Homework 8 Solutions
1. Helly’s first and second theorems were critical components of our proof of the Riesz representation theorem for continuous linear functionals on C[a, b]. You have already proved Helly’s
first theorem in a previous assignment. Prove the second one here.
Suppose that {αn } is a sequence in BV[a, b]. If αn → α pointwise on [a, b], and if Vab αn ≤ K
for all n, then α ∈ BV[a, b] and
Z
b
Z
f dαn →
b
f dα
for all f ∈ C[a, b].
a
a
Solution. Since αn → α pointwise, it follows that V (α, P ) = limn→∞ V (αn , P ) ≤ K for any
partition P . This implies that α ∈ BV[a, b] with Vab α ≤ K.
Now let f ∈ C[a, b] and > 0. Since f is uniformly continuous, there exists δ > 0 such
that
|f (x) − f (y)| <
(1)
3K
for all
|x − y| < δ, x, y ∈ [a, b].
The second conclusion of Helly’s theorem is a consequence of the following observation: For
any β ∈ BV[a, b] with Vab β ≤ K, any continuous f as above, the inequality
Z b
<
f
dβ
−
S
(f,
P,
T
)
β
3
a
(2)
holds for any partition P = {a = x0 < x1 < · · · < xn = b} such that supi (xi − xi−1 ) < δ,
and for any choice of points T = {ti }, ti ∈ [xi−1 , xi ]. We will prove the inequality (2) in a
moment. Assuming this for now, the rest of the proof is completed as follows:
Z b
Z b
Z b
Z b
f dα ≤ f dαn − Sαn (f, P, T ) + f dα − Sα (f, P, T )
f dαn −
a
a
a
a
+ |Sαn (f, P, T ) − Sα (f, P, T )|
≤ + + |Sαn −α (f, P, T )|
3 3
n
2 X
=
+
f (ti ) (αn − α)(xi ) − (αn − α)(xi−1 ) 3
i=1
2
+ ||f ||∞ V (αn − α, P )
3
<
≤
provided n is chosen large enough so that V (αn − α, P ) < 3||f||∞ . At the second step of the
string of inequalities above, we have used (2) with β replaced by αn and α respectively.
1
2
It remains to prove (2). This is established as follows,
n
Z b
Z b
X
f dα − Sα (f, P, T ) = f dα −
f (ti ) α(xi ) − α(xi−1 ) a
a
i=1
n Z xi
X
=
i=1
xi−1
n Z
X
=
≤
xi
i=1
n
X
Z
xi
f (ti )
xi−1
i=1
dα
f (x) − f (ti ) dα(x)
i=1 xi−1
n Z xi
X
f dα −
n
X
xi−1
f (x) − f (ti ) dα(x)
xi
Vxi−1 α
3K
i=1
b
=
Va α < .
3K
3
The penultimate step above uses, combined with (1), the estimate
Z d
g(x)dβ(x) ≤ ||g||∞ Vcd β
≤
c
which we proved in class. The final step uses the definition of variation of a function.
2. This problem addresses the issue of uniqueness of the integrator α in Riesz representation
theorem. Fill in the following steps.
(a) Given α ∈ BV[a, b], define β(a) = α(a), β(x) = α(x+) for a < x < b and β(b) = α(b).
Rb
Rb
Show that β is right continuous on (a, b), that β ∈ BV[a, b] and that a f dα = a f dβ
for every f ∈ C[a, b].
(b) Given α ∈ BV[a, b], show that there is a unique β ∈ BV[a, b] with β(a) = 0 such that β
Rb
Rb
is right continuous on (a, b) and a f dα = a f dβ for every f ∈ C[a, b].
(c) Given a continuous linear functional L : C[a, b] → R, we constructed a function α ∈
BV[a, b] such that
Z b
L(f ) =
f dα.
a
Argue that such α is not unique in general. However, combine the above steps to conclude
that α can be chosen to be right continuous on (a, b) with α(a) = 0, and that in this case,
α is unique.
Proof. For part (a), let us first assume that the integrator α is nondecreasing. Then the
function β defined as above is nondecreasing as well, with the additional property that
β(x) ≥ α(x) for all x ∈ [a, b].
(3)
Given any > 0 and f ∈ C[a, b], we aim to show that
Z
Z b
b
f d(α − β) = −
(α − β)df < (α(b) − α(a)) .
a
a
3
The equality above is the integration by parts formula; the terms occurring from the endpoints a and b are zero, since α and β coincide at these points. Since > 0 is arbitrary, this
would yield the stated result.
