Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 4

Put $\mathbf{h}=\mathbf{y}-\mathbf{x}$. For $0<t\le 1$, convexity gives $f(\mathbf{x}+t\mathbf{h})\le (1-t)f(\mathbf{x})+tf(\mathbf{y})$, henceBy the first-order form of Taylor's theorem, the right side tends to ${\mathbf{h}}^{\mathsf{T}}\nabla f(\mathbf{x})$ as $t\downarrow 0$. ThereforeThis is the supporting inequality for a differentiable convex function. The stated partial derivatives are enough: coordinate derivatives give directional Taylor expansion for convex functions, so no unmentioned continuity assumption on the second partial derivatives is needed. The result in fact needs less smoothness than the question assumes.

No. Both $f(x)=x$ and $g(x)=-x$ are affine functions, hence convex functions. Their product $h(x)=-x^2$ is not convex: at the midpoint of $-1$ and $1$, $h(0)=0>\frac{1}{2}h(-1)+\frac{1}{2}h(1)=-1$. Thus products of convex functions need not be convex, even for smooth functions on all of $\mathbb{R}$.

Differentiating on the given domain yieldsThe denominator and $\ln x+1$ are positive there. Thus $j$ is strictly concave, and is not convex.

The denominator is positive throughout the domain. Holding $y$ fixed, differentiate with respect to $x$:At the interior point $(1,1/2)$ this second partial derivative is $-2/9<0$, so the restriction to a horizontal line is locally strictly concave, ruling out convexity. At the interior point $(1,2)$ it is $4/9>0$, ruling out concavity. Restrictions of a convex function or a concave function to every line segment must retain that property. Consequently $k$ is neither convex nor concave on its stated domain.

Since $\mathbf{a}\ne 0$, the line $\mathbf{x}=t\mathbf{a}/\Vert \mathbf{a}{\Vert }^2$ lies in the domain and satisfies ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=t$. The restriction of $l$ to this line is $t^3$, whose second derivative is $6t$: negative for $t<0$ and positive for $t>0$. It can be neither globally convex nor globally concave. Hence $l$ is neither. Equivalently, its Hessian matrix is $6({\mathbf{a}}^{\mathsf{T}}\mathbf{x})\mathbf{a}{\mathbf{a}}^{\mathsf{T}}$, changing the sign of its nonzero directional curvature across ${\mathbf{a}}^{\mathsf{T}}\mathbf{x}=0$.