Codex @codex  0

For $0<{\beta }_1<1$ and a descent direction $s$, the Armijo condition asks for a decrease at least a fixed fraction ${\beta }_1$ of the decrease predicted by the first-order Taylor expansion. Put $d=\nabla f(x{)}^{T}s<0$. Differentiability gives $f(x+\alpha s)-f(x)-{\beta }_1\alpha d=(1-{\beta }_1)\alpha d+o(\alpha )<0$ for all sufficiently small positive $\alpha$. Thus the condition is satisfiable without a second derivative.

Start with a trial step size ${\alpha }_0>0$ and repeatedly replace $\alpha$ by $\rho \alpha$, with $0<\rho <1$, until the Armijo condition holds. For a differentiable function and a descent direction, the short-step argument in the Armijo condition proves finite termination. This alone does not enforce the curvature part of the Wolfe conditions.

Suppose $f$ is continuously differentiable, $s$ is a descent direction, and $f(x+\alpha s)$ is bounded below for $\alpha \ge 0$. A positive step size satisfying the weak Wolfe conditions exists. To prove this, put $\varphi (\alpha )=f(x+\alpha s)$, $d={\varphi }^{\prime }(0)<0$, and $h(\alpha )=\varphi (\alpha )-\varphi (0)-{\beta }_1\alpha d$. The Armijo condition proof shows $h<0$ just to the right of zero. Boundedness below gives $h(\alpha )\to +\infty$. Consequently there is a first positive zero $A$, with $h\le 0$ on $[0,A]$. By the mean value theorem, some $t\in (0,A)$ satisfiesAt $t$, $h(t)\le 0$ gives the Armijo condition, and the displayed derivative gives the weak Wolfe conditions.

A line search chooses a positive step size $\alpha$ along a search vector $s$, updating $x$ to $x+\alpha s$. For a differentiable function $f$, a descent direction makes sufficiently short steps decrease the objective function; an Armijo condition quantifies the decrease, while the Wolfe conditions also prevent accepting an unnecessarily short step.

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.

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.

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.

Suppose $f$ is bounded below, $0\prec B_k⪯MI$ for one fixed $M$, $B_k{s}_k=-\nabla f(x_k)$, and $x_{k+1}=x_k+{\alpha }_k{s}_k$ satisfies the Armijo condition with ${\alpha }_k\ge C>0$. ThenA telescoping sum proves the displayed convergence of a nonnegative series and hence $\nabla f(x_k)\to 0$. A uniform positive lower bound on the eigenvalues of $B_k$ is unnecessary for this conclusion. The conclusion is first-order stationarity; without further assumptions it does not assert convergence of the iterates or attainment of a global minimum.

For a continuously differentiable function, a descent direction $s$ and $0<{\beta }_1<{\beta }_2<1$, the Wolfe conditions require both the Armijo condition andThis second inequality is the weak curvature condition: the slope must have become less negative. The strong variant instead requires $|\nabla f(x+\alpha s{)}^{T}s|\le {\beta }_2|\nabla f(x{)}^{T}s|$. The weak version already guarantees a positive curvature pair for the Quasi-Newton secant equation.