Question 6.T.11: Let f : D → R be uniformly continuous. Then it has an extens...

Let x ∈ \bar{D}. Then there is a sequence in D such that x_{n} → x.

(x_{n}) is clearly a Cauchy sequence, and we shall now show that its image under f , (f(x_{n})), is also a Cauchy sequence. Suppose ε > 0. By the uniform continuity of f on D, there is a δ > 0 such that

x, t ∈ D, |x − t| < δ ⇒ |f(x) − f(t)| < ε.        (6.17)

Since (x_{n}) is a Cauchy sequence, there is an integer N such that

m, n ≥ N ⇒ |x_{m} − x_{n}| < δ.        (6.18)

Now (6.17) and (6.18) together imply

m, n ≥ N ⇒ |f(x_{m}) − f(x_{n})| < ε,

which means (f(x_{n})) is a Cauchy sequence, and therefore convergent, by Cauchy’s criterion.

Let

\lim f(x_{n}) = g(x).        (6.19)

To make sure that equation (6.19) defines a function on \bar{D}, we have to check that g(x) does not depend on the choice of sequence (x_{n}) that converges to x. Let (\acute{x_{n}}) be another sequence in D which converges to x. Since |\acute{x_{n}} − x_{n}| → 0, it follows from Corollary 6.9 that |f(\acute{x_{n}}) − f(x_{n})| → 0, hence f(\acute{x_{n}}) → g(x).

g is clearly an extension of f to \bar{D}, and it remains to show that g is uniformly continuous. Let ε > 0. f being uniformly continuous, there is a δ > 0 such that (6.17) is satisfied for any pair of points x, t ∈ D.

Now take x, t ∈ \bar{D} and assume |x − t| < δ/ 3. There are two sequences (x_{n}) and (t_{n}) in D such that x_{n} → x and t_{n} → t, and, by (6.19)

f(x_{n}) → g(x),  f(t_{n}) → g(t).

Therefore there is an integer M such that

n ≥ M ⇒ |x_{n} − x| < \frac{δ}{3},  |t_{n} − t| < \frac{δ}{3}

⇒ |f(x_{n}) − g(x)| < ε,  |f(t_{n}) − g(t)| < ε.        (6.20)

Consequently,

n ≥ M ⇒ |x_{n} − t_{n}| ≤ |x_{n} − x| + |x − t| + |t − t_{n}| < δ.

By (6.17),

|f(x_{n}) − f(t_{n})| < ε  for all n ≥ M.        (6.21)

Now, in view of (6.20) and (6.21), if n ≥ M, then

for all x, t ∈ \bar{D} such that |x − t| < δ/ 3. This proves g is uniformly continuous on \bar{D}.