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The function $x\mapsto \cos (1/x)$ is uniformly continuous on $[1,\infty )$. Its derivative satisfiesso the mean value theorem givesThus the function is Lipschitz, which implies uniform continuity.

Expand about $t=t_{n+2}$ and use $y^{\prime \prime }=f^{\prime }(y)f(y)$ from the chain rule. The residual of the exact solution isThe coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the multiderivative multistep method has order three.

At $h=0$, its first characteristic polynomial isIts roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the root condition for a multistep method and is zero-stable. Combining this with the defect estimate proves convergence of order three, assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $h{F}_h(y_{n+2})$, where $F_h=6f/7-2h{f}^{\prime }f/7$ has a uniform local Lipschitz continuity bound for small $h$. A bounded zero-stable impulse response and the discrete Gronwall inequality therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is convergence of a zero-stable multiderivative method, rather than a direct application of the first-derivative-only Dahlquist equivalence theorem.

Let $e_n=y^n-y(t_n)$. Subtracting the exact one-step relation from the numerical method and applying the stated Lipschitz continuity givesIf $\Vert {\tau }_{n+1}\Vert \le C{h}^{p+1}$, iteration yields the discrete Gronwall inequalityFor $nh\le t^{\ast }$,andConsequently

LetDifferentiating and using the scale equation givesSince $b$ is bounded, $h$ has global Lipschitz continuity. Thushas a pathwise unique strong solution by the standard Lipschitz existence-and-uniqueness theorem for a stochastic differential equation. Applying the deterministic inverse $g^{-1}$ gives a strong solution $X$, and uniqueness of $Y$ gives pathwise uniqueness of $X$.

For the equation actually produced by the preceding construction, namely the Bessel process equation of dimension $2-1/\delta$, the drift has Lipschitz continuity on every compact subset of $(0,\infty )$. Starting at any positive time and position, pathwise uniqueness therefore makes the time-changed process agree until its first hit of zero with the maximal local solution of a stochastic differential equation. For $\delta >1$, its dimension lies in $(1,2)$, so it can hit zero; the time-change construction then supplies further excursions, whereas the maximal local solution on $(0,\infty )$ stops at that first hit.

For $\delta =1$, $X_s=|B_s|$ is Reflected Brownian motion; away from zero it agrees with the maximal local solution of $dX=d\tilde{B}$. For the dimension-$\delta$ equation printed in the paper, the preceding construction does not agree with the maximal local solution unless $\delta =1$, for the coefficient mismatch established in part (d).

Because the Fejér kernel has normalized integral one,For $f\in \mathrm{Lip}\alpha$ this impliesOn $0\le t\le \pi$, the inequalities $\sin (t/2)\ge t/\pi$ and $|\sin (nt/2)|\le n|\sin (t/2)|$ giveSplitting the integral at $1/n$ givesFor $0<\alpha <1$, both terms are $O(n^{-\alpha })$. For $\alpha =1$, the second is $O((\log n)/n)$. Uniformly in $x$,where the constants absorb the Lipschitz continuity constant $M$.

The Euclidean projection onto a convex set is nonexpansive, and $x^{\ast }=P_C(x^{\ast })$ because the optimum is feasible. Therefore the projected subgradient method satisfiesThe subgradient inequality gives $⟨g_i,x_i-x^{\ast }⟩\ge f(x_i)-f^{\ast }$, while Lipschitz continuity of the finite convex function gives $\Vert g_i{\Vert }_2\le G$. HenceSumming this telescoping inequality for $0\le i<k$, and then bounding the smallest term by the average, yieldsWriting $D=\Vert x_0-x^{\ast }{\Vert }_2$, the right-hand side is minimized by the constant step sizeSubstitution givesIf $D=0$, the initial point is already optimal and the result is immediate.

Write $a_i^T$ for row $i$ of $A$, and define affine functionsThenThe pointwise maximum of convex functions is convex, so $f$ is a convex function. Moreover,so $f$ has Lipschitz continuity. The simpler bound $L\le \Vert c{\Vert }_2+M{\max }_i\Vert a_i{\Vert }_2$ is also valid.

