Choose $x_0$, fixed $\bar{\alpha }>0$, $\sigma ,\beta \in (0,1)$, and repeat for every $n\ge 0$: compute $g_n=\nabla f(x_n)$, choose ${\alpha }_n$ by the backtracking line search in part (b) with $s_n=-g_n$, and set $x_{n+1}=x_n-{\alpha }_n{g}_n$. At a stationary point keep the sequence constant.

Assume $f\in C^1$ has a globally Lipschitz gradient and is bounded below on its initial sublevel set $\{x:f(x)\le f(x_0)\}$. Then the steepest descent method satisfies $\boxed{\nabla f(x_n)\to 0}$. Indeed, the Armijo condition and the uniform lower step bound giveThe objective values decrease and stay in the stated sublevel set. Summing the inequality and using its lower bound $f_{\inf }$ yieldswhich proves the limit. Every cluster point is a stationary point by continuity of the gradient. If the initial sublevel set is compact, cluster points exist. This global convergence result guarantees stationarity from arbitrary initial points under these assumptions; it does not, for a general nonconvex objective, guarantee convergence of the whole iterate sequence or a global minimum.

A line search chooses a positive step size giving useful decrease along the chosen direction. Fix $\sigma ,\beta \in (0,1)$ and a trial step $\bar{\alpha }>0$. The Armijo condition isThe backtracking line search tests $\bar{\alpha },\beta \bar{\alpha },{\beta }^2\bar{\alpha },\dots$ and takes the first accepted value. A Lipschitz gradient has $\Vert \nabla f(y)-\nabla f(x){\Vert }_2\le L\Vert y-x{\Vert }_2$ with a constant $L$ independent of $x,y$.

Integrating the gradient along a segment gives the descent lemmaFor $s=-g$, every $0<\alpha \le \tau =2(1-\sigma )/L$ satisfies the Armijo condition. Thus backtracking terminates. If the trial step is accepted, $\alpha =\bar{\alpha }$; otherwise the immediately preceding step $\alpha /\beta$ failed, so it exceeded $\tau$. ConsequentlyThe same global Lipschitz constant and fixed search parameters make $c$ independent of $x$. If $g=0$, the zero direction passes the first test. If $L=0$, the gradient is constant and the first test passes, so take $c=\bar{\alpha }$.

A descent direction $s$ satisfies $g^{T}s<0$, where $g=\nabla f(x)$; the first-order Taylor expansion then gives $f(x+\alpha s)<f(x)$ for sufficiently small $\alpha >0$. To compare directions one must fix their scale. Among Euclidean unit vectors, the Cauchy-Schwarz inequality givesThis is the unit steepest descent direction. The conventional unnormalized direction used in the algorithm is the same ray, $\boxed{s=-g}$. Equivalently it is the unique minimizer of $g^{T}s+\frac{1}{2}\Vert s{\Vert }_2^2=\frac{1}{2}\Vert s+g{\Vert }_2^2-\frac{1}{2}\Vert g{\Vert }_2^2$. If $g=0$ there is no strict first-order descent direction, and the conventional update is zero.

Along a fixed nonzero branch, $\Vert x(\mu ){\Vert }_2^2=1+\mu /{\lambda }_1\to 1$ and the radial Hessian eigenvalue grows like $4{\lambda }_1^2/\mu$. The nonzero transverse eigenvalues $2({\lambda }_i-{\lambda }_1)$ stay fixed. If ${\lambda }_1$ is simple, the ratio of the largest to smallest absolute eigenvalue therefore tends to infinity. If it repeats, the Hessian is already singular because of the zero transverse directions; there is no finite spectral condition number to begin with. These conclusions concern the nonzero branches: at the stationary origin $H=2(A+\mu I)$, and the same blowup is not implied.

Ill-conditioning does not invalidate barrier minimizers as approximations to the constrained solution. Let ${\lambda }_{\min }<0$. At radius $r$, the smallest possible quadratic value is ${\lambda }_{\min }{r}^2$, so global barrier minimization reduces to minimizing ${\lambda }_{\min }t-\mu \log (1-t)$ for $0\le t<1$. For $0<\mu <-{\lambda }_{\min }$ its unique minimizing radius satisfiesThus every limiting unit vector in that eigenspace solves the original problem, with objective ${\lambda }_{\min }$. For $\mu \ge -{\lambda }_{\min }$ the barrier minimizer is the origin instead. The increasing condition number concerns the numerical difficulty of small-parameter subproblems, which can call for scaling or a primal-dual interior-point method; it does not prevent the mathematical convergence just established. This is the quadratic-ball barrier eigenvalue branch.