By the uniform continuity of f , let δ > 0 be such that |f (x)−f (y)| < whenever |x−y| < δ.
Let us fix any partition P = {a = x0 < x1 < x2 < · · · < xn = b} with (xi − xi−1 ) < δ. Then
for any choice of points T = {ti } with ti ∈ [xi−1 , xi ],
n
X
Sf (α − β, P, T ) = β(ti ) − α(ti ) ∆fi ≤
≤
i=1
n
X
i=1
n
X
α(ti +) − α(ti )
α(ti+1 ) − α(ti ) = (α(b) − α(a)) .
i=1
The second step above follows from triangle inequality and (3). The third uses the fact that
α is monotone and ti+1 ≥ ti . This proves the desired conclusion for a nondeceasing integrator
α. If α ∈ BV[a, b], we write α = α1 − α2 where each αi is nondecreasing, and observe that
this implies β = β1 − β2 , where βi is obtained from αi in the same way that β is obtained
from α. The conclusion for integrators of bounded variation therefore follows from the above
argument for nondecreasing integrators.
We turn now to part (b). Suppose that γ1 and γ2 are two right continuous functions of
bounded variation with γ1 (a) = γ2 (a) = 0 and
Z
(4)
b
Z
f dγ1 =
a
b
f dγ2
for all f ∈ C[a, b].
a
Our goal is to show that γ = γ1 − γ2 ≡ 0. We proceed to prove this by contradiction. If
γ 6≡ 0, then there exists c ∈ (a, b] such that γ(c) 6= 0. The following two cases arise:
Case 1: There does not exist c ∈ (a, b) such that γ(c) 6= 0. This means that γ(x) = 0 for
all x ∈ (a, b), and γ(b) 6= 0. Thus γ is a step function which only jumps at b. Hence choosing
any f ∈ C[a, b] that takes a nonzero value at b would give a contradiction to (4).
Case 2: There exists c ∈ (a, b) such that γ(c) 6= 0. We prove in Lemma 0.1 below that
that the variation function v(x) of γ ∈ BV[a, b] is right continuous at every point x where γ
is right continuous. This implies that there exists > 0 such that
v(c + ) − v(c) = Vcc+ γ <
|γ(c)|
.
2
Let us define a continuous function f as follows:


1
f (t) = 0

1 −
x−c
if a ≤ x ≤ c
if x ≥ c + if c ≤ x ≤ c + .
4
Then
b
Z
a
Z
c
Z
c+
f dγ dγ +
c
a
Z c+
= γ(c) +
f dγ c
Z
c+
f dγ ≥ γ(c) −
f dγ = c
|γ(c)|
≥ γ(c) − Vcc+ γ ≥
> 0,
2
which again contradicts (4).
Finally we consider part (c). The integrator α is not unique since
Z b
Z b
f dα =
f d(α + c) for all c ∈ R.
a
a
If α is assumed to be right continuous with α(a) = 0, then α is unique by parts (a) and (b).
Lemma 0.1. Suppose that γ ∈ BV[a, b], and that v(x) = Vax γ. If γ is right (respectively
left) continuous at some point c ∈ [a, b), then so is v.
Proof of lemma. Fix > 0. We aim to find δ > 0 such that v(x) − v(c) = Vcx γ < for all x
satisfying c < x < c + δ. Fix a point d > c. By definition of variation of a partition and its
monotonicity under refinements, there exists a partition P = {c = x0 < x1 < · · · < xn = d}
of [c, d] such that
(5)
0 ≤ Vcd γ − V (γ, P 0 ) < for any refinement P 0 of P .
Using the right continuity of γ at c, pick 0 < δ < x1 − c such that
for c < x < c + δ.
(6)
|γ(x) − γ(c)| <
2
Now for any x ∈ (c, c + δ), choose P 0 = P ∪ {x} = {c < x < x1 < · · · xn = d}, so that
V (γ, P 0 ) = |γ(x) − γ(c)| + |γ(x1 ) − γ(x)| +
n
X
|γ(xk ) − γ(xk−1 )|
k=2
Vxd γ,
≤ |γ(x) − γ(c)| +
and hence
(7)
V (γ, P 0 ) < + Vxd γ,
2
where the last step follows from (6). This leads to the desired estimate
Vcx γ = