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By the chain rule, $y^{\prime \prime }=f_x+f_{y}f$ along the exact solution. Applying Taylor's theorem to the exact increment and the two arguments of $f$ givesThe assumed differentiability makes the remainders uniform on $0\le x\le 1-h$. Subtracting gives the consistency errorHere $O$ denotes Big O notation. The general order of a numerical method is therefore one for $\beta \ne 1/2$ and at least two for $\beta =1/2$. For example, on $y^{\prime }=y$ with unit initial value, $y_N=(1+h+\beta h^2{)}^{1/h}$. Expanding its logarithm gives $y_N=e[1+(\beta -1/2)h+O(h^2)]$, so the first-order term is actually nonzero when $\beta \ne 1/2$.

To control accumulation, the mean value theorem and the bound on $f_y$ givePut $K_h=L(1+\beta hL)$, $q_h=1+h{K}_h$ and $e_n=y(x_n)-y_n$. The exact and numerical updates yield $|e_{n+1}|\le q_h|e_n|+h|T_n|$, with $e_0=0$. Iterating this recurrence, as in the Lipschitz one-step error bound, givesBecause $K_h\le L(1+L)$ for $h\le 1$, the prefactor is bounded independently of the mesh. When $\beta =1/2$, the consistency error is $O(h^2)$, hence the maximum global error is $O(h^2)$. This proves at least second-order numerical convergence of the explicit midpoint method.

Write $M_0={\sup }_x{u}_0(x)<\infty$ and assume $\lambda (1+p{M}_0^{p-1})\le 1$. Suprema are understood where the printed maxima on the infinite domain are not attained. We prove the invariant interval $0\le U_j^m\le M_0$ simultaneously with the stated speed bound.

Suppose the interval property holds at time $m$. By the mean value theorem, the divided differenceis between $1$ and $1+p{M}_0^{p-1}$; when the two states coincide define it as $f^{\prime }(U_j^m)$. The update is the convex combinationBoth coefficients are nonnegative and sum to one, so the same interval property holds at time $m+1$. It is true initially and hence for all times by induction. ThereforeThe propagated nonnegativity is essential; it keeps the flux increasing on the entire range of the numerical solution.

A useful differentiation theorem says: if $f(t,x)$ is integrable in $x$, differentiable in $t$ for almost every $x$, and on a neighborhood of each parameter value the derivative is bounded in absolute value by one integrable function of $x$, then the integral is differentiable and its derivative is the integral of ${\partial }_{t}f$. The proof applies dominated convergence to difference quotients, bounded by that same majorant by the mean value theorem.

Fix $t_0>0$ and choose $0<a<t_0<b$. For $t\in [a,b]$,The first part is integrable by part (i). The second is integrable by the permitted hypothesis for $h_b$; alternatively exponential decay dominates its polynomial and logarithmic factors. The theorem therefore applies locally at every $t_0$, giving

One version of dominated differentiation under the integral sign is the following. Let the parameter range be open. Suppose $f(\cdot ,y)$ is integrable, ${\partial }_{y}f(x,y)$ exists for almost every $x$, and near each parameter value its absolute value is bounded by an integrable function of $x$ independent of $y$. ThenThis follows by applying the dominated convergence theorem to difference quotients, bounded by that same integrable function through the mean value theorem.

On any compact parameter interval $0<y\le B$, the original integrand is bounded by $B/(1+x^4)$, using $|\sin (x^{2}y)|\le B{x}^2$. Its first and second parameter derivatives areTheir absolute values are bounded respectively by $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$, both integrable on $(0,\infty )$. Applying the theorem twice proves that $F$ is twice differentiable and givesThis last integral is absolutely convergent: its integrand is bounded near zero and bounded by $x^{-2}$ at infinity. Substituting $t=x\sqrt{y}$ yieldsThe expression for $c$ is finite and independent of $y$, which is all that is required.

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.

The scalar-valued multivariable mean value theorem states that, if $f$ is differentiable on an open set containing the closed line segment from $x$ to $y$, then for $x\ne y$ there exists $s\in (0,1)$ such thatThis follows by applying the one-dimensional mean value theorem to $t\mapsto f(x+t(y-x))$ on $[0,1]$. The segment condition is necessary for this formulation on a possibly nonconvex domain.

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.

The scalar-valued multivariable mean value theorem states that, if $f$ is differentiable on an open set containing the closed line segment from $x$ to $y$, then for $x\ne y$ there exists $s\in (0,1)$ such thatThis follows by applying the one-dimensional mean value theorem to $t\mapsto f(x+t(y-x))$ on $[0,1]$. The segment condition is necessary for this formulation on a possibly nonconvex domain.

We first prove the differentiation theorem. Fix $t\in (0,b)$ and take nonzero $h\to 0$ with $t+h\in (0,b)$. The measurable difference quotientsconverge pointwise to $f_t(x,t)$. The mean value theorem and the assumed derivative bound give $|q_h(x)|\le g(x)$. In particular $f_t(\cdot ,t)$ is measurable as a limit of quotients along a sequence, and its bound by $g$ proves integrability. The dominated convergence theorem, in its parameter-limit form, says that pointwise convergence with a common integrable bound permits convergence of the integrals. Applying it yieldsThe parameter-limit statement follows from the ordinary sequential theorem by applying it to every sequence of nonzero $h$ tending to zero.

