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1 Uniformly Continuity

Let f  : D → R be a function and D is a subset of R. Then we say f  is uniformly continuous on D if for anygiven > 0, we can find δ > 0 depending only on so that

|f (x)− f (y)| < , whenever |x− y| < δ, x,y ∈ D.

Hence a function is not uniformly continuous if we can find > 0 and two sequences (xn), (yn) so that|xn − yn| < 1/n but |f (xn)− f (yn)| ≥ .

Example 1.1. Let f  : R→ R be f (x) = sin x. Since for each x, y ∈ R,

| sin(x)− sin(y)| ≤ |x− y|

by mean value theorem, then f  is uniformly continuous.

Proposition 1.1. If  f  : D → R is uniformly continuous, then it is continuous on D.

Proof. The proof is based on the definition of uniformly continuity of  f .

Example 1.2. Not every continuous functions on D is uniformly continuous: Let D = (0, 1) and f  : D → R

be f (x) = 1/x. Then f  is not uniformly continuous.

Proof. Pick xn = 1/n and yn = 1/(n + 1) for any n ∈ N. Then |xn − yn| < 1/n and

|f (xn)− f (yn)| = 1, for any n ∈ N.

Theorem 1.1. Let D = [a, b] be a closed bounded interval in R and f  : D → R be a function. Then f  iscontinuous if and only if  f  is uniformly continuous.

Proof. We’ve already shown that a uniformly continuous function on D is indeed a continuous function onD. Now we want to show the converse is true if  D = [a, b]. Suppose f  : [a, b] → R is continuous but notuniformly continuous. Then we can find > 0 and two sequences (xn) and (yn) with

|xn − yn| < 1n

, and |f (xn)− f (yn)| ≥ .

Since both (xn) and (yn) are bounded sequences, then by the Bolzano-Weierstrass theorem, we can find asubsequence (xnj ) and (ynj ) so that

|xnj − ynj | <1

nj

and limj→∞xnj = x, limj→∞ ynj = y for some x, y ∈ [a, b]. Then by the continuity of  f , we find

|f (x)− f (y)| ≥ .

Since |xnj − ynj | ≤ 1/nj , then by taking j →∞, we find x− y = 0 or x = y, which leads to a contradictionthat |f (x)− f (y)| ≥ .

Theorem 1.2. Let D = (a, b) and f  : D → R be a uniformly continuous function. Then we can find acontinuous function f  : [a, b] → R so that f 

D

= f .

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Proof. Since f  is uniformly continuous, given > 0, we can find δ > 0 so that for each x, y ∈ D with|x− y| < δ, then |f (x)− f (y)| < .

a ∈ R, then we can find a sequence (xn) in D so that xn → a. Since any convergent sequence is a Cauchysequence, then for the δ mentioned above, we can find N > 0 so that for any n, m ≥ N , |xn − xm| < δ. Asa consequence, for n, m ≥ N , we have

|f (xn)− f (xm)| < .

Now we find (f (xn)) is a Cauchy sequence. Since R is complete, then we can find y so that

limn→∞

f (xn) = y.

Definef (a) = lim

n→∞f (xn).

Suppose (zn) is another sequence which is convergent to a. Then for δ > 0, we find N > 0 so that|zn − xn| < δ. Hence |f (zn) − f (xn)| < . As a consequence, limn→∞ f (zn) = limn→∞ f (xn). Thereforef (a) is well-defined. Similarly, f (b) is well-defined. For any x ∈ [a, b], let (xn) be any sequence which isconvergent to x, define

f (x) = limn→∞

f (xn).

If  x ∈ D, take xn = x for any n ∈ N, then

f (x) = limn→∞

f (xn) = limn→∞

f (x) = f (x),

which proves f D

= f . For any x, y ∈ D, pick (xn) and (yn) in D and choose N  so that for any n ≥ N ,|xn − yn| < δ and |x− y| < δ. Hence for n ≥ N , |f (xn)− f (yn)| < . As a consequence,

|f (x)− f (y)| ≤ |f (x)− f (xn)|+ |f (xn)− f (yn)|+ |f (yn)− f (y)|.

Thus |f (x)− f (y)| < if  |x− y| < δ and x, y ∈ [a, b].

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