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A function is Fréchet differentiable at $x$ when there is a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThis linear map, which is unique, is the derivative $Df(x)$.

For the chain rule, if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then $g\circ f$ is differentiable at $x$ and $D(g\circ f)(x)=Dg(f(x))Df(x)$. Indeed write $f(x+h)-f(x)=Ah+r(h)=v(h)$ and $g(f(x)+v)=g(f(x))+Bv+s(v)$, where $r(h)=o(|h|)$, $s(v)=o(|v|)$. Since $v(h)=O(|h|)$, substitution giveswhich proves the assertion, including paths on which $v(h)=0$.

For $g_1$, write $q=x^2-y^2$. Away from the two diagonal lines, $q\ne 0$ and the formula is a composition of smooth functions, hence differentiable. At the origin, $|g_1(x,y)|\le |x^2-y^2|\le x^2+y^2$, so its ratio to the displacement length tends to zero. Its derivative there is therefore zero: this is quadratic damping restores differentiability at the origin.

At a nonzero point $(a,b)$ with $a=\pm b$, necessarily $a\ne 0$. Along the $x$ direction,The denominator inside the sine is locally an invertible function of $t$, with nonzero derivative $2a$. Choose sequences tending to zero for which its reciprocal equals $2\pi n+\pi /2$ and $2\pi n+3\pi /2$. The quotients tend to $2a$ and $-2a$. Thus even this directional derivative fails to exist, and the function is not differentiable there. ConsequentlyFor $g_2$, the formula is smooth away from zero. At zero, $|g_2(x,y)|\le x^2+y^2$, so the same remainder estimate gives zero derivative. HenceNo limit of the sine factor itself is needed for either derivative at the origin.

Vary $g$ to $g+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation of the functional isThe endpoint term vanishes. Since every smooth test variation is allowed, stationarity gives the Euler-Lagrange equation

The constant prefactor in the specified energy does not affect its stationary equation. Its integrand gives $f_{{\varphi }^{\prime }}=r^2{\varphi }^{\prime }$ and $f_{\varphi }=r^2\varphi /{\lambda }^2$, soTake $\lambda >0$ as a screening length; for a nonzero real parameter of either sign its absolute value gives the same equation. Setting $u=r\varphi$ yields $u^{\prime \prime }-u/{\lambda }^2=0$, henceFor $0<R_1<R_2$ and the prescribed endpoint values, the convenient hyperbolic form isIt directly satisfies both endpoint values. The energy is a positive quadratic functional, so this stationary solution is its unique minimizer for fixed endpoints: any zero-endpoint variation adds a positive quadratic energy, with the cross term zero by the equation.

In two dimensions, cylindrical area measure replaces the radial factor $r^2$ by $r$. Up to an irrelevant common normalization, the corresponding energy and its modified Helmholtz equation areThus the two-dimensional radial equation is ${\varphi }^{\prime \prime }+r^{-1}{\varphi }^{\prime }-{\lambda }^{-2}\varphi =0$, rather than the three-dimensional equation with coefficient $2/r$ on the first derivative.

The Euler equations for an inviscid fluid with no body force is ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho$. For irrotational flow $\mathbf{u}=\nabla \varphi$, the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ removes the last term. Thus the gradient of ${\varphi }_t+|\nabla \varphi {|}^2/2+p/\rho$ is zero, giving the Unsteady Bernoulli equationThe function is spatially constant on the connected fluid region; adding a time-dependent constant to the potential changes its gauge.

For spherically symmetric incompressible radial flow, $r^2{u}_r$ is independent of $r$. The bubble boundary moves with the fluid, so $u_r(R)=\dot{R}$, giving $u_r=R^2\dot{R}/r^2$. Fix $\varphi \to 0$ at infinity, obtaining $\varphi =-R^2\dot{R}/r$ and $F=p_{\infty }/\rho$. The time derivative is taken at fixed $r$, then evaluated at the moving boundary: ${\varphi }_t(R)=-(2{\dot{R}}^2+R\ddot{R})$. With pressure continuity at the interface and no capillary or viscous correction, Bernoulli gives the Rayleigh equation for an inviscid spherical bubble

The gas law and initial pressure fix $C{M}^2=p_{\infty }{R}_0^6/2$. Multiply the radius equation by $2R^2\dot{R}$; its left side is $d(R^3{\dot{R}}^2)/dt$. Integrating from the zero-speed initial state gives the first integral of polytropic spherical-bubble motionTo find the permitted radii, put $s=R^3/R_0^3$. The bracket becomes $R_0^3(1-s)(2s-1)/(3s)$, nonnegative exactly for $1/2\le s\le 1$. ThereforeInitially $\ddot{R}=-p_{\infty }/(2\rho R_0)<0$, so the radius decreases from the upper endpoint. At the lower endpoint the gas pressure is $2p_{\infty }$, giving positive acceleration and an outward reversal. The first-integral zeros are simple, so the travel-time integrals near either endpoint behave as an integrable inverse square root. The bubble reaches both turning radii in finite time, and the autonomous undamped equation repeats this motion periodically. Thus the stated interval describes oscillation, rather than an irreversible approach to one endpoint.

