Math 360: Uniform continuity and the integraldeturck/m360/part-5-unicont-integral.pdfIntegrals We’ve gotten pretty far with our \primitive" integral of functions of bounded variation.

$Page 1: Math 360: Uniform continuity and the integraldeturck/m360/part-5-unicont-integral.pdfIntegrals We’ve gotten pretty far with our \primitive" integral of functions of bounded variation.$
Math 360: Uniform continuity and the integral

D. DeTurck

University of Pennsylvania

October 25, 2017

D. DeTurck Math 360 001 2017C: Integral/functions 1 / 28

Integrals

We’ve gotten pretty far with our “primitive” integral of functionsof bounded variation. But not all continuous functions are BV (norare all BV functions continuous), so we should define a more robustintegral so that the set of integrable functions includes both cases.

The setup is pretty much the same as for the integral we alreadyhave: To calculate ˆ b

af (x) dx

we partition the interval [a, b] with points

x0 = a < x1 < x2 < · · · < xn = b

– the new ingredient is how we choose to “sample” the function fin the interval [xi−1, xi ]. When our functions were monotonic, wechose to evaluate f either the left or right endpoint, since thatwould give the maximum or minimum value of f on the subinterval.


Riemann vs Darboux

There are two approaches to sampling:

• Riemann: In the “Riemann integral” you choose a randompoint x∗i in the interval [xi−1, xi ] for each i , and create theRiemann sum

RS(f ,P) =n∑

i=1

f (x∗i )(xi − xi−1)

If the limit of RS(f ,P) as P becomes finer and finer (n→∞and |P| → 0) exists, then that is the Riemann integral.

• Darboux: In the “Darboux integral” you make two sums: Forthe upper Darboux sum you sample [xi−1, xi ] by simply lettingf +i be the supremum of f (x) on the interval [xi−1, xi ], andyou make the lower Darboux sum by letting f −i be theinfimum of f on the subinterval.


Setup for the integral – upper and lower sums

This is going to look familiar (compare with slide set 4).Let f be any bounded function on [a, b], and P a partition of [a, b].

Definition: (upper and lower sums)

The upper sum of f corresponding to the partition P is

U(f ,P) =n∑

i=1

f +i (xi − xi−1)

and the lower sum of f corresponding to the partition P is

L(f ,P) =n∑

i=1

f −(xi − xi−1)

Since f +i = supx∈[xi−1,xi ] f (x) ≥ infx∈[xi−1,xi ] f (x) = f −i for each i ,we have U(f ,P) ≥ L(f ,P)


Setup for the integral – definition

Since there’s always a common refinement P for any partitions P1

and P2, we have

L(f ,P1) ≤ L(f ,P) ≤ U(f ,P) ≤ U(f ,P2)

where P is a common refinement for P1 and P2. So for any pair ofpartitions P1 and P2 we have

L(f ,P1) ≤ U(f ,P2).

Therefore sup L(f ,P) over all partitions P (which is called thelower Darboux integral is less than or equal to inf U(f ,P) (theupper Darboux integral). If these are equal then we say thefunction f is integrable and their common value is called theintegral: ˆ b

af (x) dx


Proving existence – regular partitions

To show that the integral exists, it is sufficient to find, for anyε > 0, a partition P such that

U(f ,P) < L(f ,P) + ε.

We did this before for monotonic functions, we can do this by usingsufficiently fine regular partitions – these are partitions having the

xi ’s evenly spaced (so xi − xi−1 =b − a

nfor all i = 1, . . . , n).

Theorem

If f : [a, b]→ R is monotonic, then

ˆ b

af (x) dx exists.

Proof: Given ε > 0, choose n so large that the regular partition P

with n steps, will have U(f ,P)− L(f ,P) =f (b)− f (a)

n< ε.


Functions of bounded variation

Proposition (Greater generality)

Then we extended the integral to functions of bounded variation:If f (x) can be written as the sum of two monotonic functions p(x)and q(x) with p increasing and q decreasing on [a, b] (such afunction is called a function of bounded variation), thenˆ b

af (x) dx exists and is equal to

ˆ b

ap(x) dx +

ˆ b

aq(x) dx .

Now, the goal is to extend the whole thing to continuous functions.


Basic properties

Because the upper and lower sums have these properties, it followsthat the integral does:

1 Linearity:ˆ b

aαf (x) + βg(x) dx = α

ˆ b

af (x) dx + β

ˆ b

ag(x) dx for

constants α, β and f and g are integrable functions on [a, b].

