The rectangle avoids the singular line $y=0$ exactly when $0<k<1$. There $y>0$, and the largest $|x|$ on $[1-h,1+h]$ is $1+h$. Since $1+1/y$ decreases with $y$, the smallest feasible uniform bound isIt is attained at $x=1+h$, $y=1-k$. For the Lipschitz constant, the mean value theorem givesThe quotients approach this upper bound as $x=1+h$ and $y,z\downarrow 1-k$, so the smallest feasible constant isSubstitution into part (a) givesThe contraction mapping restriction is automatic for every M-test-admissible $h$ precisely when $A(k)<B(k)$. Because $0<k<1$,Thereforeis exactly the range where the two proofs guarantee all the same closed symmetric intervals, including the largest M-test interval.

For completeness, set $h_M=(\sqrt{1+4A(k)}-1)/2$ and $h_C=(\sqrt{1+4B(k)}-1)/2$. The M-test allows $0<h\le h_M$; the contraction mapping theorem allows $0<h\le h_M$ with $h<h_C$. At $k=2-\sqrt{2}$, these have the same supremum $h_M=h_C$, but contraction excludes the endpoint $h=h_M$. For $2-\sqrt{2}<k<1$, it gives the strictly shorter range $0<h<h_C<h_M$. For a particular $h$, both proofs work exactly when both displayed inequalities hold, even for these larger $k$.

With the standard constant starting Picard iterate, $y_0(x)=1$, the next two Picard iterates areand, for $x>0$,ThusThe restriction $x>0$ for $y_2$ is essential for the integral from $1$ to $x$: the preceding iterate $y_1$ vanishes at $0$, where the new integrand $s+1/s$ is undefined.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.

For the Weierstrass M-test proof of the Picard-Lindelöf theorem, the required bound isTo see this, use Picard iteration starting with $y_0(x)=b$ andIf $y_n$ takes its values in $[b-k,b+k]$, then $|y_{n+1}-b|\le M|x-a|\le Mh\le k$. Thus every Picard iterate remains in the rectangle where the hypotheses hold. Successive applications of the Lipschitz condition and integration give the factorial estimate for Picard iterationFor example, the $n=0$ estimate is $M|x-a|$, and integrating the $n$th estimate multiplied by $L$ proves the next one. The sum of the uniform bounds isfor every finite $Lh$, by the exponential series. Hence the telescoping series of Picard iterates converges uniformly by the Weierstrass M-test. The uniform limit stays in the closed rectangle, and the Lipschitz condition permits passage to the limit in the integral equation. The fundamental theorem of calculus then gives the desired solution of a differential equation on $[a-h,a+h]$.

The second-order differential operator factors asTherefore setting $u=v_x$ turns the second-order partial differential equation into exactly the first-order partial differential equation in part (a). Conversely, given a sufficiently smooth solution of a differential equation $u$ of that equation on a coordinate rectangle, defineThe fundamental theorem of calculus gives $v_x=u$, so this solves the second-order equation, with an arbitrary twice continuously differentiable function $F$. This classical correspondence is local on smooth patches; for the cornered solution of part (b), it can instead be interpreted using distributional derivatives.

For $\beta =0$, the equation for the characteristic curves isThe two families are therefore $y=\text{constant}$ and the characteristic projections from part (a), $y=K{x}^2$, together with the vertical characteristic $x=0$. On either region $y>0$ or $y<0$, the hyperbolic partial differential equation admits the characteristic coordinatesIndeed $\phi$ is constant along the first family, while $(x{\partial }_x+2y{\partial }_y)\psi =0$. Their Jacobian determinant isThe coefficients of the two pure second partial derivatives in these coordinates vanish, and the coefficient of the mixed partial derivative is $2y/\sqrt{|y|}\ne 0$. Hence they give the hyperbolic characteristic normal form. They also remain regular at $x=0$, provided $y\ne 0$.

The alternative $\phi =y$, $\psi =y/x^2$ is valid on patches with $xy\ne 0$. On $y=0$, it is a parabolic partial differential equation away from the fully degenerate origin, so two independent hyperbolic characteristic coordinates are unavailable.

