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The proof is now complete.
REMARK It should be evident that the preceding theorem can easily be generalized to a piecewise smooth function $f,$ i.e., a function that is continuous on $[a,b],$ continuously differentiable on each subinterval $({t}_{i-1},{t}_{i})$ of a partition $\{{t}_{0}<{t}_{1}<...<{t}_{n}\},$ and whose derivative ${f}^{\text{'}}$ is absolutely integrable on $(a,b).$ Indeed, just apply the theorem to each of the subintervals $({t}_{i-1},{t}_{i}),$ and then carefully piece together the piecewise linear functions on those subintervals.
Now we are ready to define what a smooth curve is.
By a smooth curve from a point ${z}_{1}$ to a different point ${z}_{2}$ in the plane, we mean a set $C\subseteq C$ that is the range of a 1-1, smooth, function $\phi :[a,b]\to C,$ where $[a,b]$ is a bounded closed interval in $R,$ where ${z}_{1}=\phi \left(a\right)$ and ${z}_{2}=\phi \left(b\right),$ and satisfying ${\phi}^{\text{'}}\left(t\right)\ne 0$ for all $t\in (a,b).$
More generally, if $\phi :[a,b]\to {R}^{2}$ is 1-1 and piecewise smooth on $[a,b],$ and if $\{{t}_{0}<{t}_{1}<...<{t}_{n}\}$ is a partition of $[a,b]$ such that ${\phi}^{\text{'}}\left(t\right)\ne 0$ for all $t\in ({t}_{i-1},{t}_{i}),$ then the range $C$ of $\phi $ is called a piecewise smooth curve from ${z}_{1}=\phi \left(a\right)$ to ${z}_{2}=\phi \left(b\right).$
In either of these cases, $\phi $ is called a parameterization of the curve $C.$
Note that we do not assume that $|{\phi}^{\text{'}}|$ is improperly-integrable, though the preceding theorem might have made you think we would.
REMARK Throughout this chapter we will be continually faced with the fact that a given curve can have many different parameterizations.Indeed, if ${\phi}_{1}:[a,b]\to C$ is a parameterization, and if $g:[c,d]\to [a,b]$ is a smooth function having a nonzero derivative, then ${\phi}_{2}\left(s\right)={\phi}_{1}\left(g\left(s\right)\right)$ is another parameterization of $C.$ Since our definitions and proofs about curves often involve a parametrization, we will frequently need to prove that the results we obtain are independent of the parameterization.The next theorem will help; it shows that any two parameterizations of $C$ are connected exactly as above, i.e., there always is such a function $g$ relating ${\phi}_{1}$ and ${\phi}_{2}.$
Let ${\phi}_{1}:[a,b]\to C$ and ${\phi}_{2}:[c,d]\to C$ be two parameterizations of a piecewise smooth curve $C$ joining ${z}_{1}$ to ${z}_{2}.$ Then there exists a piecewise smooth function $g:[c,d]\to [a,b]$ such that ${\phi}_{2}\left(s\right)={\phi}_{1}\left(g\left(s\right)\right)$ for all $s\in [c,d].$ Moreover, the derivative ${g}^{\text{'}}$ of $g$ is nonzero for all but a finite number of points in $[c,d].$
Because both ${\phi}_{1}$ and ${\phi}_{2}$ are continuous and 1-1, it follows from [link] that the function $g={\phi}_{1}^{-1}\circ {\phi}_{2}$ is continuous and 1-1 from $[c,d]$ onto $[a,b].$ Moreover, from [link] , it must also be that $g$ is strictly increasing or strictly decreasing. Write ${\phi}_{1}\left(t\right)={u}_{1}\left(t\right)+i{v}_{1}\left(t\right)\equiv ({u}_{1}\left(t\right),{v}_{1}\left(t\right)),$ and ${\phi}_{2}\left(s\right)={u}_{2}\left(s\right)+i{v}_{2}\left(s\right)\equiv ({u}_{2}\left(s\right),{v}_{2}\left(s\right)).$ Let $\{{x}_{0}<{x}_{1}<...<{x}_{p}\}$ be a partition of $[a,b]$ for which ${\phi}_{1}^{\text{'}}$ is continuous and nonzeroon the subintervals $({x}_{j-1},{x}_{j}),$ and let $\{{y}_{0}<{y}_{1}<...<{y}_{q}\}$ be a partition of $[c,d]$ for which ${\phi}_{2}^{\text{'}}$ is continuous and nonzero on the subintervals $({y}_{k-1},{y}_{k}).$ Then let $\{{s}_{0}<{s}_{1}<...<{s}_{n}\}$ be the partition of $[c,d]$ determined by the finitely many points $\left\{{y}_{k}\right\}\cup \left\{{g}^{-1}\left({x}_{j}\right)\right\}.$ We will show that $g$ is continuously differentiable at each point $s$ in the subintervals $({s}_{i-1},{s}_{i}).$
Fix an $s$ in one of the intervals $({s}_{i-1},{s}_{i}),$ and let $t={\phi}_{1}^{-1}\left({\phi}_{2}\left(s\right)\right)=g\left(s\right).$ Of course this means that ${\phi}_{1}\left(t\right)={\phi}_{2}\left(s\right),$ or ${u}_{1}\left(t\right)={u}_{2}\left(s\right)$ and ${v}_{1}\left(t\right)={v}_{2}\left(s\right).$ Then $t$ is in some one of the intervals $({x}_{j-1},{x}_{j}),$ so that we know that ${\phi}_{1}^{\text{'}}\left(t\right)\ne 0.$ Therefore, we must have that at least one of ${u}_{1}^{\text{'}}\left(t\right)$ or ${v}_{1}^{\text{'}}\left(t\right)$ is nonzero. Suppose it is ${v}_{1}^{\text{'}}\left(t\right)$ that is nonzero. The argument, in case it is ${u}_{1}^{\text{'}}\left(t\right)$ that is nonzero, is completely analogous. Now, because ${v}_{1}^{\text{'}}$ is continuous at $t$ and ${v}_{1}^{\text{'}}\left(t\right)\ne 0,$ it follows that ${v}_{1}$ is strictly monotonic in some neighborhood $(t-\delta ,t+\delta )$ of $t$ and therefore is 1-1 on that interval. Then ${v}_{1}^{-1}$ is continuous by [link] , and is differentiable at the point ${v}_{1}\left(t\right)$ by the Inverse Function Theorem. We will show that on this small interval $g={v}_{1}^{-1}\circ {v}_{2},$ and this will prove that $g$ is continuously differentiable at $s.$
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