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I have this question regarding the continuity of a line integral. Suppose we have a continuous function $f:\mathbb{R}^n \to \mathbb{R}$, and a curve $C_n$ such that $C_n \to C$ pointwise as $n\to \infty$. The curves $C_n$ and $C$ can have unbounded domain. I was wondering what kind of assumptions one can make so that

$$\int_{C_n}f(s)ds \to \int_{C}f(s)ds$$

I believe some assumptions enabling the application of the dominated convergence theorem could be the answer, but I am not very knowledgeable about the topic. Any insight or reference would be much appreciated!

Ldt
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2 Answers2

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Although this doesn’t answer your question completely, I still think this is valuable to show that you’ll need some very specific or strong conditions. I’ll show by example that it isn’t strong enough to suppose that $f\in C^{\infty}(\mathbf{R}^N)$ and that the corresponding parametrizations $\psi_n$ (for $C_n$) converge uniformly to $\psi$ (for $C$).

Let $N=2$, $f(\mathbf{x})=1$ for all $\mathbf{x} \in \mathbf{R}^2$ and

$$ \forall n \in \mathbf{N} : \psi_n : [0,1] \rightarrow \mathbf{R}^2 : t \mapsto \left(t, \frac{\sin(tn)}{n}\right). $$

It’s not too difficult to show that $\psi_n \overset{[0,1]}{\rightrightarrows} \psi:[0,1]\rightarrow \mathbf{R}^2:t\mapsto(t,0)$. Clearly we have $\int_C f = 1$, but

$$ \begin{align*} I_n:=\int_{C_n} f &= \int_0^1 \sqrt{1+\cos(tn)^2}\mathrm{d}t \\&= \frac 1n \int_0^n\sqrt{1+\cos(t)^2}\mathrm{d}t. \end{align*} $$ I think it is pretty clear that this integral doesn’t have limit $1$, but for completeness’ sake I’ll provide an argument for those who aren’t bored yet. Clearly $\left\lfloor \frac n{2\pi}\right\rfloor < \frac n{2\pi}$ for all $n\in\mathbf{N}$, and since the integrand, $g(t)$, is strictly positive we have $$ \begin{align*} I_n &> \frac 1n \int_0^{2\pi \left\lfloor \frac{n}{2\pi}\right \rfloor}g(t)\mathrm{d}t=\frac 1n \sum_{i=0}^{\left\lfloor \frac n{2\pi}\right\rfloor-1} \int_{2\pi i}^{2\pi(i+1)}g(t)\mathrm{d}t\\&=\frac 1n\left\lfloor \frac n{2\pi} \right \rfloor \int_0^{2\pi}g(t)\mathrm{d}t > \frac 1n\left\lfloor \frac n{2\pi} \right \rfloor \cdot 7 \end{align*} $$ where we note that $g$ is $2\pi$ periodic, and the last inequality was given by a numerical estimate by Wolfram Alpha. Now considering that $\lim_{n\rightarrow \infty} \frac 1n\left\lfloor \frac n{2\pi} \right \rfloor = \frac 1{2\pi}$, and $2\pi<7$, we can conclude that if $\lim_{n\rightarrow \infty} I_n$ exists, it is strictly greater than $1$.

I don’t know where to go from here in determining sufficient conditions, but I still figured this might be useful.

Jan
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With a little help from my professor I was able to come up with some sufficient conditions. Assume that $f$ is continuous on some compact $K$ containing all curves $C_n$ and $C$. Moreover, assume that the parametrizations $\psi_n : [0,1] \rightarrow \mathbf{R}^n$ (for $C_n$) converge uniformly (on $[0,1]$) to the parametrization $\psi : [0,1]\rightarrow \mathbf{R}^n$ (for $C$). And, assume that also $\psi_n’$ converge uniformly to $\psi’$ on $[0,1]$. These conditions are strong enough.

We’ll need the following lemma. Let $g:\mathbf{R}^n\rightarrow \mathbf{R}$ be uniformly continuous on some $U\subseteq \mathbf{R}^n$. Let $\psi_n$ and $\psi$ be as above and their images be contained in $U$, then also $g\circ \psi_n \overset {[0,1]}{\rightrightarrows}g \circ \psi$. This can obviously be generalized quite a bit and a proof of this can be found here.

We’ll also need that we can interchange limit and integral if the integrand converges uniformly, and that the product of uniformly converging sequences converges uniformly to the product.

We note that by Heine $f$ is uniformly continuous on $K$, and thus on all curves. Also, the norm function $\Vert \cdot \Vert$ is continuous everywhere. This means we can apply the lemma twice to make sure the integrands below converge uniformly. Now we can easily see that $$ \begin{align*} \lim_{n\rightarrow +\infty} \int_{C_n} f(s)ds &= \lim_{n\rightarrow +\infty} \int_0^1 f(\psi_n(t))\Vert \psi_n’(t)\Vert dt = \\ \int_0^1 f(\psi(t))\Vert \psi’(t)\Vert dt &= \int_C f(s)ds \end{align*} $$

Jan
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