Put $d=1-x^{T}x>0$. Differentiating the barrier gradient givesAt a nonzero stationary point on the branch $Ax={\lambda }_{1}x$, $\mu /d=-{\lambda }_1$. Choose an orthonormal eigenbasis of the symmetric matrix $A$ whose first vector is $x/\Vert x{\Vert }_2$. On each of the remaining eigenvectors the rank-one term vanishes, so the corresponding Hessian eigenvalues areAlong $x$, the first two terms cancel and the remaining eigenvalue isIf ${\lambda }_1$ repeats, a transverse eigenvalue can be zero. If a lower eigenvalue than ${\lambda }_1$ exists, a transverse eigenvalue is negative, identifying this stationary branch as a saddle rather than a minimum.

The logarithmic barrier isIts gradient is $2Ax+2\mu x/(1-x^{T}x)$. Thus every nonzero stationary point satisfiesA fixed negative-eigenvalue branch exists for $0<\mu <-{\lambda }_1$. The zero point is stationary for every $\mu$ as well, and it is not an eigenvector; at zero the displayed scalar $-\mu$ need not be an eigenvalue of $A$. Thus the claimed eigenvalue conclusion requires the nonzero-point qualification. This distinction is essential for the following Hessian analysis.

Write the constraint as $c(x)=1-x^{T}x\ge 0$ and its multiplier as $y\ge 0$. The KKT conditions becomeIf $A$ is positive definite, $A+yI$ is positive definite for every $y\ge 0$, so $\boxed{x=0,y=0}$ is the only critical point.

For indefinite $A$, the origin remains a critical point, contrary to an unqualified reading of the question. Every nonzero critical point is an eigenvector with $\boxed{Ax=-yx}$, so its eigenvalue is nonpositive. A negative-eigenvalue vector must lie on the unit sphere, with multiplier equal to the negative of that eigenvalue. A nonzero vector in $\ker A$ may have any norm at most one and has multiplier zero. Positive-eigenvalue directions do not satisfy the multiplier sign condition. These first-order critical points include saddles; the global minimum selects a unit vector in the lowest-eigenvalue eigenspace.

With $L(x,y)=f(x)-y^{T}c(x)$, the Karush-Kuhn-Tucker conditions areUnder a suitable constraint qualification, such as the linear independence constraint qualification on active constraint gradients, they are necessary for a local minimum. Smoothness alone is insufficient. The sign convention here corresponds to constraints written as $c_i\ge 0$ and a minus sign in the optimization Lagrangian.

Two reasons to prefer more advanced methods for smooth, high-accuracy optimization are:

The question's blanket wording is too strong: the steepest descent method and related gradient algorithms are used in practice, especially when problem size, cheap gradients or simplicity make curvature-based alternatives costly. The objections describe limitations, not universal nonuse.

Use the uniform lower step bound from part (iv):Thus $\boxed{g(x_k)\to 0}$. This is convergence to stationarity in the usual global convergence theorem for the steepest descent method. It does not itself prove that the iterates converge, or that any stationary limit is a global minimizer of a nonconvex objective.

The Armijo condition with $p_k=-g(x_k)$ gives $f(x_k)-f(x_{k+1})\ge \beta {\alpha }_k\Vert g(x_k){\Vert }_2^2$. Summing these inequalities telescopes toIf $f\ge f_{\inf }$, these partial sums of nonnegative terms are bounded above by $(f(x_0)-f_{\inf })/\beta$. The resulting series converges, so its terms tend to zero: $\boxed{{\alpha }_k\Vert g(x_k){\Vert }_2^2\to 0}$.

For $p=-g\ne 0$, part (ii) simplifies to ${\alpha }_{\max }=2(1-\beta )/\gamma$. If the initial step is accepted, ${\alpha }_k={\alpha }_{\mathrm{init}}$. Otherwise the immediately preceding trial ${\alpha }_k/\tau$ failed and hence exceeded ${\alpha }_{\max }$. ThereforeThe fixed global Lipschitz constant and search parameters make this lower bound uniform across iterations. At $g=0$, the zero update passes the first trial and the sequence can be kept stationary.

The backtracking line search starts with ${\alpha }_{\mathrm{init}}>0$, tests the Armijo condition, and repeatedly replaces a failed step by $\tau \alpha$, with fixed $\tau \in (0,1)$. It accepts the first step that passes. Backtracking makes an overambitious trial step small enough for a local linear model to predict the change reliably; part (ii) ensures finite termination. The Armijo condition guarantees a decrease of at least $-\alpha \beta g^{T}p>0$ along a nonzero descent direction. Merely obtaining some decrease without this quantitative lower bound would not provide the convergence estimate in the following parts.