Let $I(x)=\{i:{\ell }_i(x)=f(x)\}$ be the active set. The subdifferential isthe convex hull of all active slopes. In particular, choosing any active index gives the subgradientAt a tie, every convex combination of the tied slopes is also valid.

SetThe gradient has Lipschitz continuity with constantwhere the norm is the spectral norm. The proximal gradient method is thereforeFor $\lambda >0$, part d makes the second step explicit:where $C$ is the capped simplex; when $\lambda =0$, this proximal step is the identity.

A standard fixed choice is $0<\alpha \le 1/L$; the wider interval $0<\alpha <2/L$ also gives convergence under the usual forward-backward conditions. For a general convex objective, the function-value error is $O(1/r)$. If $A$ has full column rank, the quadratic term is strongly convex and an appropriate fixed step gives a linear convergence rate.

For $s\ge 0$ and $\xi \in \mathbb{R}$, define the characteristic curve $X(t;s,\xi )$ byThe bounded derivative $F_x$ makes $F(t,\cdot )$ globally Lipschitz, uniformly in $t$. On each finite time interval, $|F(t,x)|\le |F(t,0)|+L|x|$, so Gronwall inequality prevents finite-time escape. The Picard-Lindelöf theorem therefore gives a unique trajectory for every finite $t\ge 0$. Differentiation in $\xi$ givesso the characteristic flow map is a $C^1$ increasing diffeomorphism.

Along a characteristic, the chain rule changes the equation intoTracing $(t,x)$ backward by the flow therefore givesThe regularity of the flow makes this a $C^1$ classical solution. Conversely, every classical solution obeys the same ordinary differential equation along every characteristic, so the formula also proves uniqueness.

The Lipschitz continuity of $p_1$ givesPart (d) therefore implies

By the mean value theorem, $|f(x)-f(y)|\le A|x-y|$ for all real $x,y$. Thus $f$ is a contraction mapping on the complete metric space $\mathbb{R}$. Starting from arbitrary $x_1$, define $x_{n+1}=f(x_n)$. Induction gives $|x_{n+1}-x_n|\le A^{n-1}|x_2-x_1|$. For $m>n$, the triangle inequality and a geometric series yieldHence $(x_n)$ is a Cauchy sequence and completeness of the real numbers supplies a limit $x$. The Lipschitz continuity of $f$ implies $f(x)=\lim f(x_n)=\lim x_{n+1}=x$. Thus the limit is a fixed point, independently of the initial value. If $y$ were another, $|x-y|\le A|x-y|$ would force $x=y$. This proves the contraction mapping theorem in this setting, with error estimate

A weak solution belongs to $V=H_0^1(\Omega )$ and satisfieswith the nonlinear right-hand side defining a bounded functional on $V$. Under the stated global Lipschitz hypothesis this requirement is automatic: $|f(s)|\le |f(0)|+ϵ|s|$ and boundedness of $\Omega$ give $f(u)\in L^2$ whenever $u\in V$. Continuity of $f$ alone, without growth control, would not guarantee this for an arbitrary Sobolev function.

For the semilinear Poisson contraction threshold, use the gradient norm on $V$, and let $C_P$ satisfy the Poincaré inequality $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$. For $g\in L^2$, Lax-Milgram gives the zero-Dirichlet Poisson solution $Pg\in V$. Testing with $Pg$ yields $\Vert \nabla Pg{\Vert }_2\le C_P\Vert g{\Vert }_2$. Define $T(v)=P(f(v))$, a self-map of the complete space $V$. For $v,w\in V$,Therefore $ϵC_P^2<1$ is a sufficient condition for Banach's theorem to give a unique fixed point and hence a unique weak solution. There is no need to assume $f(0)=0$ or to restrict to a small ball: the Lipschitz bound makes the map defined on all of $V$. The same estimates giveWith the optimal Poincaré constant, the threshold is $ϵ<{\lambda }_1$ for the First Dirichlet eigenvalue; the stated sufficient condition does not depend on knowing that optimal constant.