In the convention $A{v}_{xx}+2B{v}_{xy}+C{v}_{yy}=\cdots$, the coefficients of the principal symbol areThus the second-order partial differential equation isThere is no region of elliptic partial differential equation type. The terms hyperbolic, parabolic and elliptic refer to the pointwise classification of a second-order partial differential equation. At $(0,0)$ every second-order coefficient vanishes, so the principal symbol has rank zero and the equation is degenerate rather than genuinely parabolic there. Moreover the right side is $2$, so no twice continuously differentiable classical solution can satisfy the equation at the origin.

For $\beta =1$, $q=(y+x-x^2)/x^2$. On the two sides of the corner-carrying characteristic curve, the partial derivative isAt every point $(x,x^2-x)$ with $x>0$, the two one-sided partial derivatives are $x^{-2}$ and $0$. Therefore $u_y$ does not exist on this curve; its piecewise values have a jump discontinuity of magnitude $x^{-2}$. There are no other interior discontinuities. The values on the outer boundaries are interpreted from within ${\mathcal{D}}_1$.

This is exactly propagation of a derivative jump along a characteristic. Already on $x=1$, the data have slopes $1$ and $0$ on the two sides of $y=0$. The transport equation transports this corner along the characteristic curve from $(1,0)$, which is $y=x^2-x$; it provides no smoothing. This is consistent with Noncharacteristic Cauchy data and does not contradict the smooth-data theorem, whose differentiability hypothesis fails here.

Every point with $x>0$ has exactly one backward characteristic curve meeting $x=1$, at the point $(1,q(x,y))$. It meets the specified data precisely when $-1\le q\le 1$. Thus the whole region determined by these data, including the two bounding characteristic curves, isIts interior is the corresponding open domain for the partial differential equation. Uniqueness here means uniqueness in the class of continuous functions satisfying the characteristic equations for a transport equation, including the piecewise smooth function defining the solution of a differential equation above. The endpoints of the initial segment determine the two boundary values by continuity.

This region is maximal for determination by the given data: on $x>0$, arbitrary extensions of $g$ outside $[-1,1]$ give different values on the characteristic curves with $q\notin [-1,1]$. No characteristic curve from $x=1$ reaches $x\le 0$.

For $\beta =1$,The lower boundary is a straight line, the upper boundary is a parabola, and the characteristic carrying the corner is $y=x^2-x$. All three approach the excluded origin as $x\downarrow 0$.

Use the initial conditions $x(0,r)=1$, $y(0,r)=r$, $u(0,r)=g(r)$ in the characteristic equations for a transport equation. The resulting parametrized solution surface isThe Jacobian determinant of $(s,r)\mapsto (x,y)$ is $e^{3s}>0$. In fact the inverse is explicit:Consequently the explicit solution of a differential equation on the region reached from the data isTo verify it away from $q=0$, set $D=x{\partial }_x+(2y+\beta x){\partial }_y$. Direct differentiation gives $Dq=0$ and $D(x-1+2\log x)=x+2$. At $x=1$ we have $q=y$, giving precisely the prescribed boundary condition.

Across $q=0$ the formula remains continuous and locally Lipschitz, is a smooth function on each side, and solves the partial differential equation along every characteristic curve. It is also a weak solution, since its first distributional derivatives agree almost everywhere with its ordinary derivatives. It is not a classical solution throughout the region because of the inherited corner.

Parametrize the initial curve by $(x_0(r),y_0(r))=(1,r)$, $-1\le r\le 1$. Its tangent is $(0,1)$, whereas the projected characteristic vector field there is $(1,2r+\beta )$. The determinantproves that these are Noncharacteristic Cauchy data: the initial curve is everywhere transverse to the characteristic curves.

There is a regularity qualification. The prescribed function $g(r)=\min \{r,0\}$ is continuous and Lipschitz continuous, but is not differentiable at $r=0$. Thus the data are Cauchy in this geometric sense, with data given by a smooth function on each of $r<0$ and $r>0$; the usual theorem giving a global classical solution from continuously differentiable data does not apply across $r=0$.

