<< Chapter < Page | Chapter >> Page > |
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 $\u03f5>0,$ there exists a piecewise linear function $p$ for which
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 $\left|f\right(x)-f(y\left)\right|<\u03f5/2.$
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 ${\int}_{a}^{a+{\delta}^{\text{'}}}|{f}^{\text{'}}|+{\int}_{b-{\delta}^{\text{'}}}^{b}\left|{f}^{\text{'}}\right|<\u03f5/2,$ 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 $|{f}^{\text{'}}\left(x\right)-{f}^{\text{'}}\left(y\right)|<\u03f5/4(b-a)$ 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 $\u03f5/4(b-a),$ 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,
Notification Switch
Would you like to follow the 'Analysis of functions of a single variable' conversation and receive update notifications?