Assume $L$ is linear, annihilates all polynomials of degree at most $k$, is defined on the truncated-power kernels and permits interchange with the remainder integral. For $k\ge 1$, a convenient sufficient condition is that $L$ be a bounded linear functional on $C[a,b]$; more general approximation functionals need the corresponding continuity/interchange property on their domain. Taylor's integral remainder givesApplying $L$, the polynomial terms vanish, and the Peano kernel theorem givesHere $t_{+}=\max (t,0)$. This is the printed raw-kernel convention; the normalized Peano kernel includes $1/k!$ inside its definition instead, and then has no factorial outside the integral. The interchange is justified by boundedness applied to the limit of Riemann sums of the continuous kernel-valued integrand. For a kernel with $k=0$, its discontinuity requires a function space/interchange assumption accommodating the step functions as well.

For the specified Simpson's rule error, direct evaluation on $1,x,x^2,x^3$ gives zero, so $k=3$. For $0\le \theta \le 2$,Simplifying separately on the two halves givesBoth pieces are strictly negative in the open interval; they agree at the midpoint. Their integral is $-1/15$, as direct polynomial integration shows. ThusThis is optimal, not merely an upper bound. For $f(x)=x^4/24$, the fourth derivative is identically one and $L(f)=-1/90$, so the minimum constant is $\rho =1/90$.

With $\Re \alpha <0$, the causal exponential is integrable and its Fourier transform isFor the pendulum system, a transfer function is the Fourier transform of its causal impulse response, equivalently the multiplier relating the transformed forcing to the transformed zero-initial-data response. Transforming derivatives givesPartial fractions therefore giveThe exponential transform identifies the causal response as $R(t)=\frac{1}{2}{e}^{-t}\sin 2t$ for $t>0$ and zero for $t<0$. It solves the homogeneous equation for $t>0$, with $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, the jump required by a unit impulse. The forcing is understood as switched on at zero; its transform is $1/(i\omega +4)$, and convolution with $R$ gives, if the time response is wanted,This has both stated initial values equal to zero.

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

First use the algebraically closed field convention. Every line bundle on a smooth projective curve has a nonzero rational section of a line bundle: trivialize it on a nonempty open set and take its generic generator. The orders of its zeros and poles give a finite divisor $D$, and the local trivializations identify the line bundle with $\mathcal{O}(D)$. For a degree-zero line bundle writesince every closed point is a $k$-point. ThereforeOnly finitely many factors occur, and negative tensor powers mean dual line bundles. This provesFor genus zero the group is trivial, so the conclusion remains valid.

For a general field, a closed point need not be rational, and its contribution to degree is weighted by its residue degree. Consequently rational points need not generate the degree-zero Picard group of a curve. Here is an explicit counterexample to that stronger reading. Let $C$ be the smooth projective curve over ${\mathbb{F}}_2$ obtained by completing the affine equationThe polynomial is irreducible over ${\overline{\mathbb{F}}}_2(x)$: a function $r^2+r$ has even pole order at infinity when it has a pole there, while the right side has pole order five. Its affine model is smooth because the partial derivative with respect to $y$ is one. The normalized projective completion is therefore a geometrically integral smooth projective curve.

There is exactly one point $p_{\infty }$ above infinity, and it is rational. Indeed the map $x:C\to {\mathbb{P}}^1$ has degree two. At a point above infinity put $v(x)=-e$, where the ramification index $e$ is at most two. The equation forces $v(y)<0$ and $2v(y)=5v(x)=-5e$. Hence $e=2$ and $v(y)=-5$. The degree formula then leaves exactly one point of residue degree one. Thus $x$ and $y$ have pole orders two and five there. There are no affine rational points: for either $x\in {\mathbb{F}}_2$, the right side is one, while $y^2+y=0$ for every $y\in {\mathbb{F}}_2$. Hence $C({\mathbb{F}}_2)=\{p_{\infty }\}$ and the image of its Abel map of a pointed smooth projective curve is trivial.

