Past exam of the continuous optimisation course of the University of Oxford / 2025 / 1

Fix an iteration and abbreviate $x=x^k$, $s=s^k$, $d=\nabla f(x{)}^{T}s$. Since $s$ is a descent direction, $d<0$. The directional derivative givesChoose $\bar{\alpha }>0$ so that $|o(\alpha )|\le \frac{1}{2}(1-{\beta }_1)(-d)\alpha$ for $0<\alpha \le \bar{\alpha }$. ThenThis is the Armijo condition for every such positive step size, and $\alpha =0$ gives equality. In particular, a backtracking line search terminates after finitely many reductions.

Write $\varphi (\alpha )=f(x+\alpha s)$ and $h(\alpha )=\varphi (\alpha )-\varphi (0)-{\beta }_1\alpha d$. Part (a) gives $h(\alpha )<0$ just to the right of zero. If $L$ is a lower bound for $f$, thenbecause $d<0$. By continuity, there is a first positive zero $A$ of $h$, and $h(\alpha )\le 0$ on $[0,A]$. Apply the mean value theorem to $\varphi$ on this interval: for some $t\in (0,A)$,At this same $t$, $h(t)\le 0$ supplies the Armijo condition. Thus $\alpha =t>0$ satisfies both weak Wolfe conditions. The boundedness assumption prevents the objective function from following its initial negative slope forever; this is the mechanism behind Existence of a Wolfe step.

If $g=\nabla f(x)\ne 0$, the positive-definite symmetric matrix $B$ is an invertible matrix and $s=-B^{-1}g\ne 0$. Multiplying $Bs=-g$ by $s^T$ givesHence $s$ is a descent direction. If $g=0$, then $s=0$ and the algorithm has already reached a stationary point; the strict descent assertion applies before termination.

Set $p=x^{k+1}-x^k={\alpha }^k{s}^k$ and $y=\nabla f(x^{k+1})-\nabla f(x^k)$. The Quasi-Newton secant equation is $B^{k+1}p=y$. If $B^{k+1}$ is a positive-definite symmetric matrix and $p\ne 0$, it follows thatThe weak curvature part of the Wolfe conditions ensures precisely this necessary condition:The last product is positive because ${\alpha }^k>0$, ${\beta }_2-1<0$, and $s^k$ is a descent direction. This explains why a Quasi-Newton method often uses the Wolfe conditions rather than only the Armijo condition.

Interpret boundedness above uniformly: there is an $M<\infty$ such that $B^k⪯MI$ for every $k$. Since $B^k$ is a positive-definite symmetric matrix, the spectral theorem implies $(B^k{)}^{-1}⪰M^{-1}I$. ThereforeCombining the Armijo condition with ${\alpha }^k\ge C>0$ yieldsIf $f\ge L$, a telescoping sum givesThe nonnegative series converges, so its terms tend to zero:This proves first-order global convergence in the sense of stationarity from uniformly bounded quasi-Newton matrices. It does not imply that every nonconvex objective function reaches a global minimum, or that the iterates themselves converge. Every finite limit point is a stationary point, by continuity of the gradient.