Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 10D / b

Solution

The upper bound follows immediately from convexity, since

y_{0} + λ r = (1 - λ) y_{0} + λ (y_{0} + r) .

For the lower bound, write

y_{0} = \frac{1}{1 + λ} (y_{0} + λ r) + \frac{λ}{1 + λ} (y_{0} - r) .

Applying convexity and rearranging gives

(1 + λ) f (y_{0}) - λ f (y_{0} - r) \leq f (y_{0} + λ r) \leq (1 - λ) f (y_{0}) + λ f (y_{0} + r) .

Choose a fixed

r_{0} > 0

with

[y_{0} - r_{0}, y_{0} + r_{0}] \subset (a, b)

. Applying these bounds with

r = r_{0}

and

r = - r_{0}

shows that the difference

f (y_{0} + h) - f (y_{0})

is trapped between quantities of order

∣ h ∣/ r_{0}

. Both tend to zero, so every convex function on an open interval is continuous.

Now suppose

c

is a local minimum but some

x

satisfies

f (x) < f (c)

. For sufficiently small

λ > 0

, the point

z = (1 - λ) c + λ x

lies in the neighbourhood on which

c

is minimal, whereas convexity gives

f (z) \leq (1 - λ) f (c) + λ f (x) < f (c),

a contradiction. Thus

f (x) \geq f (c) for every x \in (a, b) .

Differentiability at the minimizer is unnecessary:

f (x) = ∣ x - c ∣

is convex, has its global minimum at

c

, and is not differentiable there.

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