Talk:Real Analysis/Uniform Convergence

Theorem on Uniform Convergence : Let fn : P⊆ R → R converge uniformly on P to f. Assume that fn are continuous at a ∈ X. Then f is continuous at a. Proof: To prove continuity of f at a, let ε > 0 be given. For this ε, using the uniform convergence of fn to f, there exists N such that ∀n ≥ N and x ∈ X, =>|fn(x) − f(x)| < ε/3...(1) For N as above using the continuity of fN at a, we can choose a δ > 0 such that x∈P and |x − a|<δ =>|fN (x) −fN(a)|<ε/3 Observe that for x ∈ (a − δ, a + δ) ∩ P, we have,|f(x) − f(a)| ≤ |f(x) − fN (x)| + |fN (x) − fN (a)| + |fN (a) − f(a)| < ε/3 + ε/3 + ε/3....(2) where we used (1) to estimate the first and the third term and (2) to estimate the middle term. Hence the proof. (Books on Real Analysis are available on internet )

Theorem on Uniform continuity
The statement: Let P be a closed and bounded interval. Then any continuous function f : P--> R is uniformly continuous. Proof: If f is not uniformly continuous, then there exists ε > 0 such that for all 1/n we can find an, bn ∈ P such that |an − bn| < 1/n but |f(an) − f(bn)| ≥ ε. By Bolzano-Weierstrass theorem, there exists a subsequence (ank ) such that ank →a ∈ P. Observe that: It follows that bnk → a. By continuity, f(ank) → f(a) and also f(bnk) → f(a). In particular, for all sufficiently large k, we must have |f(ank) − f(bnk)| < ε, a contradiction. Hence the theorem is proved. Mathsolution (discuss • contribs) 20:15, 21 June 2022 (UTC)
 * bnk − a| ≤ |bnk − ank| + |ank − a|.