5.8: Rectifiable Arcs. Absolute Continuity

5.8: Rectifiable Arcs. Absolute Continuity

We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

If a function (f : E^{1} ightarrow E) is of bounded variation (§7) on an interval (I=[a, b],) we can define a real function (v_{f}) on (I) by

[v_{f}(x)=V_{f}[a, x](= ext { total variation of } f ext { on }[a, x]) ext { for } x in I;]

(v_{f}) is called the total variation function, or length function, generated by (f) on (I). Note that (v_{f} uparrow) on (I.) (Why?) We now consider the case where (f) is also relatively continuous on (I,) so that the set (A=f[I]) is a rectifiable arc (see §7, Note 1 and Definition 2).

Definition 1

A function (f : E^{1} ightarrow E) is (weakly) absolutely continuous on (I=[a, b]) iff (V_{f}[I]<+infty) and (f) is relatively continuous on (I).

Theorem (PageIndex{1})

The following are equivalent:

(i) (f) is (weakly) absolutely continuous on (I=[a, b]);

(ii) (v_{f}) is finite and relatively continuous on (I ;) and

(iii) ((forall varepsilon>0) ext{ } (exists delta>0) ext{ } (forall x, y in I | 0 leq y-x


We shall show that (ii) (Rightarrow) (iii) (Rightarrow) (i) (Rightarrow) (ii).

(ii) (Rightarrow) (iii). As (I=[a, b]) is compact, (ii) implies that (v_{f}) is uniformly continuous on (I) (Theorem 4 of Chapter 4, §8). Thus

[(forall varepsilon>0) ext{ } (exists delta>0) ext{ } (forall x, y in I | 0 leq y-x


[v_{f}(y)-v_{f}(x)=V_{f}[a, y]-V_{f}[a, x]=V_{f}[x, y]]

by additivity (Theorem 1 in §7). Thus (iii) follows.

(iii) (Rightarrow) (i). By Corollary 3 of §7, (|f(x)-f(y)| leq V_{f}[x, y].) Therefore, (iii) implies that

[(forall varepsilon>0) ext{ } (exists delta>0) ext{ } (forall x, y in I| | x-y |

and so (f) is relatively (even uniformly) continuous on (I).

Now with (varepsilon) and (delta) as in (iii), take a partition (P=left{t_{0}, ldots, t_{m} ight}) of (I) so fine that


Then ((forall i) V_{f}left[t_{i-1}, t_{i} ight]

[V_{f}[I]=sum_{i=1}^{m} V_{f}left[t_{i-1}, t_{i} ight]

Thus (i) follows, by definition.

That (i) (Rightarrow) (ii) is given as the next theorem. (quad square)

Theorem (PageIndex{2})

If (V_{f}[I]<+infty) and if (f) is relatively continuous at some (p in I) (over (I=[a, b]),) then the same applies to the length function (v_{f}).


We consider left continuity first, with (a

Let (varepsilon>0.) By assumption, there is (delta>0) such that


Fix any such (x.) Also, (V_{f}[a, p]=sup _{P} S(f, P)) over ([a, p].) Thus

[V_{f}[a, p]-frac{varepsilon}{2}

for some partition

[P=left{t_{0}=a, ldots, t_{k-1}, t_{k}=p ight} ext { of }[a, p]. ext { (Why?)}]

We may assume (t_{k-1}=x, x) as above. (If (t_{k-1} eq x,) add (x) to (P. )) Then

[left|Delta_{k} f ight|=|f(p)-f(x)|

and hence

[V_{f}[a, p]-frac{varepsilon}{2}


[V_{f}[a, p]=v_{f}(p)]


[V_{f}left[a, t_{k-1} ight]=V_{f}[a, x]=v_{f}(x).]

Thus (1) yields

[left|v_{f}(p)-v_{f}(x) ight|=V_{f}[a, p]-V_{f}[a, x]

This shows that (v_{f}) is left continuous at (p).