For an active constraint with ${\lambda }_i^{\ast }>0$, the perturbed complementary slackness equation gives $c_i(x^k)\sim {\mu }_k/{\lambda }_i^{\ast }$. The Hessian matrix from part (c) therefore includesIf $\nabla c_i(x_{\ast })\ne 0$, curvature normal to that boundary grows like $1/{\mu }_k$. By contrast, a direction tangent to every active constraint annihilates these large normal outer products at the limiting point, and its curvature may remain of order one. A large spread of eigenvalues then makes the Newton linear system poorly conditioned. Under bounded remaining curvature and a positive bounded tangential curvature, the condition number typically grows like $1/{\mu }_k$; this is the usual ill-conditioning of a logarithmic barrier.

It is not a universal rate: there may be no nonzero common tangent direction, some limiting Lagrange multipliers may vanish, or all curvatures may grow together. The Hessian formula identifies the source of the difficulty without assuming these extra nondegeneracy conditions.

A precise conditional convergence statement is: with the smoothness and convexity assumptions of the question, a strict feasible point, ${\mu }_k\downarrow 0$, and barrier stationarity residualsevery convergent sequence of strictly feasible points $x^k\to x_{\ast }$ has a global minimizer as its limit. The Lagrange multiplier estimates are bounded and have a subsequence converging to multipliers satisfying the KKT conditions. The theorem does not assert convergence of the primal sequence without further assumptions.

First, continuity gives $c_i(x_{\ast })\ge 0$. For an inactive optimization constraint, $c_i(x_{\ast })>0$, soFor the active constraints, the hint permits assuming ${\lambda }_i^k\to {\lambda }_i^{\ast }$. Combining this with the inactive limits gives ${\lambda }_{\ast }\ge 0$, and passage to the limit in the residual equation proves stationarity. Since ${\lambda }_i^k{c}_i(x^k)={\mu }_k\to 0$, complementary slackness also holds. All KKT conditions follow, and part (b) proves global optimality.

The hint's multiplier assumption can in fact be weakened to selecting a subsequence. To see why the estimates are bounded, choose a strict feasible point $z$ and put $\delta ={\min }_i{c}_i(z)>0$. Concavity givesThe right side is bounded because $x^k\to x_{\ast }$ and $r_k\to 0$. This is the bounded barrier multipliers under Slater's condition argument. A convergent multiplier subsequence then gives the same KKT point, which suffices to prove that the assumed primal limit is globally optimal.

Choose a strictly feasible point, an initial barrier parameter ${\mu }_0>0$, and $0<\theta <1$. At outer iteration $k$, approximately minimize the logarithmic barrier objective $f_{{\mu }_k}$ over its strictly feasible domain of a function. For example, use damped Newton methods, choosing an inner step size by backtracking line search so that every $c_i$ stays positive and the barrier objective function decreases. Solve to a tolerance $\Vert \nabla f_{{\mu }_k}(x^k)\Vert \le {\epsilon }_k$, where ${\epsilon }_k\to 0$. Set ${\mu }_{k+1}=\theta {\mu }_k$ and use $x^k$ as the next starting point.

The associated Lagrange multiplier estimates areReducing ${\mu }_k$ makes these perturbed complementary slackness equations approach the KKT conditions. This describes the generic interior-point method; existence of a solution to every inner minimization needs appropriate additional attainment assumptions and does not follow from strict feasibility alone.

Let $G={\prod }_i{y}_i^{a_i}>0$ be the weighted geometric mean and set $x_i=y_i/G$. Since the weights sum to one,Part (i) implies ${\sum }_i{a}_i{y}_i/G\ge 1$, proving the weighted arithmetic-geometric mean inequality:Equality holds exactly when $x_i=1$ for every $i$, or equivalently when all the $y_i$ are equal.

Taking the natural logarithm of the positive product constraint gives $h(x)={\sum }_i{a}_i\log x_i=0$. Its gradient has entries $a_i/x_i>0$, so the equality constraint qualification holds everywhere on the feasible set. Use the optimization LagrangianStationarity requires $a_i-\lambda a_i/x_i=0$, so every $x_i=\lambda$. Substitution into the constraint, using ${\sum }_i{a}_i=1$, gives $\lambda =1$. Thus the only candidate for a local minimum is $x=(1,\dots ,1)$.

