A curve $\gamma$ is a continuous function from the unit interval $[0,1]$ to a (finite-dimensional, real) vector space $V$, with image $C\subset V$.

By fractal I mean that $\gamma$ is not differentiable, or that $C$ has no tangent lines.

I'm considering an integral of a continuous scalar function $f:V\to\mathbb R$ . (For more generality, $f$ may be a differential form $f(\vec x,\Delta\vec x)$ piecewise-continuous in $\vec x$ and linear in $\Delta\vec x$, with values in $\mathbb R$ or $\Lambda V$ or any space where limits make sense.)

A directed integral over $\gamma$ can be defined (or maybe not) by Riemann sums of the form

$$\sum_{k=1}^n f(\vec x_k)\,\Delta\vec x_k = \sum_{k=1}^n f\big(\gamma(t_k^*)\big)\,\big(\gamma(t_k)-\gamma(t_{k-1})\big)$$

using a partition of the unit interval into $n$ segments, where each sample point $t_k^*$ is in the $k$th segment ($t_{k-1}\leq t_k^*\leq t_k$) .

A sequence of partitions $\Delta_l$ will be called valid if the displacements $\Delta_l\vec x_k$ all converge to $\vec0$ . More precisely, for any open set $S\ni\vec0$, there exists $m\in\mathbb N$ such that, for all $l>m$ and all $k\leq n_l$ (where $n_l$ is the number of segments in the $l$th partition), $\Delta_l\vec x_k\in S$ .

Does the limit

$$\int_\gamma f(\vec x)\,d\vec x = \lim_{l\to\infty}\sum_{k=1}^{n_l} f(\vec x_k)\,\Delta_l\vec x_k$$

exist for a particular valid sequence of $\Delta_l$ and $\{t_k^*\}_l$ ? Is it unique over all such sequences?

If $\gamma$ is smooth, then this integral of a scalar function over a vector domain should equal an integral of a vector function over a scalar domain:

$$\int_\gamma f(\vec x)\,d\vec x = \int_{[0,1]} f\big(\gamma(t)\big)\gamma'(t)\,dt$$

and the integral on the right is known to exist (it's a list of scalar integrals of continuous functions). But if $\gamma$ isn't smooth, we can't use this because $\gamma'$ doesn't exist.

It is easy to see that, for the constant function $f(\vec x)=1$, every Riemann sum is exactly the displacement between the endpoints: $\gamma(1)-\gamma(0)$, so at least some functions can be integrated over arbitrary fractal curves (such as space-filling curves).

This other question is related, but it's metric-based. I don't care about length here.

"This is just general measure theory, though. As far as I know, oriented integration theory (e.g. line integrals and Stokes's theorem) hasn't yet been generalized to fractals" -- answerer Robert Haraway

Is this a reasonable approach to such a generalization? (Riemann sums seem too simple; this should have been done before if it were possible.)

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    $\begingroup$ For $\gamma $ that is Holder continuous, with large Holder exponent, this should also be definable thru approximations by smooth ones. $\endgroup$ – Behnam Esmayli Sep 28 '18 at 15:59

The answer to your question about whether the integral can be defined seems to be found on page 40, here

Let $f$ be a real-valued function on $[a, b]$ and let $P : a = x_0 < x_1 < . . . < x_n = b$ be a partition of $[a, b]$. Define $L(P) = \sum\limits_{i=1}^n [(x_{i} − x_{i-1})^2 + (f(x_{i}) − f(x_{i−1}))^2]^{\frac{1}{2}}$. The function $f$ is of bounded variation on $[a, b]$, denoted $f\in BV [a, b]$, if and only if $\sup\limits_{P} L(P) <\infty$. If $f\in BV [a, b]$, then, by Jordan’s Theorem, $f$ can be written as $f = g − h$, where $g$ and $h$ are monotone functions. Since Lebesgue’s Theorem states that a monotone function on $(a, b)$ is differentiable almost everywhere on $(a, b)$, $f'(x)$ exists almost everywhere on $(a, b)$ if $f\in BV [a, b]$. Since nowhere differentiable functions are not differentiable at any point in their domains, they cannot be of bounded variation. Thus, the graph of a continuous nowhere differentiable function must have infinite length.