2 Monotonicity: If f (x) ≥ g(x) for all x ∈ [a, b] then´ ba f (x) dx ≥

´ ba g(x) dx .

3 If f (x) is integrable on [a, b], then so is |f (x)| and∣∣∣∣ˆ b

af (x) dx

∣∣∣∣ ≤ ˆ b

a|f (x)| dx .


More basic properties

4 If a < b < c then

ˆ c

af (x) dx =

ˆ b

af (x) dx +

ˆ c

bf (x) dx .

5 If m ≤ inf{f (x) | x ∈ [a, b]} and M ≥ sup{f (x) | x ∈ [a, b]}then

(b − a)m ≤ˆ b

af (x) dx ≤ (b − a)M.

6 Mean value theorem for integrals: If f is continuous andintegrable (soon we can remove this as an assumption) on[a, b] then there is a c with a < c < b such that

f (c) =1

b − a

ˆ b

af (x) dx .


Fundamental Theorem – integrals of derivatives

Fundamental theorem of calculus I: Integrals of derivatives

Let F (x) be a differentiable function on [a, b] with derivativeF ′(x), and suppose F ′ is an integrable function on [a, b]. Then

ˆ b

aF ′(x) dx = F (b)− F (a).

For any interval (xi−1, xi ) in the partition P we have

F ′−i ≤F (xi )− F (xi−1)

xi − xi−1≤ F ′+i

by the mean value theorem (for derivatives), since the middlequantity is F ′(x) for some x in the interval.


Fundamental Theorem – proof conclusion

But then

L(F ′,P) =n∑

i=1

F ′−i (xi − xi−1)

≤n∑

i=1

F (xi )− F (xi−1)

≤n∑

i=1

F ′+i (xi − xi−1) = U(F ′,P).

But the middle sum telescopes to F (xn)− F (x0) = F (b)− F (a).

Now F (b)− F (a) is trapped between supP

L(F ′,P) and

infP

U(F ′,P), both of which are equal to the integral

ˆ b

aF ′(x) dx

because F ′ is integrable.


Second fundamental theorem – derivatives of integrals

Fundamental theorem of calculus II: Derivatives of integrals

Let f (x) be a continuous (and integrable) function on [a, b].Define the function F (x) via

F (x) =

ˆ x

af (t) dt.

Then F is differentiable and F ′(x) = f (x).

First, if h > 0 we have

F (x + h)− F (x) =

ˆ x+h

xf (t) dt = hf (c)

for some c between x and x + h by properties 4 and 6 (mean valuetheorem for integrals).


Second fundamental theorem – proof conclusion

Therefore the difference quotient

F (x + h)− F (x)

h= f (c) where x < c < x + h

But since f is continuous, f (c)→ f (x) as h→ 0+.

For h < 0 we use that F (x + h)− F (x) = −ˆ x

x+hf (t) dt = hf (c)

for some c between x + h and x , and the proof goes through as forthe h > 0 case.


Uniform continuity

To show that continuous functions on closed intervals areintegrable, we’re going to define a slightly stronger form ofcontinuity:

Definition (uniform continuity):

A function f (x) is uniformly continuous on the domain D if forevery ε > 0 there is a δ > 0 that depends only on ε and not onx ∈ D such that for every x , y ∈ D with |x − y | < δ, it is the casethat |f (x)− f (y)| < ε.

A uniformly continuous function is necessarily continuous, but onnon-compact sets (i.e., sets that are not closed and bounded) acontinuous function need not be uniformly so.


Not uniformly continuous

To help understand the import of uniform continuity, we’ll reversethe definition:

Definition (not uniformly continuous):

A function f (x) is not uniformly continuous on D if there is someε > 0 such that for every δ > 0, no matter how small, it is possibleto find x , y ∈ D with |x − y | < δ but |f (x)− f (y)| > ε.

For instance f (x) = x2 is not u.c. on the set [0,∞) because wecan choose ε = 1 and then for any δ > 0, we have(x + δ)2 − x2 = 2xδ + δ2 and we can choose x > 1/(2δ) so that2xδ > 1. So there is no δ that works for every x in the infiniteinterval.

Likewise, g(x) = 1/x is not u.c. on (0, 1] because, again usingε = 1, for any δ > 0 we’ll pick an integer n > 0 so that 1/n < δThen let x = 1/n and y = 1/(n + 1) and |x − y | < δ but|1/x − 1/y | = 1.