For the specific function, $|f(x,t)|\le e^{-x^2}$, which is integrable by the allowed fact. It is differentiable in $t$, withThe last majorant has integral one on $[0,\infty )$, using the substitution $u=x^2$. All three hypotheses therefore hold. The indicator of the parameter domain is one at every point under consideration and does not affect this derivative.

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.

A sufficient differentiation under the integral sign theorem is the following local domination criterion. Suppose $u(t,\cdot )$ is measurable for each $t$, $u(t_0,\cdot )$ is integrable at one point of a neighborhood $J$ of $t_0$, and for almost every $x$ the function $u(\cdot ,x)$ is differentiable on $J$. Assume an integrable function $g_J$ satisfies $|{\partial }_{t}u(t,x)|\le g_J(x)$ for $t\in J$ and almost every $x$. Then $I$ is differentiable on the interior of $J$ andThe mean value theorem bounds the difference quotients by $g_J$, and the dominated convergence theorem passes their pointwise limit through the integral. Applying this criterion locally throughout $(a,b)$ proves differentiability there. One may impose the stronger condition that every $u(t,\cdot )$ is integrable from the outset.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.

The partial derivatives satisfy a global bound of $1$ in absolute value. For example,because $2|x|\le 1+x^2$ and $1+x^2\ge 1$; the same argument applies to ${\partial }_{y}f$. Apply the one-dimensional mean value theorem first along a horizontal segment and then along a vertical segment:Thus $f$ is Lipschitz with constant $1$ for the coordinate-sum norm.

Define the scalar function $H(s)=⟨f(x+s(y-x)),z⟩$ on $[0,1]$. It is continuous there and differentiable inside, with derivative $⟨df(x+s(y-x))(y-x),z⟩$ by the chain rule. The one-dimensional mean value theorem gives $H(1)-H(0)=H^{\prime }(\theta )$ for some $0<\theta <1$, proving the displayed vector-paired version. No claim of a single intermediate point giving a vector equality without pairing is required.

For the final assertion, use the scalar mean value theorem directly on $s\mapsto g(sh)$: it is continuous at $s=0$ and differentiable for $s>0$. Thusfor some $0<{\theta }_h<1$. The assumed limits of all partial derivatives imply $\Vert \nabla g(x)\Vert \to 0$ as $x\to 0$, $x\ne 0$. Consequently $|g(h)-g(0)|/\Vert h\Vert \le \Vert \nabla g({\theta }_{h}h)\Vert \to 0$. Hence $\boxed{dg(0)=0}$ and $g$ is differentiable there. This argument uses only differentiability on the interior of the line segment, so it does not assume the conclusion at its endpoint.

A sufficient differentiation theorem is the following local version. Suppose $u(t,\cdot )$ is integrable, $u(\cdot ,x)$ is continuously differentiable for almost every $x$, and for each compact parameter interval there is an integrable $g(x)$ bounding $|{\partial }_{t}u(t,x)|$ uniformly there. Then $I^{\prime }(t)={\int }_E{\partial }_{t}u(t,x)dx$. The mean value theorem bounds the difference quotients by $g$, and dominated convergence proves the formula. Integrability at one parameter value together with the derivative bound also suffices locally.

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.

Fix $x\in U$. Since $U$ is open, choose an open ball $B(x,\epsilon )\subset U$. For any $y$ in this ball the segment $x+t(y-x)$ remains in it. For each coordinate $G_j$, the chain rule givesThe ordinary mean value theorem therefore gives $G_j(y)=G_j(x)$ for every coordinate, hence $G(y)=G(x)$. Thus $G$ is a locally constant function.

For each $z\in {\mathbb{R}}^m$, its preimage is a union of such open balls, so $G^{-1}(\{z\})$ is open in ${\mathbb{R}}^n$. Its complement in $U$ is the union of the other fibers and is also open. If $U$ is nonempty and connected, the nonempty fiber through any chosen point is both open and closed in $U$, and must be all of $U$. ThereforeThe empty-domain case is vacuous. Notice that continuity of $DG$ was never needed.

With $G$ a standard normal random variable,For $f\in C_b^1$, in the usual sense that $f$ and its continuous derivative are bounded, the mean value theorem bounds the difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$. Dominated convergence theorem therefore allows differentiation inside the integral:This also follows pathwise from $X(t,x+h)-X(t,x)=a(t)h$. If the textual description were read as assuming only bounded differentiability, some control of $f^{\prime }$ is needed for the displayed right side; the stated $C_b^1$ hypothesis supplies it. Gaussian smoothing itself still differentiates a bounded measurable $f$, using the derivative of the Normal distribution instead.

Write the diffusion and drift coefficients asBoth are continuously differentiable, withThese derivatives are bounded. For example $2|x|\le 1+x^2$ bounds both absolute values by one. The mean value theorem therefore makes both coefficients globally Lipschitz. They are also bounded, so satisfy the linear growth condition for an SDE.

The global existence theorem for stochastic differential equations with Lipschitz coefficients consequently givesIn particular, there is no finite explosion time. The solution has continuous paths, so all subsequent level-hitting and stopping arguments apply.