One form of Schauder's theorem says that if $M$ is a nonempty closed bounded convex subset of a Banach space and $T:M\to M$ is continuous with relatively compact image, then $T$ has a fixed point.

For a Schauder solution of a small Hammerstein integral equation, take $X=C([0,1])$ with the supremum norm and $M=\{u:\Vert u{\Vert }_{\infty }\le R\}$. This set is nonempty, closed, bounded and convex. PutBoth maxima are finite by continuity on compact sets. Define $(T_{\mu }u)(x)=\mu {\int }_0^{1}G(x,y)f(y,u(y))dy$. Continuity and uniform continuity of the kernel give $T_{\mu }u\in X$, and for $0<\mu \le {\mu }_0$,so it is a self-map. This choice of ${\mu }_0$ also handles the case when either maximum is zero without division by zero.

If $u_n\to u$ uniformly in $M$, uniform continuity of $f$ on $[0,1]\times [-R,R]$ makes $f(\cdot ,u_n(\cdot ))\to f(\cdot ,u(\cdot ))$ uniformly. The integral estimate then proves continuity of $T_{\mu }$ in the supremum norm. Moreoverwhich tends to zero with $|x-x^{\prime }|$ uniformly in $u\in M$. Thus the image is uniformly bounded and equicontinuous. Arzelà-Ascoli states that such a family on a compact interval has compact closure in the supremum norm. Hence $T_{\mu }(M)$ is relatively compact, and Schauder gives a fixed point for each $\mu$ in the specified interval. Therefore a continuous solution with supremum norm at most $R$ exists. The argument imposes no Lipschitz hypothesis on $f$ and makes no uniqueness assertion.

For $0\le x\le 1$ and every real $z$,where $2|z|\le 1+z^2$ was used. The mean value theorem therefore proves Lipschitz continuity in $y$ with constant $1$.

Let $e_n=y(x_n)-y_n$. The explicit Euler method and the exact-solution identity from part (i) giveUsing the requested local truncation error bound $|{\tau }_n|\le h/2$, including at $n=0$, yieldsIteration and the finite geometric series giveSince $1+h\le e^h$ and $Nh=1$,Keeping the stronger bound $|{\tau }_n|\le h/4$ from part (i) improves the same proof to $(e-1)h/4$. The requested inequality follows already from the weaker bound.

The descent lemma follows by integrating the gradient along the line and using its Lipschitz continuity:This upper bound is at most the right side of the Armijo condition when $\alpha [(1-\beta )g^{T}p+\gamma \alpha \Vert p{\Vert }_2^2/2]\le 0$. Since $g^{T}p<0$, it holds forThe zero step gives equality; every positive step in this interval satisfies the prescribed sufficient-decrease test.

The coefficients of this linear multistep method are ${\alpha }_0=b$, ${\alpha }_1=-a$, ${\alpha }_2=1$ and ${\beta }_0={\beta }_1=0$, ${\beta }_2=c$. Substitution into the order of a numerical method conditions givesThe second equation gives $c=2-a$; substituting this into the third gives $-2+3a/2=0$. Consequently there is exactly one second-order choice,The resulting scheme is the BDF2 method,For a $C^4$ exact solution, Taylor's theorem in the original normalization gives the exact leading order defectIn particular the coefficient $-2/9$ is nonzero: the method has formal order exactly two, rather than three. Bounded third derivatives already suffice for the uniform estimate $T_n(h)=O(h^2)$; the stronger smoothness above only makes the displayed remainder immediate. Multiplying the recurrence by $3/2$ instead gives the common derivative normalization with leading defect $-h^2{y}^{\prime \prime \prime }(x_n)/3$.

The first of the characteristic polynomials of a linear multistep method isBoth roots are simple and lie in the closed unit disk, so the root condition for a multistep method proves zero-stability.