Right continuity is proved similarly on noting that

[v_{f}(x)-v_{f}(p)=V_{f}[p, b]-V_{f}[x, b] ext { for } p leq x

Thus (v_{f}) is, indeed, relatively continuous at (p.) Observe that (v_{f}) is also of bounded variation on (I,) being monotone and finite (see Theorem 3(ii) of §7).

This completes the proof of both Theorem 2 and Theorem 1. (quad square)

We also have the following.

Corollary (PageIndex{1})

If (f) is real and absolutely continuous on (I=[a, b]) (weakly), so are the nondecreasing functions (g) and (h(f=g-h)) defined in Theorem 3 of §7.

Indeed, the function (g) as defined there is simply (v_{f}.) Thus it is relatively continuous and finite on (I) by Theorem 1. Hence so also is (h=f-g.) Both are of bounded variation (monotone!) and hence absolutely continuous (weakly).

Note 1. The proof of Theorem 1 shows that (weak) absolute continuity implies uniform continuity. The converse fails, however (see Problem 1(iv) in §7).

We now apply our theory to antiderivatives (integrals).

Corollary (PageIndex{2})

If (F=int f) on (I=[a, b]) and if (f) is bounded (left(|f| leq K in E^{1} ight)) on (I-Q) ((Q) countable), then (F) is weakly absolutely continuous on (I.)

(Actually, even the stronger variety of absolute continuity follows. See Chapter 7, §11, Problem 17).


By definition, (F=int f) is finite and relatively continuous on (I,) so we only have to show that (V_{F}[I]<+infty.) This, however, easily follows by Problem 3 of §7 on noting that (F^{prime}=f) on (I-S) ((S) countable). Details are left to the reader. (quad square)

Our next theorem expresses arc length in the form of an integral.

Theorem (PageIndex{3})

If (f : E^{1} ightarrow E) is continuously differentiable on (I=[a, b]) (§6), then (v_{f}=intleft|f^{prime} ight|) on (I) and

[V_{f}[a, b]=int_{a}^{b}left|f^{prime} ight|.]


Let (a

[Delta v_{f}=v_{f}(x)-v_{f}(p)=V_{f}[p, x] . quad ext {(Why?)}]

As a first step, we shall show that

[frac{Delta v_{f}}{Delta x} leq sup _{[p, x]}left|f^{prime} ight|.]

For any partition (P=left{p=t_{0}, ldots, t_{m}=x ight}) of ([p, x],) we have

[S(f, P)=sum_{i=1}^{m}left|Delta_{i} f ight| leq sum_{i=1}^{m} sup _{left[t_{i-1}, t_{i} ight]}left|f^{prime} ight|left(t_{i}-t_{i-1} ight) leq sup _{[p, x]}left|f^{prime} ight| Delta x.]

Since this holds for any partition (P,) we have

[V_{f}[p, x] leq sup _{[p, x]}left|f^{prime} ight| Delta x,]

which implies (2).

On the other hand,

[Delta v_{f}=V_{f}[p, x] geq|f(x)-f(p)|=|Delta f|.]

Combining, we get

[left|frac{Delta f}{Delta x} ight| leq frac{Delta v_{f}}{Delta x} leq sup _{[p, x]}left|f^{prime} ight|<+infty]

since (f^{prime}) is relatively continuous on ([a, b],) hence also uniformly continuous and bounded. (Here we assumed (a


[| | f^{prime}(p)|-| f^{prime}(x)| | leqleft|f^{prime}(p)-f^{prime}(x) ight| ightarrow 0 quad ext { as } x ightarrow p,]

so, taking limits as (x ightarrow p,) we obtain

[lim _{x ightarrow p} frac{Delta v_{f}}{Delta x}=left|f^{prime}(p) ight|.]