A Lipschitz gradient with constant $\gamma >0$ gives the descent lemmaFor $\alpha >0$, this upper bound is at most $f(x)+\alpha \beta g^{T}p$ whenever $\frac{\gamma }{2}\alpha \Vert p{\Vert }_2^2\le (\beta -1)g^{T}p$. Since $p$ is a descent direction, the right side is positive. Thus the Armijo condition holds throughoutAt $\alpha =0$ it holds by equality. If the gradient Lipschitz constant can be taken as zero, the objective is affine and every positive step along a descent direction passes this sufficient-decrease test.

For $g=\nabla f(x)\ne 0$, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ by the Cauchy-Schwarz inequality. The usual unnormalized direction is $\boxed{p=-g}$; choosing its scale separately as a line-search step gives the same ray. It satisfies $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$, so it is a descent direction. The first-order Taylor expansion then guarantees decrease for sufficiently small positive steps. The subsequent bounds use this unnormalized negative-gradient convention.

Suppress the iteration subscript. From the quadratic subproblem, $As=-c$ and $g=-Bs+A^{T}y$. ThereforeWhen $|c^{T}y|>0$, the stated bound on $\mu$ makes ${\mu }^{-1}\Vert c{\Vert }_p^p\ge |c^{T}y|$. Consequentlyusing positive definiteness of $B$. Thus every nonzero such SQP step is a descent direction for the penalty function. If $c^{T}y=0$, the quotient bound is unnecessary and the same conclusion follows for every $\mu >0$. A zero step has zero directional derivative and is not a strict descent direction; it corresponds to an already feasible stationary point. This nonzero-step qualification is implicit in the requested conclusion.

Since $p$ is even, differentiating $c_i^p$ givesAt a stationary point define $y_i=-c_i(x{)}^{p-1}/\mu$. Then $\nabla f(x)-Dc(x{)}^{T}y=0$, precisely the stationarity equation for the optimization Lagrangian. Bounded multipliers giveThus along any subsequence of stationary points with $x_{\mu }\to x_{\ast }$, a further subsequence of bounded multipliers converges to $y_{\ast }$, and continuity yields $c(x_{\ast })=0$ and $\nabla f(x_{\ast })-Dc(x_{\ast }{)}^T{y}_{\ast }=0$. These are the equality-constrained first-order conditions. Bounded multipliers alone do not establish convergence of the primal points; the precise conclusion concerns their finite cluster points. This is the power penalty multiplier limit.

Stationarity of the equality-constrained quadratic subproblem gives $B_k{s}_k+g_k-A_k^T{y}_{k+1}=0$. Together with feasibility, this isMultiplying the bottom block row by $-1$ gives a symmetric indefinite saddle-point matrix, with right side $(-g_k,c_k{)}^T$.

One solution strategy uses the Schur complement. From $s_k=B_k^{-1}(A_k^T{y}_{k+1}-g_k)$ obtainThis is an $m\times m$ positive-definite system when $A_k$ has full row rank; factor $B_k$ and solve with it rather than forming its inverse.

A second strategy uses a nullspace basis $Z$ for $\ker A_k$. Choose one particular $s_p$ with $A_k{s}_p=-c_k$, write $s_k=s_p+Zz$, and solve the reduced systemThe reduced Hessian is positive definite. These range-space and nullspace reductions exploit different relative sizes of $m$ and $n-m$.

The sequential quadratic programming step minimizes a quadratic local objective model while enforcing the first-order constraint expansion $c(x_k+s)\approx c_k+A_{k}s=0$. When the iterate and multipliers are close to a regular solution and $B_k$ approximates the Lagrangian Hessian, the resulting correction approximates a Newton step for the nonlinear KKT conditions.

Full row rank of $A_k$ ensures that $A_{k}s=-c_k$ is solvable for every $c_k$ and that the multiplier of the quadratic subproblem is unique. A positive-definite $B_k$ makes the objective strictly convex and coercive on the affine feasible set, giving a unique global subproblem minimizer. In fact positivity on $\ker A_k$, rather than on every vector, is enough for uniqueness and boundedness on that set; full positive definiteness is a convenient stronger assumption.

Use the optimization Lagrangian $L(x,y)=f(x)-y^{T}c(x)$ and let $A=Dc$ be the $m\times n$ constraint Jacobian matrix. Under the standard regularity hypothesis $\mathrm{rank}A(x_{\ast })=m$, a local constrained minimizer satisfies, for some multiplier $y_{\ast }$,The second-order necessary condition isRegularity makes each such tangent vector the velocity of a feasible smooth curve. Twice differentiating the objective along that curve and using stationarity and the differentiated constraints gives the displayed nonnegative second derivative.

The regularity hypothesis is needed for these multiplier conditions to be necessary in this form. For example minimizing $f(x)=x$ subject to $c(x)=x^2=0$ has the unique feasible minimizer zero, but $f^{\prime }(0)-y{c}^{\prime }(0)=1$ for every $y$. The question does not explicitly state the qualification, so it must be supplied when using the usual KKT conditions.