Homework 7 Solutions - Math 321, Spring 2015
1. Recall Jordan’s theorem: a function f : [a, b] → R is of bounded variation if and only if f
can be written as the difference of two non-decreasing functions g and h.
(a) Show that the decomposition f = g − h is by no means unique, and that there are
uncountably many ways of writing f in this form.
(b) The following decomposition of f is often useful. Define the positive and negative variations of f by
1
1
v(x) + f (x) − f (a) ,
n(x) = v(x) − f (x) + f (a) ,
2
2
x
where v(x) = Va f is the variation function defined in class. Show that p and n are
nondecreasing functions on [a, b] and use this to give an alternative representation of f
as the difference of nondecreasing functions.
p(x) =
(c) The relevance of p and n is that it injects a certain amount of uniqueness into the
Jordan decomposition of f , in the following sense. If g and h are any two non-decreasing
functions on [a, b] such that f = g − h, then
Vxy p ≤ Vxy g
and
Vxy n ≤ Vxy h for all x < y in [a, b].
Prove this.
Solution. If f = g−h is a decomposition of the appropriate type, then so is f = (g+c)−(h+c),
where c is any constant. Thus the decomposition, without further specifications, is highly
non-unique.
Given any x ∈ [a, b], and a partition P = {a = x0 < x1 < x2 < · · · < xn = x} of [a, x], let
us define
V+ (f, P ) =
n
X
max f (xi ) − f (xi−1 ), 0
and
V− (f, P ) = −
i=1
n
X
min f (xi ) − f (xi−1 ), 0 .
i=1
It is clear that V± (f, P ) ≥ V± (f, P 0 ) whenever P is a refinement of P 0 , and that
f (x) − f (a) = V+ (f, P ) − V− (f, P )
V (f, P ) = V+ (f, P ) + V− (f, P ),
for any partition P . Setting
v+ (x) = sup V (f, P ),
v− (x) = sup V− (f, P ),
P
P
and choosing increasingly finer partitions P of [a, x], we find that
v+ (x) + v− (x) = v(x)
v+ (x) − v− (x) = f (x) − f (a).
and
Hence p(x) = v+ (x) and n(x) = v− (x). Since v± are both non-decreasing functions of x
by definition, the result in part (b) follows.
Suppose now that g, h are nondecreasing functions such that f = g − h. Then for any
x < y in [a, b],
1
2
Vxy p = v+ (y) − v+ (x) = sup
Q
= sup
Q
≤ sup
Q
=
n
X
n
X
max f (qi ) − f (qi−1 ), 0
i=1
max g(qi ) − g(qi−1 ) − h(qi ) − h(qi−1 ) , 0
i=1
n
X
n
X
max g(qi ) − g(qi−1 ), 0 = sup
g(qi ) − g(qi−1 )
Q
i=1
i=1
Vxy g.
Here Q = {x = q0 < q1 < · · · < qn = y} denotes an arbitrary partition of [x, y]. The
inequality for Vxy n and Vxy h is similarly obtained.
2. We stated the “integration by parts” formula in class, using it to highlight the interchangability of integrand and integrator. The purpose of this problem is to fill in the details of its
proof. Throughout this problem f and α denote arbitrary real-valued functions on [a, b].
(a) Given any partition P = {a = x0 < x1 < x2 < · · · < xn = b} and a collection of points
T = {t1 , · · · , tn } with tj ∈ [xj−1 , xj ], prove the following identity:
Sf (α, P, T ) = f (b)α(b) − f (a)α(a) − Sα (f, P 0 , T 0 ).
Here P 0 = {a = t0 , t1 , · · · , tn , tn+1 = b} and T 0 = P .
(b) Use part (a) to show that f ∈ Rα [a, b] if and only if α ∈ Rf [a, b]. Show that in either
case,
Z
b
Z
b
α df = f (b)α(b) − f (a)α(a).
f dα +
a
a
Note that this is one of those rare instances where one implication implies the other!
Solution. For any partition P = {a = x0 < x1 < · · · < xn = b} and any choice of points
T = {t1 , · · · , tn } with ti ∈ Ii = [xi − 1, xi ], we can arrange the sum representing Sf (α, P, T )
as follows:
Sf (α, P, T ) =
=
=
n
X
i=1
n
X
α(ti ) [f (xi ) − f (xi−1 )]
f (xi )α(ti ) −
n
X
i=1
i=1
n
X
n−1
X
f (xi )α(ti ) −
i=1
=−
n
X
α(ti )f (xi−1 )
f (xi )α(ti+1 )
i=0
f (xi ) [α(ti+1 ) − α(ti )] − f (x0 )α(t0 ) + f (xn )α(tn+1 )
i=0
= f (b)α(b) − f (a)α(a) − Sα (f, P 0 , T 0 ).
