Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 1

Choose $M$ with $|f|\le M$ on $U$, and fix $r=(1-\Vert x\Vert )/4>0$. We prove a local Lipschitz continuity bound, which gives more than continuity. Let $u,v\in B(x,r)$, put $d=\Vert v-u\Vert$, and suppose $d>0$. The point $w=u+(2r/d)(v-u)$ lies in $B(x,3r)\subset U$, while $d<2r$ andBy convexity, $f(v)-f(u)\le [d/(2r)](f(w)-f(u))\le Md/r$. Interchanging $u,v$ givesThe case $d=0$ is immediate. Every point has such a neighborhood, so $f$ is locally Lipschitz and therefore continuous throughout $U$. This is the bounded convex functions are locally Lipschitz on an open ball argument; no finite-dimensional assumption is needed.

Fix $y\in E$, and take $t>0$ sufficiently small that $x+ty\in U$. For $0<s<t$, convexity along the line segment givesThus the positive secant slopes are nondecreasing in their parameter, and have a limit as the parameter decreases to zero. The local Lipschitz continuity from part (a) bounds their absolute values by $L\Vert y\Vert$, where $L=M/r$. Hence the directional derivative exists as a finite real number andFor $y=0$ it is zero.

Write $p(y)=D_{y}f(x)$. Rescaling the parameter in the directional derivative gives $p(ay)=ap(y)$ for $a>0$, and $p(0)=0$ handles $a=0$. For sufficiently small $t>0$, convexity givesSubtract $f(x)$, divide by $t$, and let $t\downarrow 0$. The finite limits from part (b) giveThese are exactly the defining conditions of a sublinear functional. In particular, a directional derivative of a convex function is sublinear; it need not be linear.

The dominated real form of the Hahn-Banach theorem extends the zero linear functional on $\{0\}$ to a linear functional $l:E\to \mathbb{R}$ with $l(y)\le p(y)$ everywhere. If $x+z\in U$, convexity of the ball puts the whole segment in $U$. Its increasing secant slopes giveThereforeThis constructs a supporting affine minorant at $x$. Part (e) shows that it is a continuous supporting functional for a locally bounded convex function.

The bound in part (b) and domination $l\le p$ give $l(y)\le L\Vert y\Vert$. Applying the same bound to $-y$, and using linearity, gives $-l(y)=l(-y)\le L\Vert y\Vert$. ConsequentlyA bounded linear functional is continuous, so $l$ belongs to the continuous dual space $E^{\prime }$. The domination by a sublinear functional alone was algebraic; the local Lipschitz continuity supplies the needed boundedness.

By part (a), $f$ is a bounded continuous function, hence is Borel measurable and integrable for the given probability measure. By part (e), $l\in E^{\prime }$, and its restriction to the unit ball is also bounded and integrable. For every $u\in U$, the supporting inequality says $f(u)\ge f(x)+l(u)-l(x)$. Taking expectations and using the barycentre hypothesis for this particular $l$ givesThus $\boxed{\mathbb{E}f\ge f(x)}$, the Jensen inequality. This proof only uses the stated weak barycentre identity; a vector-valued integral for the identity map is not needed.