# 6.1 Smooth curves in the plane  (Page 2/4)

Recall that a function $f$ that is continuous on a closed interval $\left[a,b\right]$ and continuously differentiable on the open interval $\left(a,b\right)$ is called a smooth function on $\left[a,b\right].$ And, if there exists a partition $\left\{{t}_{0}<{t}_{1}<...<{t}_{n}\right\}$ of $\left[a,b\right]$ such that $f$ is smooth on each subinterval $\left[{t}_{i-1},{t}_{i}\right],$ then $f$ is called piecewise smooth on $\left[a,b\right].$ Although the derivative of a smooth function is only defined and continuous on the open interval $\left(a,b\right),$ and hence possibly is unbounded, it follows from part (d) of [link] that this derivative is improperly-integrable on that open interval.We recall also that just because a function is improperly-integrable on an open interval, its absolute value may not be improperly-integrable.Before giving the formal definition of a smooth curve, which apparently will be related to smooth or piecewise smooth functions,it is prudent to present an approximation theorem about smooth functions. [link] asserts that every continuous function on a closed bounded interval is the uniform limit of a sequence of step functions.We give next a similar, but stronger, result about smooth functions. It asserts that a smooth function can be approximated “almost uniformly” by piecewise linear functions.

Let $f$ be a smooth function on a closed and bounded interval $\left[a,b\right],$ and assume that $|{f}^{\text{'}}|$ is improperly-integrable on the open interval $\left(a,b\right).$ Given an $ϵ>0,$ there exists a piecewise linear function $p$ for which

1.   $|f\left(x\right)-p\left(x\right)|<ϵ$ for all $x\in \left[a,b\right].$
2.   ${\int }_{a}^{b}|{f}^{\text{'}}\left(x\right)-{p}^{\text{'}}\left(x\right)|\phantom{\rule{0.166667em}{0ex}}dx<ϵ.$

That is, the functions $f$ and $p$ are close everywhere, and their derivatives are close on average in the sense that theintegral of the absolute value of the difference of the derivatives is small.

Because $f$ is continuous on the compact set $\left[a,b\right],$ it is uniformly continuous. Hence, let $\delta >0$ be such that if $x,y\in \left[a,b\right],$ and $|x-y|<\delta ,$ then $|f\left(x\right)-f\left(y\right)|<ϵ/2.$

Because $|{f}^{\text{'}}|$ is improperly-integrable on the open interval $\left(a,b\right),$ we may use part (b) of [link] to find a ${\delta }^{\text{'}}>0,$ which may also be chosen to be $<\delta ,$ such that ${\int }_{a}^{a+{\delta }^{\text{'}}}|{f}^{\text{'}}|+{\int }_{b-{\delta }^{\text{'}}}^{b}|{f}^{\text{'}}|<ϵ/2,$ and we fix such a ${\delta }^{\text{'}}.$

Now, because ${f}^{\text{'}}$ is uniformly continuous on the compact set $\left[a+{\delta }^{\text{'}},b-{\delta }^{\text{'}}\right],$ there exists an $\alpha >0$ such that $|{f}^{\text{'}}\left(x\right)-{f}^{\text{'}}\left(y\right)|<ϵ/4\left(b-a\right)$ if $x$ and $y$ belong to $\left[a+{\delta }^{\text{'}},b-{\delta }^{\text{'}}\right]$ and $|x-y|<\alpha .$ Choose a partition $\left\{{x}_{0}<{x}_{1}<...<{x}_{n}\right\}$ of $\left[a,b\right]$ such that ${x}_{0}=a,{x}_{1}=a+{\delta }^{\text{'}},{x}_{n-1}=b-{\delta }^{\text{'}},{x}_{n}=b,$ and ${x}_{i}-{x}_{i-1} for $2\le i\le n-1.$ Define $p$ to be the piecewise linear function on $\left[a,b\right]$ whose graph is the polygonal line joining the $n+1$ points $\left(a,f\left({x}_{1}\right)\right),$ $\left\{\left({x}_{i},f\left({x}_{i}\right)\right)\right\}$ for $1\le i\le n-1,$ and $\left(b,f\left({x}_{n-1}\right)\right).$ That is, $p$ is constant on the outer subintervals $\left[a,{x}_{1}\right]$ and $\left[{x}_{n-1},b\right]$ determined by the partition, and its graph between ${x}_{1}$ and ${x}_{n-1}$ is the polygonal line joining the points $\left\{\left({x}_{1},f\left({x}_{1}\right)\right),...,\left({x}_{n-1},f\left({x}_{n-1}\right)\right)\right\}.$ For example, for $2\le i\le n-1,$ the function $p$ has the form

$p\left(x\right)=f\left({x}_{i-1}\right)+\frac{f\left({x}_{i}\right)-f\left({x}_{i-1}\right)}{{x}_{i}-{x}_{i-1}}\left(x-{x}_{i-1}\right)$

on the interval $\left[{x}_{i-1},{x}_{i}\right].$ So, $p\left(x\right)$ lies between the numbers $f\left({x}_{i-1}\right)$ and $f\left({x}_{i}\right)$ for all $i.$ Therefore,

$|f\left(x\right)-p\left(x\right)|\le |f\left(x\right)-f\left({x}_{i}\right)|+|f\left({x}_{i}\right)-l\left(x\right)|\le |f\left(x\right)-f\left({x}_{i}\right)|+|f\left({x}_{i}\right)-f\left({x}_{i-1}\right)|<ϵ.$

Since this inequality holds for all $i,$ part (1) is proved.

Next, for $2\le i\le n-1,$ and for each $x\in \left({x}_{i-1},{x}_{i}\right),$ we have ${p}^{\text{'}}\left(x\right)=\left(f\left({x}_{i}\right)-f\left({x}_{i-1}\right)\right)/\left({x}_{i}-{x}_{i-1}\right),$ which, by the Mean Value Theorem, is equal to ${f}^{\text{'}}\left({y}_{i}\right)$ for some ${y}_{i}\in \left({x}_{i-1},{x}_{i}\right).$ So, for each such $x\in \left({x}_{i-1},{x}_{i}\right),$ we have $|{f}^{\text{'}}\left(x\right)-{p}^{\text{'}}\left(x\right)|=|{f}^{\text{'}}\left(x\right)-{f}^{\text{'}}\left({y}_{i}\right)|,$ and this is less than $ϵ/4\left(b-a\right),$ because $|x-{y}_{i}|<\alpha .$ On the two outer intervals, $p\left(x\right)$ is a constant, so that ${p}^{\text{'}}\left(x\right)=0.$ Hence,

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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