For integration with respect to signed (the closest standard example of oriented integration as defined in your question) measures the convention is that the integral is only defined if the complex (say) measure has total variation finite (see Rudin's Real and Complex Analysis for more on the development of complex measures). So some sort of regularity (differentiability) of $\gamma$ is needed here.

As an aside, stochastic integrals handle integration over Brownian motions (here a Brownian motion is a stochastic process rather than a simple curve). There is the theory of stochastic calculus which is designed to handle such questions.

  • $\begingroup$ Is $V$ finite dimensional? It is not clear what non-differentiability means in your problem if $V$ is not a metric space. $\endgroup$ – asd Sep 22 '18 at 2:20
  • $\begingroup$ There is topology... I'll have to see the definition of differentiability again. $\endgroup$ – mr_e_man Sep 22 '18 at 2:27
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    $\begingroup$ The limiting behavior of sums of products of values of a function and numbers for sub-intervals has a natural interpretation as integration with respect to a measure. Complex measures were mentioned as they are the simplest example of such a vector valued measure. $\endgroup$ – asd Sep 22 '18 at 2:28
  • $\begingroup$ Shouldn't it be sufficient to say that a curve is differentiable iff its coordinates are differentiable, for any fixed basis? (This doesn't involve partial derivatives.) $\endgroup$ – mr_e_man Sep 22 '18 at 2:38
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    $\begingroup$ No, $\vec{\gamma}(t_k)−\vec{\gamma}(t_{k-1})=\int\limits_{t_{k-1}}^{t_{k}} \vec{h}(t) dt$ gives a representation of $\gamma$ as a measure, $d \gamma=h dt$. $\endgroup$ – asd Sep 22 '18 at 3:20

Perhaps this could be simplified or clarified by reducing everything to scalars, even though that $\gamma'$ formula won't work. Using a basis $\{\vec e_j\}$, the curve is $\gamma(t)=\sum_j\gamma_j(t)\vec e_j$, the function is $f\circ\gamma:[0,1]\to\mathbb R$ (which is continuous), and the Riemann sum is

$$\sum_{k=1}^n f(\vec x_k)\,\Delta\vec x_k = \sum_{k=1}^n f\circ\gamma(t_k^*)\sum_j\big(\gamma_j(t_k)-\gamma_j(t_{k-1})\big)\vec e_j$$

$$= \sum_j\bigg(\sum_{k=1}^n f\circ\gamma(t_k^*)\big(\gamma_j(t_k)-\gamma_j(t_{k-1})\big)\bigg)\vec e_j$$

$$\int_\gamma f(\vec x)\,d\vec x = \sum_j\bigg(\lim_{l\to\infty}\sum_{k=1}^{n_l} f\circ\gamma(t_{k,l}^*)\big(\gamma_j(t_{k,l})-\gamma_j(t_{k-1,l})\big)\bigg)\vec e_j$$

These coefficients look like integrals $\int_{[0,1]}f\circ\gamma\,d\mu_j$ with respect to signed measures $\mu_j$, which apply to subintervals of $[0,1]$ by

$$\mu_j[a,b] = \gamma_j(b)-\gamma_j(a)$$

and apply additively to unions of intervals. As @asd answered, because $\gamma$ is not differentiable, these measures have infinite total variation... and some sets (infinite unions of intervals where $\gamma_j$ increases) will have measure ${^+}\infty$, and others ${^-}\infty$. So these are not acceptable signed measures.

But the integrals can still be interpreted as Stieltjes integrals or something similar. Then any function $f$ is integrable if $f\circ\gamma$ and $\gamma_j$ are Holder continuous with exponents whose sum exceeds $1$.


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