ 0 s.t. § If d_X (x,p) δ, then d_Y (f(x),f(p)) ε ○ So x∈f^(−1) (V) if d_X (x,p) δ ○ This shows that p is an interior point of f^(−1) (V) ○ Therefore f^(−1) (V) is open in X • Proof (⟸) ○ Suppose f^(−1) (V) is open in X for every open set V⊂Y ○ Fix p∈X, ε 0 ○ Let V≔{y∈Y│d_Y (y,f(p)) ε } ○ V is open, so f^(−1) (V) is also open ○ Thus, ∃δ 0 s.t. if d_X (p,x) δ, then x∈f^(−1) (V) ○ But if x∈f^(−1) (V), then f(x)∈V so d_Y (f(x),f(p)) ε ○ So, f:X→Y is continuous at p ○ Since p∈X was arbitrary, f is continuous on X • Corollary ○ Given metric spaces X,Y ○ f:X→Y is continuous on X if and only if ○ f^(−1) (V) is closed in X for every closed set V in Y • Proof ○ A set is closed if and only if its complement is open ○ Also, f^(−1) (E^c )=[f^(−1) (E)]^c, for every E⊂Y](https://shawnzhong.com/wp-content/uploads/2018/04/img_5add053c62b0b.png)