The equations $y=0$, $x^2+x+1=0$ define a degree-two closed point $D$, since $x^5+x+1$ vanishes modulo $x^2+x+1$. The divisor $D-2p_{\infty }$ has degree zero but is not principal. To prove this, a function with poles bounded by $2p_{\infty }$ is regular on the affine curve and hence has a unique expression $A(x)+B(x)y$. The possible pole orders of its two terms are $2\mathrm{deg}A$ and $2\mathrm{deg}B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\mathrm{deg}A\le 1$. But no nonzero $a+bx$ over ${\mathbb{F}}_2$ vanishes at $D$, whose residue class of $x$ has irreducible quadratic minimal polynomial. Thus no function has principal divisor $D-2p_{\infty }$, proving its class is nonzero. This establishes the rational points need not generate the degree-zero Picard group qualification with a complete obstruction, while the generation proof above supplies the intended algebraically closed-field answer.

Over ${\overline{\mathbb{F}}}_3$, take the projective cubicIn characteristic three its partial derivatives are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the Jacobian criterion proves smoothness everywhere, including infinity.

It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in ${\overline{\mathbb{F}}}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. Gauss lemma for polynomials then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the smooth plane cubic in characteristic three. In particular, choosing a Fermat cubic here would fail, because its cubic partial derivatives vanish in characteristic three.

Take the plane quartic with exactly two nodesPut $d=X^2-Z^2$. Its partial derivatives areOn $Y=0$, the equation gives $X^2=Z^2$, producing exactly $P_{+}=[1:0:1]$ and $P_{-}=[-1:0:1]$, and their gradients vanish.

There are no singular points with $Y\ne 0$. Normalize $Y=1$. The derivative equations give $X=4dZ$ and $(16d^2+1)Z=0$. If $Z=0$, then $X=0$ and $F_Y=4\ne 0$, so $Z\ne 0$ and $d^2=-1/16$. The remaining equation gives $XZ=-2$, hence $Z^2=-1/(2d)$. But $d=X^2-Z^2=-2Z^2=1/d$, forcing $d^2=1$, a contradiction. Thus the singular set is exactly $\{P_{+},P_{-}\}$.

In the chart $Z=1$ near $P_{\epsilon }$, put $X=\epsilon +u$, $Y=v$ with $\epsilon =\pm 1$. The lowest-degree part is $4u^2+\epsilon v^2$, which splits into two distinct complex linear factors. Each singularity is therefore an ordinary double point, with two smooth transverse branches.

It remains to prove irreducibility. A repeated polynomial factor would make every point on that component singular, contradicting the finite singular set, so $F$ is reduced. If it were reducible, partition its irreducible factors into two coprime factors of degrees $r$ and $4-r$, $1\le r\le 3$. Every intersection of the two components would be singular on $C$, hence one of the two ordinary double points. At an ordinary double point shared by them, the two branches are transverse, so its intersection multiplicity is one. Their total intersection multiplicity would therefore be at most two. The Bézout theorem instead makes it $r(4-r)\ge 3$. This contradiction proves the displayed quartic is irreducible and has exactly two singular points.

Choose the utility function primitive $U(c)=c^{1-R}/(1-R)$. Any other primitive with the same derivative differs by a constant $C$; this adds $C/\rho$ to the value function and does not alter the policy. Put $p=1-R$ and let ${\pi }_t$ be the stock fraction. The self-financing portfolio equation, including consumption withdrawals, isFor a time-independent infinite-horizon value function $V$, the Hamilton-Jacobi-Bellman equation isScaling initial portfolio wealth and all consumption rates by the same positive number preserves admissibility and scales utility function by its $p$th power. This homogeneity gives $V(w)=K{w}^p/p$. In particular,The consumption first-order condition $c^{-R}=V^{\prime }(w)$ and stock first-order condition $w(\mu -r)V^{\prime }(w)+{\sigma }^2\pi w^2{V}^{\prime \prime }(w)=0$ giveThe second derivatives of the optimized expressions are negative, so these are their global maximizers. Writing ${\gamma }_M=K^{-1/R}$ and substituting into the Hamilton-Jacobi-Bellman equation givesThereforeThe given inequality is precisely ${\gamma }_M>0$, so the Merton consumption-investment problem value is $V(w)={\gamma }_M^{-R}{w}^{1-R}/(1-R)$. This derives the requested policy; no additional verification argument is needed here.