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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.

Use a local first-derivative Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi ,x)$ with sufficient differentiability, and let variations have compact support or vanish at the spacetime boundary. If the phrase “a function of the field” were read literally as forbidding derivatives, the equation below would reduce to $\partial \mathcal{L}/\partial \varphi =0$; a propagating field requires derivative dependence. Varying the action givesIntegration by parts and arbitrary interior variations yield the Euler-Lagrange field equation,The boundary condition is part of this derivation; if boundary variations are allowed, their separate boundary equations must also be imposed.

A variational symmetry of a Lagrangian density is an infinitesimal field change $\delta \varphi =\epsilon X$ for which $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$ off shell. Equality to zero is sufficient but is not necessary: a divergence changes only the boundary contribution to the action. For fixed-coordinate field variations,Equating the two expressions proves the scalar-field form of the Noether theorem:Thus every differentiable one-parameter variational symmetry gives a conserved current, and its Noether charge $Q=\int d^{3}x{j}^0$ is constant when the spatial boundary flux vanishes. This is the Noether current for a first-derivative scalar field; the same identity applies to several real components by summing over them. Identically conserved improvement terms can change its local expression without changing the charge under the same boundary assumptions.

For an active spacetime translation, ${\varphi }^{\prime }(x)=\varphi (x-a)$ and $\delta \varphi =-a^{\nu }{\partial }_{\nu }\varphi$. If the Lagrangian density has no explicit coordinate dependence, $\delta \mathcal{L}=-a^{\nu }{\partial }_{\nu }\mathcal{L}$, so $K^{\mu }=-a^{\mu }\mathcal{L}$. The Noether current is $j^{\mu }=-a^{\nu }{T}^{\mu }{}_{\nu }$, whereThese four translation currents are the canonical stress-energy tensor, also called the energy-momentum tensor. The sign in $j=-aT$ follows from the chosen active translation; the translation charges themselves can be labelled by $P_{\nu }=\int T^0{}_{\nu }{d}^{3}x$.

For the free real scalar field, take the standard kinetic term and mass term. Its Lagrangian density and Klein-Gordon equation areHere ${\Pi }^{\mu }={\partial }^{\mu }\varphi$, so $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Raising the charge index gives the four-momentum of a free real scalar field,The first is total energy, and the three components of the second are physical spatial momentum. With signature $(+,-,-,-)$, $P_i=-P^i$; this explains the opposite sign if the conserved quantities are instead written with lower spatial indices. For example, a plane wave proportional to $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density along $\mathbf{k}$, confirming the sign. All charges require convergence of the integrals and vanishing boundary flux, or periodic boundary conditions in a finite box.

For the complex scalar field, treat $\psi$ and ${\psi }^{\ast }$ as independent variables when varying, equivalently use their two real components. The global phase symmetry of a complex scalar field is $\psi \mapsto e^{-i\alpha }\psi$, ${\psi }^{\ast }\mapsto e^{i\alpha }{\psi }^{\ast }$ for constant $\alpha$. Both the kinetic term and $V(|\psi {|}^2)$ are invariant. This is a global internal symmetry of a classical field theory with circle group $U(1)$; a spacetime-dependent phase would require a gauge connection.

Choosing this orientation for the phase parameter gives the Noether charge of a complex scalar field:To check conservation directly, the Euler-Lagrange field equations are $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and their complex conjugates, assuming a real differentiable potential. Hence ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$. Reversing the phase-parameter orientation reverses the current and charge, which is merely a convention. For a scalar carrying electric charge $q$, the physical electric charge is $qQ$ after electromagnetic coupling. In a neutral theory the same global charge can instead label an internal conserved quantum number; the density alone does not identify it automatically with electricity.

Let $u(t)=\gamma (t)/|\gamma (t)|$, a continuous map into the unit circle. By uniform continuity, partition $[0,1]$ so finely that on each interval beginning at $t_j$, $|u(t)/u(t_j)-1|<1$. This ratio lies in the right half-plane and has a continuous complex argument with value zero at $t_j$. Starting from ${\theta }_0$, add these local arguments successively, matching endpoint values. This constructs a continuous real lift with $u(t)=e^{i\theta (t)}$.

Any two such lifts with the same initial value differ by a continuous function taking values in $2\pi \mathbb{Z}$, and hence coincide. Since the curve closes, $e^{i(\theta (1)-\theta (0))}=1$. Define its winding number byChanging the chosen initial argument adds a constant multiple of $2\pi$ to the whole lift, so the winding number is independent of that choice. No differentiability of the curve is required.