Thus (v_{f}) is differentiable at each (p) in ((a, b),) with (v_{f}^{prime}(p)=left|f^{prime}(p) ight|.) Also, (v_{f}) is relatively continuous and finite on ([a, b]) (by Theorem 1). Hence (v_{f}=intleft|f^{prime} ight|) on ([a, b],) and we obtain

[int_{a}^{b}left|f^{prime} ight|=v_{f}(b)-v_{f}(a)=V_{f}[a, b], ext { as asserted.} quad square]

Note 2. If the range space (E) is (E^{n}) (*or (C^{n})), (f) has (n) components

[f_{1}, f_{2}, ldots, f_{n}.]

By Theorem 5 in §1, (f^{prime}=left(f_{1}^{prime}, f_{2}^{prime}, ldots, f_{n}^{prime} ight),) so

[left|f^{prime} ight|=sqrt{sum_{k=1}^{n}left|f_{k}^{prime} ight|^{2}},]

and we get

[V_{f}[a, b]=int_{a}^{b} sqrt{sum_{k=1}^{n}left|f_{k}^{prime} ight|^{2}}=int_{a}^{b} sqrt{sum_{k=1}^{n}left|f_{k}^{prime}(t) ight|^{2}} d t quad ext {(classical notation).}]

In particular, for complex functions, we have (see Chapter 4, §3, Note 5)

[V_{f}[a, b]=int_{a}^{b} sqrt{f_{mathrm{re}}^{prime}(t)^{2}+f_{mathrm{im}}^{prime}(t)^{2}} d t.]

In practice, formula (5) is used when a curve is given parametrically by

[x_{k}=f_{k}(t), quad k=1,2, ldots, ext{ }n,]

with the (f_{k}) differentiable on ([a, b].) Curves in (E^{2}) are often given in nonparametric form as

[y=F(x), quad F : E^{1} ightarrow E^{1}.]

Here (F[I]) is (n o t) the desired curve but simply a set in (E^{1}.) To apply (5) here, we first replace "(y=F(x))" by suitable parametric equations,

[x=f_{1}(t) ext { and } y=f_{2}(t);]

i.e., we introduce a function (f : E^{1} ightarrow E,) with (f=left(f_{1}, f_{2} ight).) An obvious (but not the only) way of achieving it is to set

[x=f_{1}(t)=t ext { and } y=f_{2}(t)=F(t)]

so that (f_{1}^{prime}=1) and (f_{2}^{prime}=F^{prime}.) Then formula (5) may be written as

[V_{f}[a, b]=int_{a}^{b} sqrt{1+F^{prime}(x)^{2}} d x, quad f(x)=(x, F(x)).]


Find the length of the circle


Here it is convenient to use the parametric equations

[x=r cos t ext { and } y=r sin t,]

i.e., to define (f : E^{1} ightarrow E^{2}) by

[f(t)=(r cos t, r sin t),]

or, in complex notation,

[f(t)=r e^{t i}.]

Then the circle is obtained by letting (t) vary through ([0,2 pi].) Thus (5) yields

[V_{f}[0,2 pi]=int_{a}^{b} r sqrt{cos ^{2} t+sin ^{2} t} d t=r int_{a}^{b} 1 d t=rleft.t ight|_{0} ^{2 pi}=2 r pi.]

Note that (f) describes the same circle (A=f[I]) over (I=[0,4 pi].) More generally, we could let (t) vary through any interval ([a, b]) with (b-a geq 2 pi.) However, the length, (V_{f}[a, b],) would change (depending on (b-a)). This is because the circle (A=f[I]) is not a simple arc (see §7, Note 1), so (ell A) depends on (f) and (I,) and one must be careful in selecting both appropriately.

Watch the video: Τόξο κατασκευή εύκολο (July 2022).


  1. Kurt

    The excellent and duly answer.

  2. Salmaran

    I believe you have been misled.

  3. Gail

    I'm sorry, but I think you are making a mistake. Let's discuss. Email me at PM, we'll talk.

  4. Tojagami


Write a message