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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-Lindelof 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

A continuous finite-dimensional vector field has a local solution through every initial point. The theorem asserts existence without uniqueness; local Lipschitz continuity supplies uniqueness through the Picard-Lindelof theorem.

For an explicit Runge-Kutta method with coefficients $A,b$, put $c=A1$ and let $c^2$ mean componentwise squaring. The conditions $b^{T}1=1$, $b^{T}c=1/2$, $b^T{c}^2=1/3$ and $b^{T}Ac=1/6$ match the exact flow expansion through degree three: the two cubic differentials are $f^{\prime \prime }[f,f]$ and $(f^{\prime }{)}^{2}f$. For sufficiently smooth vector fields these give local error $O(h^4)$ and, under the usual finite-interval Lipschitz continuity assumptions, global error $O(h^3)$.

The diffusion coefficient obtained from a scale transform for an additive-noise diffusion need only be defined on an open interval. If it is Lipschitz, its limit at a finite endpoint is zero: a positive limit would bound the derivative $1/h$ of the inverse scale and prevent the inverse from diverging. Extending it by zero beyond finite endpoints therefore preserves global Lipschitz continuity.