Integrating continuous functions

Our goal is:

Theorem

If f (x) is a continuous function on the closed, bounded interval[a, b], then f is integrable on [a, b].

We’ll accomplish this in two jumps:

Lemma 1

If f (x) is a uniformly continuous function on the closed, boundedinterval [a, b], then f is integrable on [a, b].

Lemma 2

If f (x) is continuous on the closed, bounded interval [a, b] then fis uniformly continuous on [a, b].

It’s easy to see how the theorem follows from the lemmas.


Lemma 1

Lemma 1

If f (x) is a uniformly continuous function on the closed, boundedinterval [a, b], then f is integrable on [a, b].

Proof: Given ε > 0, let δ > 0 be such that|f (x)− f (y)| < ε/(b − a) whenever x , y ∈ [a, b] and |x − y | < δ.Now choose n so that (b − a)/n < δ, and let P be the regularpartition of [a, b] into n subintervals. We’ll have xi − xi−1 < δ forall i , so f +i − f −i < ε/(b − a) for all i . Then

U(f ,P)− L(f ,P) =n∑

i=1

(f +i − f −i )b − a

n<

n∑i=1

ε

b − a

b − a

n= ε.

so U(f ,P) < L(f ,P) + ε, proving the integrability of f .


Lemma 2

Lemma 2

If f (x) is continuous on the closed, bounded interval [a, b] then fis uniformly continuous on [a, b].

We (almost!) did the proof of this in class usingBolzano-Weierstrass, so here’s the whole thing.

Proof by contrapositive: Assume that f is not uniformlycontinuous on [a, b], and we’ll find a point of discontinuity of f .To do it, recall that since f is not u.c., there is an ε > 0 such thatfor any δ > 0, no matter how small, there is a pair of pointsp, q ∈ [a, b] with |p − q| < δ but |f (p)− f (q)| > ε.


Proof of lemma 2, continued

Construct two sequences {pn} and {qn} where all of the points ofeach sequence are in [a, b], and such that |pn− qn| < 1/n (so we’remaking δ smaller and smaller) but |f (p)− f (q)| > ε. Since {pn} iscontained in [a, b] it is bounded and by Bolzano Weierstrass it hasa convergent subsequence {pni}, with limit L ∈ [a, b].

Because |pn − qn| < 1/n, the subsequence {qni} must converge toL as well, since

|qni − L| < |qni − pni |+ |pni − L|

and the first of these terms can be made arbitrarily small since it isless than 1/ni and the second can be made small because pni → L.


Proof of lemma 2, concluded

Now we’ll show that f is not continuous at L. If it were, then forany E > 0 there would be a D > 0 such that if x ∈ [a, b] with|x − L| < D, then |f (x)− f (L)| < E . Let E = ε/2, so we have thecorresponding D > 0. We know that for i sufficiently large we willhave

|pni − L| < D and |qni − L| < D

because both (sub)sequences converge to L. But then

|f (pni )− f (qni )| ≤ |f (pni )− f (L)|+ |f (qni )− f (L)| < 2E < ε

contradicting the fact that |f (pn)− f (qn)| > ε for all n.


Integration by substitution

We still have the substitution rule:

Integration by subsitution

Suppose f : [a, b]→ R and g : [c , d ]→ R are integrable, and thatthe derivative of g exists and is bounded and continuous on (c , d).Further, assume that the image of g : (c , d)→ R is contained inthe interval (a, b). Then

ˆ d

cf (g(x))g ′(x) dx =

ˆ g(d)

g(c)f (x) dx .

Proof: Apply FTC to the function

H(x) =

ˆ x

cf (g(t)) dt −

ˆ g(x)

g(c)f (t) dt.


Integration by parts

Integration by parts

Suppose f and g are differentiable functions on [a, b], with f ′ andg ′ continuous as well. Then

ˆ b

af (x)g ′(x) dx = f (x)g(x)

∣∣∣∣bx=a

−ˆ b

af ′(x)g(x) dx .

To prove this, consider the function

H(x) =

ˆ x

af (t)g ′(t) dt − f (t)g(t)

∣∣∣∣xt=a

+

ˆ x

af ′(t)g(t) dt.

Using the fundamental theorem and the product rule forderivatives, we have

H ′(x) = f (x)g ′(x)−(f (x)g ′(x) + f ′(x)g(x)

)+ f ′(x)g(x) = 0

so H(x) is constant. But H(a) = 0, so the constant is zero. Thefact that H(b) = 0 is the formula in the box.


