 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/4/b/ii/solution

Take

g (x) = 1 + ∣ x ∣, f (x) = (1 + ∣ x ∣)^{2} .

Both functions have values in

[1, \infty)

∣ g (x) - g (y) ∣ \leq ∣ x - y ∣

, so

g

For

x_{n} = n

and

y_{n} = n + 1/ n

, the distances satisfy

∣ x_{n} - y_{n} ∣ = 1/ n \to 0

, whereas

∣ f (y_{n}) - f (x_{n}) ∣ = 2 \frac{n + 1}{n} + \frac{1}{n ^{2}} > 2.

In particular tolerance

ε = 1

defeats every proposed

δ

: choose

n

with

1/ n < δ

. Thus $f$ does not have uniform continuity on $R$ .

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