 Solution

ID: past-exam-of-the-calculus-of-variations-course-of-the-university-of-oxford/2009/1/a/ii/solution

A strictly convex function satisfies the same inequality strictly whenever

x \neq = y

and

0 < t < 1

For the unlabelled Hessian criterion, convexity implies

F (z_{0} + t h) + F (z_{0} - t h) - 2 F (z_{0}) \geq 0

. Divide by

t^{2}

and let

t \to 0

to obtain

h^{T} D^{2} F (z_{0}) h \geq 0

for every

h

Conversely, if every Hessian is positive semidefinite, Taylor's integral formula with

h = z - z_{0}

gives

F (z) - F (z_{0}) - D F (z_{0}) \cdot h = \int_{0}^{1} (1 - t) h^{T} D^{2} F (z_{0} + t h) h d t \geq 0.

The allowed supporting-plane characterization then proves

F convex ⟺ D^{2} F (z) positive semidefinite for all z .

For the integral inequality put

m = (b - a)^{- 1} \int_{a}^{b} u

. The supporting tangent at

m

gives

F (u (x)) \geq F (m) + F^{'} (m) (u (x) - m)

almost everywhere. Integrating cancels the linear term, so

F (\frac{1}{b - a} \int_{a}^{b} u) \leq \frac{1}{b - a} \int_{a}^{b} F (u) .

This is Jensen's inequality. It remains valid with value

+ \infty

on the right: the affine lower bound has integrable negative part, so the extended integral is well-defined.

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