The Dahlquist equivalence theorem states that, for fixed-coefficient linear multistep methods on the usual class of initial value problems with a right-hand side continuous in $x$ and uniformly Lipschitz continuous in $y$, convergence of a numerical method for every such problem and every convergent set of starting values is equivalent to consistency of a numerical method and zero-stability. The statement is on fixed finite intervals where the exact solution exists, with implicit updates well defined. Its order estimate says that a zero-stable method with $T_n(h)=O(h^p)$ and starting errors $O(h^p)$ has global error $O(h^p)$ under the corresponding solution smoothness and Lipschitz continuity assumptions; merely convergent starting values do not ensure the order-$p$ rate.

For this application, sufficient assumptions are a unique solution $y\in C^3([x_0,x_0+T])$, a uniform Lipschitz continuity bound $L$ in $y$ on a neighborhood of its graph, and the nearby solution branch of each implicit update. For sufficiently small $h$, the mapis a contraction mapping with factor at most $2hL/3<1$ on a suitable small closed neighborhood. The Banach fixed-point theorem supplies local solvability of the nearby update. With starting errorsthe order estimate givesThus the BDF2 method is second-order convergent under these standard hypotheses. Exact $Y_0=y(x_0)$ together with one Euler method step for $Y_1$ already gives the required starting error when $y^{\prime \prime }$ is bounded, since only one starting step is taken.

The question's assumption of continuity of $f$ alone does not establish uniqueness, the needed solution smoothness, or this uniform convergence estimate. The added Lipschitz continuity condition is a sufficient hypothesis for the theorem used here; its failure does not imply that every particular initial value problem is nonunique or that its numerical approximations cannot converge.

Fix $x$ and put $g={\nabla }_{\mathcal{B}}f(x)$, the block coordinate gradient. For $t\ge 0$, define the one-variable function $q(t)=f(x-tg)$. By the chain rule,The last equality holds because the coordinates of $g$ outside $\mathcal{B}$ are zero. The fundamental theorem of calculus and the Cauchy-Schwarz inequality therefore giveFor $\alpha \ge 0$, the supplied condition (6), applied at the fixed starting point $x$ with parameter $t$, bounds the integrand by $L_{\mathcal{B}}t\Vert g{\Vert }^2$. Integration yieldsThis argument needs only the given directional Lipschitz continuity and continuous differentiability; convexity is not needed for this part. If the block coordinate gradient is zero, the inequality is equality.

The subgradient inequality for the differentiable convex function $f$ givesThus $y$ is a global minimizer of $\varphi$ with value zero. Subtracting an affine function preserves convexity.

We must also justify using part (a) for $\varphi$, since its partial derivative can differ from that of $f$. First show that condition (6) for a singleton block implies the full coordinate Lipschitz continuity estimateIf ${\partial }_{i}f(z)\ne 0$, choose $\alpha =-t/{\partial }_{i}f(z)$ in (6). If it is zero but ${\partial }_{i}f(z+t{e}_i)\ne 0$, apply the same argument starting at $z+t{e}_i$ and reverse the displacement. If both partial derivatives are zero, the estimate is immediate. The availability of all real $\alpha$ in (6) justifies these choices.

Now ${\partial }_i\varphi (z)={\partial }_{i}f(z)-{\partial }_{i}f(y)$, so differences of this partial derivative along coordinate lines are unchanged. Hence $\varphi$ has the same coordinate Lipschitz continuity constant $L_i$ and satisfies the singleton version of (6). Apply part (a) to $\varphi$ with step size $1/L_i$:Combining with its already established nonnegativity provesThe coordinate estimate survives subtraction of the supporting affine function. This is the step that allows the arbitrary comparison point $y$ to be exploited in part (c).

For every $z\in B$, $z_3+3>2$ and $|z_1+z_2|<2$. Consequently each of the three partial derivatives in the preceding gradient has absolute value below $1/2$, soThe segment joining $x,y\in B$ lies inside $B$. The mean value theorem and Cauchy-Schwarz inequality therefore giveFor $x=y$ both sides vanish. This proves the required Lipschitz continuity uniformly on the ball.