In this problem, I know that the hypothesis of Green's theorem must ensure that the simple closed curve is smooth, but what is smooth? Could you give a definition and an intuitive explanation?

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    $\begingroup$ Intuitive: "no corners". Rigorous: "a sufficient number of its derivatives are continuous." $\endgroup$ Nov 27, 2010 at 6:05
  • $\begingroup$ @J.M.: What is "sufficient"? $\endgroup$
    – Jichao
    Nov 27, 2010 at 6:08
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    $\begingroup$ Sivaram elaborated a bit; the working definition I have is "whatever the application needs." $\endgroup$ Nov 27, 2010 at 6:40
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    $\begingroup$ @J.M. The "no corners" characterization is good for a regular curve (where the tangent vector doesn't vanish). However, for a smooth curve, as defined in the accepted definition, we can have cusps, as in the deltoid, for example en.wikipedia.org/wiki/Deltoid_curve $\endgroup$
    – yasmar
    Nov 27, 2010 at 16:11
  • $\begingroup$ @yas: So, "piecewise smooth" then. :) $\endgroup$ Nov 28, 2010 at 1:22

4 Answers 4


There are many ways you can characterize the smoothness of a curve.

Typically, people use the notation $C^{(n)}(\Omega)$ where $n \in \mathbb{N}$.

So when we say $f(x) \in C^{(n)}(\Omega)$, we mean that $f(x)$ has $n$ derivatives in the entire domain ($\Omega$ denotes the domain of the function) and the $n^{th}$ derivative of $f(x)$ is continuous i.e. $f^{n}(x)$ is continuous.

Also by convention, if $f(x)$ is just continuous, then we say $f(x) \in C^{(0)}(\Omega)$.

Also, $f(x) \in C^{(\infty)}$ if the function is differentiable any number of times. For instance, $e^{x} \in C^{(\infty)}$

An example to illustrate is to consider the following function $f: \mathbb{R} \rightarrow \mathbb{R}$. $$f(x) = \begin{cases}0, &\mbox{if }x \leq 0 \\ x^2, &\mbox{if }x>0\end{cases}$$

This function is in $C^{(1)}(\mathbb{R})$ but not in $C^{(2)}(\mathbb{R})$.

When the domain of the function is the largest set over which the function definition makes sense, we omit $\Omega$ and write that $f \in C^{(n)}$, the domain being understood as the largest set over which the function definition makes sense.

Note that $C^{(n)} \subseteq C^{(m)}$ whenever $n>m$.


In case of Green's theorem, when we apply the formula $$\oint_c (L\,dx + M\,dy) = \iint_D \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\,dx\,dy$$ we need $L,M \in C^1{(\Omega)}$, where $\Omega$ is a domain containing the curve and the interior of the curve viz $D$.

The simple closed curve $C$ should be piecewise smooth or more generally the curve $C$ should be in $C^{(0)}$.

We say that the curve $C$ is piecewise smooth curve when the two conditions below are satisfied:

(i) $C \in C^{(0)}$

(ii) The domain over which the curve is defined can be partitioned into disjoint subsets such that the curve is in $C^{(\infty)}$ (or sufficiently smooth i.e. the curve is in $C^{(n)}$ for some $n$ till which we are interested) over each of these subsets.

  • $\begingroup$ @Sivaram: Do you mean if curve $L$ is smooth, then $L \in C^{(0)}$? $\endgroup$
    – Jichao
    Nov 27, 2010 at 7:10
  • $\begingroup$ @Jichao: In the Green's theorem, you want the curve over which you are integrating i.e. the curve $C$ to be piecewise smooth. It depends on to what extent of smoothness you want. In the case of Green's theorem, you want the functions $L$ and $M$ to be smooth up to their first derivative i.e. $L,M \in C^{(1)}$ and the curve over which you are integrating i.e. $C$ to be smooth up to being continuous i.e. $C \in C^{(0)}$. $\endgroup$
    – user17762
    Nov 27, 2010 at 7:14
  • $\begingroup$ @Sivaram: So smoothness is a relative concept and the smoothest curve should belong to $C^{(\infty)}$? $\endgroup$
    – Jichao
    Nov 27, 2010 at 7:17
  • $\begingroup$ @Jichao: Exactly. Smoothness is a relative concept and is problem specific. $C^{(\infty)}$ is as smooth as smooth can be. In applications, when you say the curve is smooth it means till the derivatives you are interested in the curve has to be continuous. So for instance in Green's theorem, smoothness would mean the functions $L,M \in C^{(1)}$ and the curve $C \in C^{(0)}$. $\endgroup$
    – user17762
    Nov 27, 2010 at 7:24
  • $\begingroup$ @Sivaram: So considering parametric representation of 2-dimensional curve $C$, $C \in C^{(n)}$ means $x(t) \in C^{(n)}$ and $y(t) \in C^{(n)}$ ? $\endgroup$
    – Jichao
    Nov 27, 2010 at 7:44

I stumbled upon this old question and I'd like to add something: there is a difference of perspective on smoothness depending whether you look at the geometric object or its parametrization.

Look at the standard example: the real cusp. It is a curve in the real plane parametrized $f:t\to (t^2,t^3)$. Of course, the mapping $f$ is smooth (of any order), and the graph of $f$ is a smooth manifold in $\mathbb{R}^3$, but its image is singular: it is the zero set $x^3=y^2$. It is "worse than a corner"!

So you need to be always clear what you want: do you need only differentiability of the parametrization or do you want the image to be a differentiable manifold (typically in such a case you would assume that the derivative of $f$ does not vanish).


Consider the following curve in the plane, $(x(t),y(t))$, this curve is called smooth if the functions $x(t)$ and $y(t)$ are smooth, which simply means that for all $N$, the derivatives $\frac{d^Nx}{dt^N}$ and $\frac{d^Ny}{dt^N}$ exist.

  • $\begingroup$ for any sign of any set correct? Given that Abs(x) == -x ? $\endgroup$
    – Jay
    Aug 14, 2019 at 16:35
  • $\begingroup$ Re: stackoverflow.com/questions/57497171/… $\endgroup$
    – Jay
    Aug 14, 2019 at 16:39
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    $\begingroup$ This is incorrect - consider $(t^2, t^3)$. $\endgroup$
    – user785085
    Feb 20, 2022 at 20:03

Smooth means differentiable (at least once). In other words, smooth means continuous and without "corners".

Edit: changed "edges" to "corners" after J.M.

  • $\begingroup$ So smooth is different with piecewise smooth? $\endgroup$
    – Jichao
    Nov 27, 2010 at 6:07
  • $\begingroup$ A piecewise smooth curve could have corners. Think of a square. The individual sides are smooth, but the square itself is a nonsmooth curve at the corners. $\endgroup$
    – TKN
    Nov 27, 2010 at 6:14
  • $\begingroup$ This is wrong. A smooth curve must be defined by a function which has infinite derivatives. $\endgroup$ Nov 27, 2010 at 6:18
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    $\begingroup$ @Matt: It depends on the author. Some authors define "smooth" to mean $C^\infty$, others define it to mean $C^1$, and still others define it to mean $C^1$ and has non-vanishing derivative. $\endgroup$ Jan 31, 2011 at 0:52

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