Do line integrals along non-piecewise-smooth curves exist?

Question

This article at Wolfram Mathworld has the following theorem on conservative vector fields:

Theorem. The following conditions are equivalent for a conservative vector field $ \mathbf{F} $ defined on an open subset $ U $ of $ \mathbb{R}^{n} $:

1. For any oriented, simple and closed curve $ C $ whose image lies in $ U $, the contour integral of $ \mathbf{F} $ along $ C $ equals $ 0 $, i.e., $ \displaystyle \oint_{C} \mathbf{F} \cdot \mathrm{d}{\mathbf{s}} = 0 $.

2. For any two oriented and simple curves $ C_{1} $ and $ C_{2} $ with the same endpoints whose images lie in $ U $, we have $ \displaystyle \int_{C_{1}} \mathbf{F} \cdot \mathrm{d}{\mathbf{s}} = \int_{C_{2}} \mathbf{F} \cdot \mathrm{d}{\mathbf{s}} $.

3. There exists a differentiable function $ f: U \to \mathbb{R} $, called the scalar potential function of $ \mathbf{F} $, such that $ \mathbf{F} = \nabla f $.

However, I have read in other places that the curves $ C $, $ C_{1} $ and $ C_{2} $ must be piecewise smooth. It seems like a necessary restriction on otherwise arbitrary curves.

Hence, my question is:

Question. Is the theorem valid for non-piecewise-smooth curves? What about non-piecewise-differentiable curves? I think that the line integral might not be defined then.

Answer 1

The general contour integral is defined as follows.

• Let $ (X,d) $ be a metric space.
• Let $ C: [0,1] \to X $ be a simple closed curve in $ X $.
• As $ X $ may not be a differentiable manifold, the smoothness of $ C $ may not make sense.
• Let $ F: X \to \mathbb{R} $ be a (not necessarily continuous) function.
• For all sequences $ (t_{n,0},\ldots,t_{n,n};s_{n,1},\ldots,s_{n,n})_{n \in \mathbb{N}} $ of tagged partitions of $ [0,1] $ whose corresponding sequence of meshes goes to $ 0 $, suppose that the limit $$ \lim_{n \to \infty} \sum_{i = 0}^{n - 1} F(C(s_{n,i + 1})) \cdot d(C(t_{n,i + 1}),C(t_{n,i})) $$ exists and is the same.
• We then denote this limit by $ \displaystyle \oint_{C} F $ and call it the contour integral of $ F $ with respect to the simple closed curve $ C $.

If $ F $ is continuous and $ C $ is rectifiable, then $ \displaystyle \oint_{C} F $ exists. The concept of smoothness is not mentioned at all.

Smoothness ― more generally, differentiability ― only comes into the picture when $ X $ is a differentiable manifold, e.g. when $ X $ is a Euclidean space or a Riemannian manifold (actually, every differentiable manifold possesses a Riemannian metric, but there is no canonical one).

Answer 2

This is in response to tipshoni’s request above.

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Let us first introduce some conventions and definitions.

• For each $ n \in \mathbb{N} $, we define $ [n] \stackrel{\text{df}}{=} \{ i \in \mathbb{N}_{0} \mid 0 \leq i \leq n - 1 \} $.

• Fix a metric space $ (X,d) $.

• Let $ \gamma: [0,1] \to X $ be a (continuous) curve in $ X $.

• A partition of $ [0,1] $ is defined as a finite subset of $ [0,1] $ containing $ 0 $ and $ 1 $.

• We denote the set of all partitions of $ [0,1] $ by $ \mathbf{P} $. Observe that $ (\mathbf{P},\subseteq) $ is a directed set.

• If $ \mathcal{P} \in \mathbf{P} $ has cardinality $ n $ and $ t^{\mathcal{P}}: [n] \to \mathcal{P} $ denotes the order-preserving enumeration of $ \mathcal{P} $, then we define $ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] $ by $$ {L_{\gamma}}(\mathcal{P}) \stackrel{\text{df}}{=} \sum_{i \in [n]} d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). $$

• If $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality applied to the metric $ d $, we obtain $$ {L_{\gamma}}(\mathcal{P}) \leq {L_{\gamma}}(\mathcal{Q}). $$

• Finally, we say that $ \gamma $ is rectifiable if and only if $$ L_{\gamma} \stackrel{\text{df}}{=} \sup(\{ {L_{\gamma}}(\mathcal{P}) \in [0,\infty] \mid \mathcal{P} \in \mathbf{P} \}) < \infty. $$

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Let $ \gamma $ be a rectifiable curve and $ F: X \to \mathbb{R} $ a bounded function (not assumed to be continuous yet).

