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

Recall that a function $f$ that is continuous on a closed interval $[a,b]$ and continuously differentiable on the open interval $(a,b)$ is called a smooth function on $[a,b].$ And, if there exists a partition $\{t_0<t_1<...<t_n\}$ of $[a,b]$ such that $f$ is smooth on each subinterval $[t_{i-1},t_i],$ then $f$ is called piecewise smooth on $[a,b].$ Although the derivative of a smooth function is only defined and continuous on the open interval $(a,b),$ 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 $[a,b],$ and assume that $|f^{\text{'}}|$ is improperly-integrable on the open interval $(a,b).$ Given an $ϵ>0,$ there exists a piecewise linear function $p$ for which

1.   $\left|f\right(x)-p(x\left)\right|<ϵ$ for all $x\in [a,b].$
2.   ${\int }_a^b|f^{\text{'}}\left(x\right)-p^{\text{'}}\left(x\right)|dx<ϵ.$

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

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

Now, because $f^{\text{'}}$ is uniformly continuous on the compact set $[a+{\delta }^{\text{'}},b-{\delta }^{\text{'}}],$ there exists an $\alpha >0$ such that $|f^{\text{'}}\left(x\right)-f^{\text{'}}\left(y\right)|<ϵ/4(b-a)$ if $x$ and $y$ belong to $[a+{\delta }^{\text{'}},b-{\delta }^{\text{'}}]$ and $|x-y|<\alpha .$ Choose a partition $\{x_0<x_1<...<x_n\}$ of $[a,b]$ such that $x_0=a,x_1=a+{\delta }^{\text{'}},x_{n-1}=b-{\delta }^{\text{'}},x_n=b,$ and $x_i-x_{i-1}<min(\delta ,\alpha )$ for $2\le i\le n-1.$ Define $p$ to be the piecewise linear function on $[a,b]$ whose graph is the polygonal line joining the $n+1$ points $(a,f\left(x_1\right)),$ $\left\{(x_i,f\left(x_i\right))\right\}$ for $1\le i\le n-1,$ and $(b,f\left(x_{n-1}\right)).$ That is, $p$ is constant on the outer subintervals $[a,x_1]$ and $[x_{n-1},b]$ determined by the partition, and its graph between $x_1$ and $x_{n-1}$ is the polygonal line joining the points $\{(x_1,f\left(x_1\right)),...,(x_{n-1},f\left(x_{n-1}\right))\}.$ For example, for $2\le i\le n-1,$ the function $p$ has the form

on the interval $[x_{i-1},x_i].$ 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,

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

Next, for $2\le i\le n-1,$ and for each $x\in (x_{i-1},x_i),$ we have $p^{\text{'}}\left(x\right)=(f\left(x_i\right)-f\left(x_{i-1}\right))/(x_i-x_{i-1}),$ which, by the Mean Value Theorem, is equal to $f^{\text{'}}\left(y_i\right)$ for some $y_i\in (x_{i-1},x_i).$ So, for each such $x\in (x_{i-1},x_i),$ 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 $ϵ/4(b-a),$ 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,