Math 521 - 4/20

Uniform Continuity • Let X,Y be metric spaces, f:X→Y • f is uniformly continuous on X if for every ε 0, ∃δ 0 s.t. • If p,q∈X and d_X (p,q) δ, then d_Y (f(p),f(q)) ε Theorem 4.19 • Statement ○ Let X,Y be metric spaces, X compact ○ If f:X→Y is continuous, then f is also uniformly continuous • Proof ○ Let ε 0 be given ○ Since f is continuous, ∀p∈X, ∃ϕ(p) s.t. § If q∈X, and d_X (p,q) ϕ(p), then d_Y (f(p),f(q)) ε/2 ○ Let J(p)≔{q∈X│d_X (p,q) 1/2 ϕ(p) } § p∈J(p),∀p∈X, so {J(p)} is an open cover of X § Since X is compact, {J(p)} has a finite subcover § There exists finite set of points p_1,…,p_n∈X s.t. § X⊂J(p_1 )∪…∪J(p_n ) ○ Let δ=1/2 min⁡{ϕ(p_1 ),…,ϕ(p_n )} 0 § Given p,q∈X s.t. d_X (p,q) δ § Since X⊂J(p_1 )∪…∪J(p_n ), § ∃m∈{1,2,…,n} s.t. p∈J(p_m ) ○ Hence § d_X (p,p_m ) 1/2 ϕ(p_m ) ϕ(p_m ) § d_X (q,p_m )≤d_X (p,q)+d_X (p,p_m ) δ+1/2 ϕ(p_m )≤ϕ(p_m ) ○ By the triangle inequality and definition of ϕ(p), ○ d_Y (f(p),f(q))≤d_Y (f(p),f(p_m ))+d_Y (f(p_m ),f(q)) ε/2+ε/2=ε Theorem 4.20 • Definition ○ Let E be noncompact set in R ○ Then there exists a continuous function f on E s.t. (a) f is not bounded (b) f is bounded but has no maximum (c) E is bounded, but f is not uniformly continuous • Proof : If E is bounded ○ Since E is noncompact, E must be not closed ○ So there exists a limit point x_0∈E s.t. x_0∉E ○ f(x)≔1/(x−x_0 ) establishes (c) § f is continuous by Theorem 4.9 § f is clearly unbounded § f is not uniformly continuous □ Let ε 0 and δ 0 be arbitrary □ Choose x∈E s.t. |x−x_0 | δ □ Taking t close to x_0 □ We can make |f(t)−f(x)| ε, but |t−x| δ □ Since δ 0 is arbitrary ○ g(x)≔1/(1+(x−x_0 )^2 ) establishes (b) § g is continuous by Theorem 4.9 § g is bounded, since 0 g(x) 1 § g has no maximum, since sup┬(x∈E)⁡g(x)=1, but g(x) 1 • Proof: If E is not bounded ○ f(x)≔x establishes (a) ○ h(x)≔x^2/(1+x2 ) establishes (b) Example 4.21 • Let X=[0,2π) • Let f:X→Y given by f(t)=(cos⁡t,sin⁡t ) • Then f is continuous, and bijective • But f^(−1) is not continuous at f(0)=(1,0)