Approximating piecewise smooth curves by smooth ones

Question

I have been told that any piecewise smooth curve in the plane can be approximated by a sequence of smooth curves whose length converges to the one of the original curve. As long as I know it is done by removing the vertices and making them smooth. However, I was not able to find a proper proof of this fact.

In order to give some context, by piecewse smooth curve I refer to a curve $\alpha \colon [a,b] \rightarrow \mathbb{R}^2$ such that it is $\mathcal{C}^\infty$ everywhere except in a collection of points $a=t_0<t_1<\dots<t_{k-1}<t_k=b$ where the rigth and left first derivative exist eventhough they might not be the same. The points $\alpha(t_i)$ are called vertices and the curves $\alpha_{|[t_i, t_i+1]} \colon [t_i, t_i+1] \rightarrow \mathbb{R^2}$ are called edges. The legth of the curve $\alpha$ is just the sum of the length of the edges.

I don't know the datails about in which way do the smooth curves approximate the original one, neither anything formal about this fact, sorry for that. Can anyone help me and send apropiate references?

Answer 1

In fact, suppose merely that $\alpha$ is piecewise $C^1$. Let $v=\alpha’$ be “the” velocity of $\alpha$; this is piecewise continuous (it’s not necessarily defined at the vertices, but that doesn’t matter, assign it whatever value you wish there). Due to density of $C^{\infty}([a,b])$ in $L^1([a,b])$, there exists a sequence of $C^{\infty}$ maps $\beta_j:[a,b]\to\Bbb{R}^n$ such that $\|v-\beta_j\|_{L^1([a,b])}\to 0$ as $j\to\infty$. Now, define $\gamma_j:[a,b]\to\Bbb{R}^n$ as $\gamma_j(t)=\alpha(a)+\int_a^t\beta_j(s)\,ds$, which is smooth. Then, by the FTC and triangle and reverse-triangle inequality, we have for each $s\in [a,b]$, \begin{align} |\gamma_j(s)-\alpha(s)|+ \left|L(\gamma_j)-L(\alpha)\right|&=\left|\int_a^s(\gamma’(t)-\alpha’(t))\,dt\right|+\left|\int_a^b|\gamma_j’(t)|\,dt-\int_a^b|\alpha’(t)|\,dt\right|\\ &\leq \int_a^b|\gamma_j’(t)-\alpha’(t)|\,dt+\int_a^b\left||\gamma_j’(t)|-|\alpha’(t)|\right|\,dt\\ &\leq 2\int_a^b|\gamma_j’(t)-\alpha’(t)|\,dt\\ &=2\int_a^b|\beta_j(t)-v(t)|\,dt, \end{align} So, taking the supremum over all $s\in [a,b]$, we see that \begin{align} \|\gamma_j-\alpha\|_{C^0([a,b])}+|L(\gamma_j)-L(\alpha)|\leq 2\|\beta_j-v\|_{L^1([a,b])}. \end{align} By hypothesis, this last term vanishes as $j\to\infty$. Thus, we found a sequence of smooth curves $\gamma_j$ which converge to $\alpha$ uniformly, and such that their arclengths approach that of $\alpha$.

Note that due to continuity of $\alpha$, the FTC can be applied to it. Also, we added $\alpha(a)$ as the constant of integration to $\gamma_j(t)$’s definition so that both $\alpha$ and $\gamma_j$ have the same starting point (hence justifying our very first equal sign).

---

If you don’t want to appeal to this standard density result, then we can be more hands-on as follows. It suffices to consider only one vertex (the general case follows by induction). Hence, we may suppose that $v:[a,b]\to\Bbb{R}^n$ is such that for some $c\in (a,b)$, which WLOG we may take to be $c=0$, we have $v|_{(a,0)}$ and $v|_{(0,b)}$ extend to smooth functions on open neighbourhoods of $[a,0]$ and $[0,b]$ respectively.

Now, let $\phi_{\delta}:[a,b]\to\Bbb{R}$ be a smooth bump function taking values in $[0,1]$, and such that it is identically $0$ on $[-\delta,\delta]$ and identically $1$ outside $[-2\delta,2\delta]$. Then, the function $\beta_{\delta}=\phi_{\delta}\cdot v$ will be smooth on $[a,b]$, and \begin{align} \int_a^b|\beta_{\delta}(t)-v(t)|\,dt&=\int_a^b|\phi_{\delta}(t)-1||v(t)|\,dt\leq\|v\|_{\infty}\int_{-2\delta}^{2\delta}|\phi_{\delta}(t)-1|\,dt\leq \|v\|_{\infty}\cdot (1+1)\cdot (4\delta). \end{align} The $1+1$ is because $|\phi_{\delta}-1|\leq |\phi_{\delta}|+1\leq 1+1$. Hence, as $\delta\to 0$ this vanishes, showing that we found a sequence of smooth functions which converge to $v$ in the $L^1([a,b])$-norm.

Answer 2

$\newcommand\R{\mathbb{R}}$An explicit way to approximate a compactly supported continuous function is convolution. Recall that the convolution of two functions $f, g: \R\rightarrow \R $ is given by $$ (f\star g)(x) = \int_{-\infty}^{\infty} f(x-y)g(y)\,dy = \int_{-\infty}^{\infty} f(y)g(x-y)\,dy. $$ Suppose $g$ is smooth, nonnegative and satisfies $$ \int_{-\infty}^{\infty}g(y)\,dy = 1. $$ For example, $g$ could be the Gaussian density function with zero mean and unit variance. For each $\delta > 0$, let $$ g_\delta(x) = \delta^{-1}g(\delta^{-1}x). $$ Given a piecewise $C^1$ function $\alpha: [a,b] \rightarrow \mathbb{R}^2$, extend it to a function on $\R$ by setting it to zero outside $[a,b]$. For each $\delta > 0$, let $$ \alpha_\delta(x) = \alpha\star g_\delta. $$ It is straightforward to check that $\alpha_\delta$ is smooth and that as $\delta\rightarrow 0$, $$ \lim_{\delta\rightarrow 0} \int_a^b \alpha_\delta(x)\,dx = \int_a^b \alpha(x)\,dx. $$