For each $ \mathcal{P} \in \mathbf{P} $, define \begin{align} \mathcal{L}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))); \\ \mathcal{U}(\gamma,\mathcal{P};F) & \stackrel{\text{df}}{=} \sum_{i \in [\mathsf{card}(\mathcal{P})]} \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))). \end{align}

Define also \begin{align} \mathcal{L}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}; \\ \mathcal{U}(\gamma;F) & \stackrel{\text{df}}{=} \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F), \quad \text{if the limit exists in $ \mathbb{R} $}. \end{align}

Here, the limits are to be taken in the sense of a net. For example, if $ A \in \mathbb{R} $, then we write $$ A = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) $$ if and only if for every $ \epsilon > 0 $, there exists a $ \mathcal{P}_{0} \in \mathbf{P} $ such that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ |A - \mathcal{L}(\gamma,\mathcal{P};F)| < \epsilon. $$

If $ \mathcal{L}(\gamma;F) $ and $ \mathcal{U}(\gamma;F) $ both exist and are equal, then we define the line integral of $ F $ along $ \gamma $ by $$ \int_{\gamma} F \stackrel{\text{df}}{=} \mathcal{L}(\gamma;F) = \mathcal{U}(\gamma;F). $$

If $ X = \mathbb{R}^{2} $ and $ \gamma $ is a Jordan curve, then we usually denote the line integral of $ F $ along $ \gamma $ by $ \displaystyle \oint_{\gamma} F $, if it exists.

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Now, assume that $ \gamma $ is a rectifiable curve as before and that $ F $ is continuous.

Claim: The line integral of $ F $ along $ \gamma $ exists.

Proof of Claim

As $ F \circ \gamma $ is a continuous function defined on the closed and bounded interval $ [0,1] $, its range lies in the interval $ [- M,M] $ for some finite $ M > 0 $. Then we clearly have $$ \forall \mathcal{P} \in \mathbf{P}: \quad |\mathcal{L}(\gamma,\mathcal{P};F)| \leq \sum_{i \in [\mathsf{card}(\mathcal{P})]} M \cdot d(\gamma({t^{\mathcal{P}}}(i)),\gamma({t^{\mathcal{P}}}(i + 1))) \leq M \cdot L_{\gamma}. $$

Furthermore, observe that if $ \mathcal{P},\mathcal{Q} \in \mathbf{P} $ and $ \mathcal{P} \subseteq \mathcal{Q} $, then by the Triangle Inequality, we get $$ \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma,\mathcal{Q};F). $$ Hence, the net $ \{ \mathbf{P} \to \mathbb{R}; \mathcal{P} \mapsto \mathcal{L}(\gamma,\mathcal{P};F) \} $ is bounded and monotone, which implies that $$ \mathcal{L}(\gamma;F) = \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) = \sup_{\mathcal{P} \in \mathbf{P}} \mathcal{L}(\gamma,\mathcal{P};F) \in \mathbb{R}. $$

Next, we will show that $$ \lim_{\mathcal{P} \in \mathbf{P}} \mathcal{U}(\gamma,\mathcal{P};F) = \mathcal{L}(\gamma;F). $$

Let $ \epsilon > 0 $. As $ F \circ \gamma $ is uniformly continuous on $ [0,1] $, we can find a $ \mathcal{P}_{0} \in \mathbf{P} $ sufficiently fine so that for all $ \mathcal{P} \in \mathbf{P} $ with $ \mathcal{P}_{0} \subseteq \mathcal{P} $, we have $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{L}(\gamma,\mathcal{P};F) \leq \mathcal{L}(\gamma;F) \qquad (\clubsuit) $$ and $$ \forall i \in [\mathsf{card}(\mathcal{P})]: \quad \left[ \sup_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] - \left[ \inf_{t \in [{t^{\mathcal{P}}}(i),{t^{\mathcal{P}}}(i + 1)]} F(\gamma(t)) \right] < \frac{\epsilon}{L_{\gamma} + 1}. $$ Let $ \mathcal{P} \in \mathbf{P} $ satisfy $ \mathcal{P}_{0} \subseteq \mathcal{P} $. It follows immediately from the preceding inequality that $$ 0 \leq \mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma,\mathcal{P};F) \leq \frac{\epsilon}{L_{\gamma} + 1} \cdot L_{\gamma} < \epsilon. \qquad (\spadesuit) $$ Adding the inequalities $ (\clubsuit) $ and $ (\spadesuit) $ yields $$ \mathcal{L}(\gamma;F) - \epsilon < \mathcal{U}(\gamma,\mathcal{P};F) < \mathcal{L}(\gamma;F) + \epsilon, $$ or equivalently, $$ |\mathcal{U}(\gamma,\mathcal{P};F) - \mathcal{L}(\gamma;F)| < \epsilon. $$ Therefore, as $ \epsilon $ is arbitrary, we obtain $ \mathcal{U}(\gamma;F) = \mathcal{L}(\gamma;F) $. This concludes the proof. $ \quad \blacksquare $

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Concluding remarks: As it is not required for $ X $ to be equipped with a smooth structure, it is not necessary to invoke the notion of ‘smoothness’ in order to discuss line integrals. However, in the context of $ X = \mathbb{R}^{2} $, where a canonical smooth structure exists, it would certainly be profitable to restrict one’s attention to the class of positively oriented and piecewise-smooth Jordan curves so as to exploit powerful results like Green’s Theorem.