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The Frechet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves differentiability.

The converse is false. Set $f(x,y)=x^2\sin (1/x)$ for $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Frechet derivative is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, the putative first partial derivative would require the limit $y_0\sin (1/x)$, which does not exist. Consequently the exact set of differentiability is

First eliminate the density-dependent definition of the auxiliary $\mathbf{g}$. The Newtonian gravitational stress tensor can be written without derivatives of its normalization:Here ${\delta }_{ij}$ is the Kronecker delta, and repeated indices are summed. Differentiating givesThe first and third terms cancel because mixed partial derivatives commute; the final step uses the Poisson equation for Newtonian gravity. Therefore the gravitational force density is a stress divergence:The identity holds for spatially varying mass density: substituting the definition of $\mathbf{g}$ before differentiating prevents erroneous extra density-gradient terms.

For the gravitational stress contribution to angular-momentum transport, the anisotropic term $\rho g_i{g}_j$ supplies angular momentum transport. In cylindrical components its radial–azimuthal entry isIn the momentum conservation equation, $r{T}_{r\varphi }$ is the radial stress contribution to the angular-momentum flux, whose sign depends on the correlated radial and azimuthal field components. It vanishes for a perfectly axisymmetric potential but can be nonzero for spiral disturbances.

The other term is isotropic and acts as an effective negative pressure, $P_g=-\rho g^2/2=-|\nabla \Phi {|}^2/(8\pi G)$. Its force contribution is $-\nabla P_g=+\nabla (\rho g^2/2)$, modifying normal compression and force balance. It has no off-diagonal shear component and therefore no direct radial angular momentum transport. In an axisymmetric averaged disk it supplies no azimuthal torque. The complete symmetric Newtonian gravitational stress tensor also expresses conservation of angular momentum without an internal couple.

In Einstein summation convention, the curl isApply another curl and use the contraction of two Levi-Civita symbols:Smoothness allows the partial derivatives to commute. This proves the curl of the curl identity:

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

Write derivatives with respect to $u$. The arc-length condition is $f^{\prime 2}+g^{\prime 2}=1$. The tangent vectors areTheir scalar products give the first fundamental formChoose the unit normal vector $N=(-g^{\prime }\cos v,-g^{\prime }\sin v,f^{\prime })$, since ${\sigma }_u\times {\sigma }_v=fN$ and $f>0$. Scalar products of $N$ with the second partial derivatives giveThe mixed coefficient is zero. Reversing $N$ reverses the second fundamental form but not the Gaussian curvature. Its formula isDifferentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }=0$. Substitute it directly, without dividing by $g^{\prime }$ (which may vanish), to obtain

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.

Write the minimal surface equation for a graph as $\mathrm{div}F(Du)=0$ in its weak formulation, where $F(p)=p/\sqrt{1+|p{|}^2}$. For negative increments use the same difference quotient convention, soFor $\zeta \in C_c^1({\Omega }^{\prime \prime })$, the function ${\delta }_{\ell ,h}\zeta$ is an admissible test function in $\Omega$. Commuting the first partial derivative with translation, then applying discrete integration by parts, givesThe fundamental theorem of calculus along a line segment gives the averaged linearization of a nonlinear divergence-form equationThus the backward difference quotient solves a linear equation in divergence form:To verify the ellipticity of the minimal surface flux, computeand henceThe eigenvalue parallel to $p$ is $(1+|p{|}^2{)}^{-3/2}$; every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. The same lower and upper bounds pass to the averaged symmetric matrix $A_h$ whenever both endpoint gradients have norm at most $M$:The printed $C^1(\Omega )$ hypothesis supplies such an $M$ on each relatively compact subset, uniformly for sufficiently small $h$. It therefore establishes a locally uniformly elliptic operator. A single lower bound on all of ${\Omega }^{\prime \prime }$ additionally requires bounded gradients there and at the shifted points; this is automatic for bounded $\Omega$ and fixed $h$, but is not supplied on an arbitrary open set.

Take ${\Omega }^{\prime }$ nonempty and compactly contained in $\Omega$. The argument is a compactness proof of a nonnegative elliptic solution estimate, with the global Schauder estimate retaining control all the way to the boundary. The operator and the Hölder space exponent are fixed throughout.

Suppose the estimate fails. For each integer $k$ there is an admissible nonnegative solution $u_k$, with forcing $f_k$, such thatPut $v_k=u_k/M_k$ and $g_k=f_k/M_k$. Then $0\le v_k\le 1$, ${\max }_{\overline{\Omega }}{v}_k=1$, $v_k=0$ on the boundary, $L{v}_k=g_k$, andGlobal elliptic regularity and the global Schauder estimate for zero Dirichlet boundary condition giveThis estimate includes the $C^0$ term, so it does not require invertibility of $L$ or a sign restriction on $c$. The compact embedding of Hölder spaces, or the Arzela-Ascoli theorem applied through second partial derivatives, gives a subsequence converging in $C^2(\overline{\Omega })$ to $v$. Thus $v\ge 0$, $Lv=0$, $v=0$ on the boundary, and $\max v=1$.

Choose $x_k\in {\Omega }^{\prime }$ with $v_k(x_k)\le {\inf }_{{\Omega }^{\prime }}{v}_k+1/k$. A further subsequence has $x_k\to x_{\ast }\in \overline{{\Omega }^{\prime }}\subset \Omega$. Uniform convergence yields $v(x_{\ast })=0$. Part (b) says that a nonnegative homogeneous solution on the connected domain is either zero everywhere or strictly positive inside. Both alternatives contradict, respectively, $\max v=1$ and $v(x_{\ast })=0$. ThereforeThe interior infimum and the forcing control the global supremum. Global boundary control is crucial here: an interior-only compactness argument could lose the normalized maximum at the boundary.

For fixed $\gamma$, differentiation under the integral sign is justified by bounded $P$ and $X$. At the unconstrained minimizer $H_{\gamma }$, the first-order condition isDefine the strictly positive, bounded pricing kernelThe first-order condition says $\mathbb{E}[Z_{\gamma }P]=p$, so $Z_{\gamma }\in \mathcal{Z}$. When taking the requested partial derivative, hold $H$ fixed and only afterwards set $H=H_{\gamma }$. Writing $A_{\gamma }=H_{\gamma }\cdot p-x$, we obtainThe assumed dual bound applies to this particular pricing kernel. ThusNo derivative of the map $\gamma \mapsto H_{\gamma }$ is needed; confusing a partial derivative with a derivative along the minimizing path would add an unnecessary hypothesis.

The first partial derivatives areBoth vanish at $(\lambda ,\lambda )$, so every positive diagonal point is a stationary point. The Hessian matrix there isIts eigenvectors $(1,1)$ and $(1,-1)$ have respective eigenvaluesThe zero eigenvalue reflects the entire line of nonisolated stationary points: $f(\lambda ,\lambda )=2$. In fact $f-2=(x-y{)}^2[1/(xy)-1/a^2]$, showing a non-strict local minimum when $\lambda <|a|$, a non-strict local maximum when $\lambda >|a|$, and values on both sides of two near $\lambda =|a|$. At that last point the whole Hessian matrix vanishes, so its signs alone cannot classify the point.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.

The chain rule for Frechet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.