To prove global minimality without assuming the inequality to be established, set $u_i=\log x_i$. The constraint becomes the affine hyperplane ${\sum }_i{a}_i{u}_i=0$, and the objective function becomes $F(u)={\sum }_i{a}_i{e}^{u_i}$. Its Hessian matrix $\mathrm{diag}(a_i{e}^{u_i})$ is positive-definite, so $F$ is strictly convex. The elementary tangent inequality $e^t\ge 1+t$, with equality only at zero, givesEquality forces every $u_i=0$. ThereforeThis point is the unique global minimizer and the only local minimizer. The logarithm change of variables also explains why a multiplicative constraint leads to a convex optimization argument.

The Cauchy point provides both an inexpensive fallback step and a standard against which a better step can be judged. Requiring a fixed fraction of Cauchy decrease givesConsequently an exact global minimum of each trust-region subproblem is unnecessary. A step from any inner solver is adequate if it meets this bound.

The convergence argument combines this decrease with model accuracy. For example, suppose the gradient is uniformly Lipschitz continuous on a neighborhood of the visited level set, the model Hessian matrices are uniformly bounded, and the usual acceptance and radius safeguards are used. If $\Vert \nabla f(x^k)\Vert \ge ϵ>0$, the Taylor expansion gives a model error bounded by $K\Vert s_k{\Vert }^2$. For sufficiently small ${\Delta }_k$, division by the Cauchy decrease bound makes $|1-{\rho }_k|$ small. Such trials are successful and cannot keep shrinking the radius. With the gradient bounded away from zero, the radius therefore has a positive lower bound, and each accepted step decreases $f$ by a fixed positive amount. There must be infinitely many accepted steps, since repeated rejection would force the radius below the small-radius acceptance threshold. This contradicts boundedness below of $f$ and proves ${\mathrm{liminf}}_k\Vert \nabla f(x^k)\Vert =0$.

Stronger standard conclusions can be obtained with further regularity and safeguards. Merely being twice continuously differentiable and bounded below does not supply uniform Hessian bounds on an unbounded level set; those uniform assumptions must be stated when using this argument.

At $x^k$, choose $a=f(x^k)$, $g=\nabla f(x^k)$, $H={\nabla }^{2}f(x^k)$, and the current radius $\Delta ={\Delta }_k>0$. A bounded symmetric matrix approximation to the Hessian matrix can also be used. Stop if the gradient is sufficiently small; otherwise approximately solve the trust-region subproblem to obtain $s_k$ with $\Vert s_k\Vert \le {\Delta }_k$ and at least a fixed fraction of Cauchy decrease. Its predicted decrease is then positive, so calculate the trust-region acceptance ratioFix $0<{\eta }_1<{\eta }_2<1$, $0<{\gamma }_d<1<{\gamma }_u$, and a maximum radius ${\Delta }_{\max }$. One valid update is:

The trust region controls the reliability of the model; the ratio decides whether that reliability was borne out by the actual objective function.

The Cauchy point minimizes the model along the negative gradient ray inside the trust region. With $q=g^{T}Hg$, writeIf $q\le 0$, this last quadratic function decreases throughout the interval. If $q>0$, its unconstrained minimum occurs at $t=\Vert g{\Vert }^2/q$. ThusThe additive constant $a$ has no effect on the Cauchy point. To obtain the Cauchy decrease bound, first suppose $q>0$. Since $\tau q\le \Vert g{\Vert }^2$,The spectral norm inequality $q\le \Vert H\Vert \Vert g{\Vert }^2$ gives $\tau \ge \min \{\Delta /\Vert g\Vert ,1/\Vert H\Vert \}$. HenceIf $q\le 0$, the decrease is at least $\Delta \Vert g\Vert$, which also implies the displayed bound. When $H=0$, interpret $\Vert g\Vert /\Vert H\Vert =+\infty$. Thus the universal choice is $c=1/2$.

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.