Line Integrals and the Gradient Theorem

Inserting one point into an admissible partition only splits one smooth piece. By [L2], the original scalar summand equals the two new scalar summands, and the same holds for the vector summand. Hence either total sum is unchanged.

The union of the two finite partitions is a finite common refinement. Repeatedly applying step 1.1 shows that each original sum equals the sum over this refinement.

Therefore the two original scalar sums agree, and the two original vector sums agree.

A partition with no interior breakpoints already equals its own refinement. For a constant path on a nondegenerate interval every derivative extension is zero, while on a singleton interval both integrals are zero by [L1]. Thus the conclusion includes both boundary cases.

Refine at the breakpoints of $h$ and at their preimages of the breakpoints of $\gamma$. On each resulting interval, [L5] gives $(\gamma \circ h{)}^{\mathrm{\prime }}=({\gamma }^{\mathrm{\prime }}\circ h)h^{\mathrm{\prime }}$. The refinements do not alter either line integral by [L3].

For the scalar integrand, step 1.1 gives $f(\gamma (h(t)))\parallel (\gamma \circ h{)}^{\mathrm{\prime }}(t){\parallel }_2=f(\gamma (h(t)))\parallel {\gamma }^{\mathrm{\prime }}(h(t)){\parallel }_2\mathrm{\mid }{h}^{\mathrm{\prime }}(t)\mathrm{\mid }\mathrm{.}$

For the vector integrand, step 1.1 and bilinearity give $⟨F(\gamma (h(t))),(\gamma \circ h{)}^{\mathrm{\prime }}(t)⟩=⟨F(\gamma (h(t))),{\gamma }^{\mathrm{\prime }}(h(t))⟩h^{\mathrm{\prime }}(t)\mathrm{.}$

When $h$ is increasing, $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }=h^{\mathrm{\prime }}$ and [L4] identifies the sum of these integrals with ${\int }_{\gamma }fds$. When $h$ is decreasing, $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }=-h^{\mathrm{\prime }}$ and the reversal of the oriented substitution limits supplies the second minus sign. Thus the scalar equality holds in both cases.

Applying [L4] piece by piece to step 2.2 gives the same oriented integral when $h(c)=a,h(d)=b$, and its negative when $h(c)=b,h(d)=a$. These are respectively the orientation-preserving and orientation-reversing cases in [L1].

Steps 3.1 and 3.2 prove all three formulas. The nondegenerate-interval, nonzero-derivative, and bijectivity hypotheses in [L1] rule out singleton reparametrizations, pauses, and multiple traversals.

If $a=b$, [L5] makes the two line integrals over both $\gamma$ and ${\gamma }^{-}$ zero. If $a<b$, the affine map $t\mapsto a+b-t$ is orientation-reversing, so applying [L2] to the reversal in [L1] proves the two formulas.

Split the concatenation at $1\mathrm{/}2$. The first half is the orientation-preserving affine reparametrization $t\mapsto 2t$ of $\alpha$, and the second is the orientation-preserving affine reparametrization $t\mapsto 2t-1$ of $\beta$.

By [L2], each half-integral in step 1.2 equals the corresponding integral over $\alpha$ or $\beta$. By [L3], [L4], and [L5], the sum of the two half-integrals is the integral over $\alpha \ast \beta$. This proves both concatenation formulas.

The join point is an allowed partition point, so no derivative match is required there. If either path is constant, its derivative and both of its line-integral contributions are zero, and the formulas remain valid.

On each smooth piece, $-M\parallel v_i(t){\parallel }_2\le f(\gamma (t))\parallel v_i(t){\parallel }_2\le M\parallel v_i(t){\parallel }_2\mathrm{.}$

By [L2] and the bound on $F$, $-M\parallel v_i(t){\parallel }_2\le ⟨F(\gamma (t)),v_i(t)⟩\le M\parallel v_i(t){\parallel }_2\mathrm{.}$

Integrate the two inequalities in step 1.1 using [L3], sum them using [L1], and identify the speed sum with [L4]. This gives $-ML(\gamma )\le {\int }_{\gamma }fds\le ML(\gamma ),$ hence the scalar estimate.