3
This completes the proof of part (a).
For part (b), let us assume that f ∈ Rα [a, b] and fix > 0. We choose a parititon P ∗ so
that
Sα (f, P, T ) −
Z
b
f dα < for all P ⊃ P ∗ and any choice of T.
a
Since P 0 as described above need not be a refinement of P , we consider the natural
modification P 00 = {x0 , t1 , x1 , t2 , · · · } ⊃ P , with the accompanying choice of points being
T 00 = {x0 , x0 , x1 , x1 , · · · , xn , xn }, and observe that
0
0
Sα (f, P , T ) =
n
X
f (xi ) α(ti+1 ) − α(ti )
i=0
=
n
X
n
X
f (xi ) α(ti+1 ) − α(xi ) +
f (xi ) α(xi ) − α(ti )
i=0
i=0
00
00
= Sα (f, P , T ).
Now for any P ⊃ P ∗ , we appeal to part (a) of the problem, obtaining
Z b
i Z b
h
f dα − Sα (f, P 00 , T 00 ) < .
f dα = Sf (α, P, T ) − f (b)α(b) − f (a)α(a) −
a
a
Since is arbitrary, this shows that α ∈ Rf [a, b] and
Z b
Z b
α df = f (b)α(b) − f (a)α(a) −
f dα.
a
a
∗
Note: It is not enough just to find a partition P satisfying
Z b
h
i
∗
f dα < Sf (α, P , T ) − f (b)α(b) − f (a)α(a) −
a
We also need its refinements to satisfy this.
3. Let α ∈ BV[a, b] and let β(x) = Vax α. Recall that both β and β − α are increasing. Show
that Rα [a, b] = Rβ [a, b] ∩ Rβ−α [a, b]. This identity was instrumental to our conclusion that
Rα [a, b] is a vector space, an algebra and a lattice. (Hint: Argue that it suffices to only
show that Rα [a, b] ⊆ Rβ [a, b]. )
Solution. Let β(x) = Vax α. First we show that Rα [a, b] ⊆ Rβ [a, b]. Let f ∈ Rα [a, b], > 0
then ∃I, P ∗ , ∀P ⊃ P ∗ and choice of T corresponding to P, we have
|Sα (f, P, T ) − I| < Pick P = {a = x0 < · · · < xn = b} ⊃ P ∗ such that |V (α, P ) − Vab α| < . Let,
Mi =
sup
x∈[xi−1 ,xi ]
f (x), mi =
inf
x∈[xi−1 ,xi ]
f (x), ∆αi = α(xi ) − α(xi−1 )
Now pick two sets of sample points S = {s1 , · · · , sn }, T = {t1 , · · · , tn } where si , ti ∈ [xi−1 , xi ]
such that if ∆αi ≥ 0,
Mi − < f (si ) ≤ Mi , mi ≤ f (ti ) < mi +
2
2
4
and if ∆αi < 0,
Mi −
< f (ti ) ≤ Mi , mi ≤ f (si ) < mi +
2
2
This implies
(Mi − mi )|∆αi | ≥ (f (si ) − f (ti ))∆αi
Pn
Using
i=1
|∆αi | ≤ Vab α then
n
n
n
X
X
X
(Mi − mi )|∆αi | ≥
(f (si ) − f (ti ))∆αi >
(Mi − mi )|∆αi | − Vab α
i=1
i=1
Hence
|
n
X
i=1
n
X
(Mi − mi )|∆αi | −
(f (si ) − f (ti ))∆αi | < Vab f
i=1
i=1
Now by our choice of P, |Sα (f, P, S) − I| < , |Sα (f, P, T ) − I| < . Hence
|
n
X
(f (si ) − f (ti ))∆αi | = |Sα (f, P, S) − Sα (f, P, T )| < 2
i=1
Hence
X
X
X
n
n
X
n
n
≤
+
(M
−
m
)|∆α
|
(M
−
m
)|∆α
|
−
(f
(s
)
−
f
(t
))∆α
(f
(s
)
−
f
(t
))∆α
i
i
i
i
i
i
i
i
i
i
i
i
i=1
<
i=1
Vab α
i=1
i=1
+ 2
Now we are ready to show that f ∈ Rβ [a, b]. Since f is BV it is bounded by some M and so
|Mi − mi | ≤ 2M . Also
i
∆βi = Vxxi−1
α ≥ |α(xi ) − α(xi−1 )| = |∆αi |
Hence
n
X
|Uβ (f, P ) − Lβ (f, P )| = |
(Mi − mi )∆βi |
i=1
n
N
X
X
=
(Mi − mi )(∆βi − |∆αi |) + (Mi − mi )|∆αi |
<
i=1
n
X
i=1
|Mi − mi |(∆βi − |∆αi |) + (Vab α + 2)
i=1
≤ 2M
=
n
X
(∆βi − |∆αi |) + (Vab α + 2)
i=1
2M (Vab α
− V (α, P )) + (Vab α + 2)
< 2M + (Vab α + 2)
So f ∈ Rβ [a.b].
