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I keep getting confused with the definition of a smooth path. Here is a definition from William T. Shaw's Complex Analysis with Mathematica:

A path $\phi$ is a continuous mapping from a segment of the real axis into the complex numbers; i.e. $\phi:[a,b]\rightarrow C$.

A path $\phi$ is smooth if it is a differentiable path, and furthermore, the derivative map $\phi':[a,b]\rightarrow C$ is continuous.

OK, now here is my path: $\phi:[-2,2]\rightarrow C$ by $\phi(t) = t^2 + i t^3$. Now, I believe it is differentiable:

$$\phi'(t)=2t+i 3t^2$$

And I believe that $\phi'$ is continuous on $[-2,2]$. However, here is the image of the path:

enter image description here

See the sharp cusp at (0,0)? This is a smooth path?

I am obviously missing a subtle point.

D.

David
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    Note, that is not the graph of the function in the same way that we talk about graphs from ordinary calculus. Think about it - the actual graph would be 3-dimensional. This is just the image of the function, and does not determine the function. – Thomas Andrews Feb 17 '13 at 19:16
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    The curve's "speed" tends to zero as it reaches the cusp, so it is a sufficiently gentle turnaround that it remains continuous (and even continuously differentiable). Your intuition would probably be better captured with the term regular curve (where the tangent vector never jerks around suddenly or takes a rest stop). – anon Feb 17 '13 at 19:18
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    In many books, there is an extra condition on smooth curves, namely that $\phi'(t) \neq 0$ everywhere. (This would make your example non-smooth.) – mrf Feb 17 '13 at 20:24

2 Answers2

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It is smooth by the definition you gave. Smooth path are probably used to make integration along them... cusps are not a problem. The important thing is to find a tangent vector almost everywhere.

Emanuele Paolini
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  • Almost everywhere? The definition requires everywhere. – Thomas Andrews Feb 17 '13 at 19:17
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    In the cusp (as in any point where the derivative is $0$) you don't have a tangent line. – Emanuele Paolini Feb 17 '13 at 19:19
  • There is no tangent line at $0$, but there is a tangent vector (or perhaps velocity vector), namely the zero vector. Integration over a curve involves its tangent vector, not its tangent line. –  Feb 18 '13 at 02:27
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In short, a smooth curve $\vec{c}: [a,b]\rightarrow C$ must:

  1. map distinct points on $(a,b)$ to distinct values
  2. have a $\vec{c}'(t)$ that exists, is continuous, and $\neq\vec{0}$ at any point in $(a,b)$

So no, the example you have given is not smooth, because at $(0,0)$, $\vec{c}'(t)=\vec{0}$ so that it can turn around, which violates condition 2.

(I was going to leave this as a comment for posterity's sake, but unfortunately do not have the reputation to :/)

Matias
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