Repeating step 2.1 with step 1.2 gives the vector estimate.

If $M=0$ or $L(\gamma )=0$, either two-sided bound has both endpoints equal to zero, so the corresponding integral is zero and the asserted estimate still holds.

On a nondegenerate interval, [L1] gives ${\int }_{\gamma }1ds={\sum }_{i<m}{\int }_{t_i}^{t_{i+1}}\parallel v_i(t){\parallel }_{2}dt\mathrm{.}$

By [L2], the right-hand side of step 1.1 is $L(\gamma )$.

On a singleton interval both sides are zero by [L1] and [L2]. A constant path on a nondegenerate interval has zero speed, so step 2.1 gives zero on both sides there as well.

The speed $v(t):=\parallel {\gamma }^{\mathrm{\prime }}(t){\parallel }_2$ is continuous because ${\gamma }^{\mathrm{\prime }}$ and the Euclidean norm are continuous.

By [L1] and [L2], for every $t\in [a,b]$, $s_{\gamma }(t)={\int }_a^{t}v(u)du,$ including $t=a$.

Apply [L3] to step 2.1. It gives $s_{\gamma }^{\mathrm{\prime }}(t)=v(t)=\parallel {\gamma }^{\mathrm{\prime }}(t){\parallel }_2$ throughout $[a,b]$, with the asserted endpoint interpretation.

If $a=b$, [L1] makes the line integral zero and the two endpoint values agree. Assume henceforth that $a<b$. On the interior of the $i$th smooth piece, [L2] and [L3] give $\frac{d}{dt}\varphi (\gamma (t))=⟨\mathrm{\nabla }\varphi (\gamma (t)),{\gamma }^{\mathrm{\prime }}(t)⟩\mathrm{.}$

The continuous derivative extension on that piece is the integrand in [L1]. Applying [L4] gives ${\int }_{t_i}^{t_{i+1}}⟨\mathrm{\nabla }\varphi (\gamma (t)),v_i(t)⟩dt=\varphi (\gamma (t_{i+1}))-\varphi (\gamma (t_i))\mathrm{.}$

Sum step 2.1 over the finite partition. All interior endpoint values cancel, leaving $\varphi (\gamma (b))-\varphi (\gamma (a))$, and [L1] identifies the left side with the line integral.

For a constant path the integrand is zero and the endpoints coincide, so both sides are zero. The same conclusion holds whenever merely $\gamma (a)=\gamma (b)$.

Choose a potential $\varphi$ as in [L1]. If $\alpha$ and $\beta$ have the same initial point $x$ and terminal point $y$, then [L2] gives ${\int }_{\alpha }F\cdot d\mathbf{r}=\varphi (y)-\varphi (x)={\int }_{\beta }F\cdot d\mathbf{r}\mathrm{.}$

If $\gamma$ is closed, then its two endpoint values agree, and [L2] gives ${\int }_{\gamma }F\cdot d\mathbf{r}=0$.

Hence $F$ is path-independent by [L1].

The closed-loop conclusion does not require connectedness: it is an endpoint calculation for each path that exists.

Assume condition 1, and let $\gamma$ be closed at $x$. The path $\gamma$ and the constant path at $x$ have the same endpoints, so [L1] and [L3] give ${\int }_{\gamma }F\cdot d\mathbf{r}=0$. Thus condition 2 holds.

Conversely, assume condition 2. Let $\alpha$ and $\beta$ be paths from $x$ to $y$. The increasing affine bijection of $[0,1]$ onto a path's domain is an orientation-preserving oriented reparametrization, so by [L4] we may replace each path by its reparametrization on $[0,1]$ without changing either integral. With both domains $[0,1]$, the concatenation in [L2] is defined and $\alpha \ast {\beta }^{-}$ is closed.

By condition 2 and [L2], $0={\int }_{\alpha \ast {\beta }^{-}}F\cdot d\mathbf{r}={\int }_{\alpha }F\cdot d\mathbf{r}-{\int }_{\beta }F\cdot d\mathbf{r}\mathrm{.}$

Hence the two integrals agree, and [L1] gives path independence.