Finally, using Question 2, f ∈ Rα [a, b] ⇒ f ∈ Rβ [a, b] ⇒ α, β ∈ Rf [a, b] ⇒ β − α ∈ Rf [a, b]
⇒ f ∈ Rβ−α [a, b] ⇒ f ∈ Rβ−α [a, b] ∩ Rβ [a, b].
f ∈ Rβ−α [a, b] ∩ Rβ [a, b] ⇒ β, α − β ∈ Rf [a, b] ⇒ α ∈ Rf [a, b]. ⇒ f ∈ Rα [a, b].
5
4. Let {fn } be a bounded sequence in BV[a, b], i.e., suppose that ||fn ||BV ≤ K for all n.
Show that fn admits a pointwise convergent subsequence whose limit f lies in BV[a, b] with
||f ||BV ≤ K. This is known as Helly’s first theorem. (Hint: First try out the case when all
the functions fn are non-decreasing, then adapt it for functions of bounded variation.)
Proof. Assume first that fn is non-decreasing. Since kfn kBV ≤ K, we have
|fn (x)| ≤ |fn (x) − fn (a)| + |fn (a)| ≤ Vab fn + |fn (a)| = kfn kBV ≤ K
So f is bounded by K. Hence for each x ∈ [a, b], fn (x) has a convergent subsequence.
Since Q is countable. By cantor’s diagonal argument we can find a subsequence gk = fnk
that converges pointwise on [a, b]. Define g : [a, b] → R.
(
limk→∞ gk (x)
if x ∈ Q
g(x) =
supq<x,q∈Q∩[a,b] g(q) if x ∈
/ Q.
Since gk is nondecreasing, g is nondecreasing on Q and then by the definition, g is nondecreasing on [a.b].
Claim: gk actually converges to g at every point that g is continuous.
Proof of claim: Suppose g is continuous at x. Choose δ > 0 such that |x − y| < δ =⇒
|g(x) − g(y)| < . Since Q is dense we can find p, q ∈ Q ∩ [a, b] such that p, q ∈ [x − 2δ , x + 2δ ]
and since gk converges at p, q then if k is sufficiently large,
|gk (p) − gk (q)| ≤ |gk (p) − g(p)| + |g(p) − g(q)| + |gk (q) − g(q)| ≤ 3
Since gk is nondecreasing, gk (p) ≤ gk (x) ≤ gk (q) then
|gk (x) − g(x)| ≤ |gk (x) − gk (p)| + |gk (p) − g(p)| + |g(p) − g(x)|
≤ |gk (q) − g(q)| + |gk (p) − g(p)| + |g(p) − g(x)|
< 3 + + So we have the claim. Now g is non-decreasing, it has at most a countable number of
discontinuities. As gk is uniformly bounded, we can again use the diagonal argument to
extract another subsequence of gk the converges at every point on [a, b]. We are done in case
that gn is nondecreasing.
Now for any fn ∈ BV [a, b], kfn kBV ≤ K we can decompose fn = gn − hn where gn (x) =
Vax fn , hn = gn − fn . we know that both are nondecreasing. We can see
|gn (x)| ≤ K, |hn (x)| ≤ |fn (x)| + |gn (x)| ≤ 2K
By the previous case, gn has a subsequence gnk that converges at every point in [a, b]. Apply
the argument again to hnk then there is a subsequence, call it gnk , hnk that both of them
converge everywhere in [a, b]. i.e. fnk converges pointwise to some function f on [a, b]. We
will show that kf kBV ≤ K. Fix a partition P ,
V (fnk , P ) + |fnk (a)| ≤ Vab fnk + |fnk (a)| ≤ K
Since there is only finitely many terms in V (fnk , P ), let k → ∞ then
V (f, P ) + |fnk (a)| ≤ K
Take supremum over P we have kf kBV ≤ K as required.