Trapezoidal rule

You probably recall the trapezoidal rule for estimating integralsfrom elementary calculus:

Trapezoidal rule

If f is a twice-differentiable function on [a, b], then

ˆ b

af (x) dx ≈ h

2(f0 + 2f1 + 2f2 + · · ·+ 2fn−1 + fn) ,

where fi = f (xi ), and xi = a + hi for i = 0, . . . , n andh = (b − a)/n

We’ll derive the trapezoidal rule and then find an estimate for theerror (difference between the approximation on the right side andthe actual integral on the left).


Derivation of the trapezoidal rule

To derive the trapezoidal rule, we’ll look at a single “panel” fromxi−1 to xi . Via the change of variables x = xi−1 + t we can writeˆ xi

xi−1

f (x) dx =

ˆ h

0f (xi−1 + t) dt.

Now letf (xi−1 + t) = c0 + c1t + c2t

2 + · · ·be the beginning of the Taylor expansion of f around the pointt = 0 (which is x = xi−1). Then f (xi−1) = c0,f (xi ) = c0 + c1h + c2h

2 + · · · , andˆ h

0f (xi−1 + t) dt = c0h + c1

h2

2+ · · · .

Therefore,

Bothh

2(f (xi−1) + f (xi )) and

ˆ h

0f (xi−1 + t) dt = c0h+ c1

h2

2+ · · ·


Error in the trapezoidal rule

So we have

ˆ h

0f (xi−1 + t) dt − h

2(f (xi−1) + f (xi )) = · · ·

where · · · hides terms that have powers of h beginning withh3 andderivatives of f beginning with f ′′. So we would expect that themagnitude of this difference can be expressed in terms of these.

Also, when the trapezoidal rule is implemented withe many (n)“panels” or steps, the right endpoint of each panel is the leftendpoint of the next. That’s where all the 2’s come from in therule. In the worst case, the errors in each panel will reinforce eachother rather than cancel, so the error will be n times somethingthat involves h3 and f ′′.


Single-step error

We use integration by parts to estimate the error incurred in asingle panel (u = f , dv = dt, say v = t + A for a tbd constant A):

ˆ h

0f (xi−1+t) dt = (t + A)f (xi−1 + t)

∣∣∣∣ht=0

−ˆ h

0(t+A)f ′(xi−1+t) dt

First concentrate on

(t + A)f (xi−1 + t)

∣∣∣∣ht=0

= (h + A)f (xi )− Af (xi−1)

We choose A = −h/2 so that this is the trapezoidal formula. Sowe have

ˆ h

0f (xi−1+t) dt−h

2(f (xi−1)+f (xi )) = −

ˆ h

0

(t − h

2

)f ′(xi−1+t) dt


Single-step error

So far, our expression for the single-step error is

ET = −ˆ h

0

(t − h

2

)f ′(xi−1 + t) dt, so we integrate by parts

again:

ET = −(t2

2− h

2t + B

)f ′(xi−1 + t)

∣∣∣∣h0

+

ˆ h

0

(t2

2− h

2t + B

)f ′′(xi−1 + t) dt

We want to choose B so that the term on the first line is zero.That term, expanded, is B(f ′(xi−1)− f ′(xi )) so we should chooseB = 0. And so now:

ET =

ˆ h

0

1

2(t(t − h))f ′′(xi−1 + t) dt.


Single-step error

So far, ET =

ˆ h

0

1

2(t(t − h))f ′′(xi−1 + t) dt.. If we were to try and

integrate by parts again, we would find that we cannot choose theintegration constant for dv (i.e., the “C”) to make the first partzero again. So this expression is the best we can do.

Turn this into an estimate by letting M = maxx∈[xi−1,xi ]

|f ′′(x)|, so

|ET | ≤ˆ h

0|12

(t(t−h))||f ′′(xi−1+t)| dt ≤ M

2

ˆ h

0t(t−h) dt =

Mh3

12

This is the standard formula for the single-step error in thetrapezoidal rule. Multiplying this by n for the global estimate, andrecalling that h = (b − a)/n gives the standard global error:

ET (f , [a, b]) ≤ (b − a)3M

12n2where M = max

x∈[a,b]|f ′′(x)|.


Math 360: Uniform continuity and the integraldeturck/m360/part-5-unicont-integral.pdfIntegrals We’ve gotten pretty far with our \primitive" integral of functions of bounded variation.

Documents