Step 1.1 proves the forward direction, and steps 1.2, 2.1, and 3.1 prove the reverse direction. Piecewise-$C^1$ path-connectedness guarantees that the comparison paths relevant to condition 1 exist between any two points of $U$.

By [L1], the displayed formula defines one real number $\varphi (x)$ for every $x\in U$. Choosing the constant path at $a$ and using [L2] gives $\varphi (a)=0$.

Fix $x\in U$ and a coordinate $j$. Since $U$ is open, there is $r>0$ such that $x+q{e}_j\in U$ whenever $\mathrm{\mid }q\mathrm{\mid }<r$. Take any path from $a$ to $x$ and reparametrize it by the increasing affine bijection of $[0,1]$ onto its domain; this is an orientation-preserving oriented reparametrization, so [L4] leaves its integral unchanged. Both it and the coordinate segment ${\sigma }_q(t)=x+tq{e}_j$, $0\le t\le 1$, now have domain $[0,1]$, so the concatenation in [L2] is defined; append ${\sigma }_q$.

Path independence and [L2] give, for $0<\mathrm{\mid }q\mathrm{\mid }<r$, $\varphi (x+q{e}_j)-\varphi (x)={\int }_{{\sigma }_q}F\cdot d\mathbf{r}=q{\int }_0^1{F}_j(x+tq{e}_j)dt\mathrm{.}$

Divide step 2.1 by $q$. For every $\epsilon >0$, continuity of $F_j$ at $x$ makes $\mathrm{\mid }{F}_j(x+tq{e}_j)-F_j(x)\mathrm{\mid }<\epsilon$ uniformly for $0\le t\le 1$ when $q$ is sufficiently small. Therefore the quotient tends to $F_j(x)$, so ${\mathrm{\partial }}_j\varphi (x)=F_j(x)$.

Since $x$ and $j$ were arbitrary, all partial derivatives of $\varphi$ are the continuous components of $F$. By [L3], $\varphi$ is totally differentiable everywhere with $\mathrm{\nabla }\varphi =F$, and these derivatives vary continuously; hence $\varphi$ is $C^1$.

Thus $\varphi$ is the normalized potential asserted in the Statement, and $F$ is conservative.

Condition 1 implies condition 2, and also condition 3, by [L1].

Conditions 2 and 3 imply each other by [L2].

Condition 2 implies condition 1 by [L3], which also supplies the displayed normalized potential.

Thus each of the three conditions implies the other two, proving their equivalence and the final assertion.

Let $x,y$ lie in one piecewise-$C^1$ path component, and choose a path $\gamma$ from $x$ to $y$ as in [L1].

Since $\mathrm{\nabla }(\varphi -\psi )=0$, [L2] gives $0={\int }_{\gamma }0\cdot d\mathbf{r}=(\varphi -\psi )(y)-(\varphi -\psi )(x)\mathrm{.}$

Thus $(\varphi -\psi )(y)=(\varphi -\psi )(x)$. Since $x,y$ were arbitrary within the component, the difference is constant there.

No equality of the constants on distinct components is asserted, because [L1] supplies no path joining such points.

On $U_1\cap U_2$, both gradients equal $F$. Since this intersection is nonempty and piecewise-$C^1$ path-connected, [L1] gives a single constant $C$ such that ${\varphi }_1-{\varphi }_2=C$ throughout the overlap.

Therefore the two clauses in the displayed definition of $\varphi$ agree at every point of $U_1\cap U_2$, so $\varphi$ is well-defined.

Every point of $U_1\cup U_2$ has a neighbourhood on which $\varphi$ equals either the $C^1$ function ${\varphi }_1$ or the $C^1$ function ${\varphi }_2+C$. Hence $\varphi$ is $C^1$ on the union.

On those same neighbourhoods, $\mathrm{\nabla }\varphi$ equals $\mathrm{\nabla }{\varphi }_1=F$ or $\mathrm{\nabla }({\varphi }_2+C)=F$. Thus $\mathrm{\nabla }\varphi =F$ on all of $U_1\cup U_2$.

Steps 2.1, 3.1, and 4.1 prove the gluing assertion. Nonemptiness permits a comparison constant, and path-connectedness makes one adjustment valid on the whole overlap.

If $i=j$, the displayed equality has identical sides. Assume henceforth that $i\mathrm{\ne }j$. Since $U$ is open, choose a ball about $x$ contained in $U$, and choose nonzero $h,k$ small enough that the coordinate rectangle with vertices $x$, $x+h{e}_i$, $x+k{e}_j$, and $x+h{e}_i+k{e}_j$ lies in that ball.

Divide the rectangular difference $R(h,k):=\varphi (x+h{e}_i+k{e}_j)-\varphi (x+h{e}_i)-\varphi (x+k{e}_j)+\varphi (x)$ by $hk$. Apply [L2] on the ordered endpoint intervals first in direction $i$ and then in direction $j$; this covers either sign of $h$ and $k$. Using [L1], there are points $\xi$ and $\eta$ inside the corresponding coordinate intervals such that $\frac{R(h,k)}{hk}={\mathrm{\partial }}_j{\mathrm{\partial }}_i\varphi (x+\xi e_i+\eta e_j)\mathrm{.}$

Applying [L2] in the opposite order gives interior points ${\xi }^{\mathrm{\prime }}$ and ${\eta }^{\mathrm{\prime }}$ such that the same quotient equals $\frac{R(h,k)}{hk}={\mathrm{\partial }}_i{\mathrm{\partial }}_j\varphi (x+{\xi }^{\mathrm{\prime }}{e}_i+{\eta }^{\mathrm{\prime }}{e}_j)\mathrm{.}$

Let $h$ and $k$ tend to zero through nonzero values. All four intermediate displacements tend to zero. Continuity of the two second partials in steps 2.1 and 2.2 therefore gives ${\mathrm{\partial }}_j{\mathrm{\partial }}_i\varphi (x)={\mathrm{\partial }}_i{\mathrm{\partial }}_j\varphi (x)\mathrm{.}$

Together with the $i=j$ case in step 1.1, this proves the assertion for every pair of indices.

Choose the $C^2$ potential $\varphi$ from [L1]. For every $i,j$, ${\mathrm{\partial }}_j{F}_i={\mathrm{\partial }}_j{\mathrm{\partial }}_i\varphi ={\mathrm{\partial }}_i{\mathrm{\partial }}_j\varphi ={\mathrm{\partial }}_i{F}_j,$ where the middle equality is [L2].

The equalities in step 1.1 are precisely the closedness condition in [L1], so $F$ is closed.

By [L1], the integrand defining $\varphi (x)$ is defined for every $t\in [0,1]$; it is continuous, so the integral exists. Fix $x\in U$ and a coordinate $j$. Openness and [L1] provide a small closed coordinate interval about $x$ whose radial segments from $a$ remain in $U$.

On that interval, [L3] differentiates the defining integral with respect to $x_j$ and gives ${\mathrm{\partial }}_j\varphi (x)={\int }_0^1(F_j(z_t)+t{\sum }_i{\mathrm{\partial }}_j{F}_i(z_t)(x_i-a_i))dt,$ where $z_t=a+t(x-a)$. The integrand and its parameter derivative are continuous because $F$ is $C^1$.

By closedness in [L2], ${\mathrm{\partial }}_j{F}_i={\mathrm{\partial }}_i{F}_j$. Thus the integrand in step 2.1 is $F_j(z_t)+t{\sum }_i{\mathrm{\partial }}_i{F}_j(z_t)(x_i-a_i)=\frac{d}{dt}(t{F}_j(z_t))\mathrm{.}$

Apply [L4] to step 3.1. The endpoint at $t=0$ is $0\cdot F_j(a)=0$, and the endpoint at $t=1$ is $F_j(x)$, so ${\mathrm{\partial }}_j\varphi (x)=F_j(x)$.

Since this holds for every $x$ and $j$, the partial derivatives of $\varphi$ are the $C^1$ functions $F_j$. In particular they are continuous, so [L5] gives $\mathrm{\nabla }\varphi =F$, and their first partials are continuous; hence $\varphi$ is $C^2$.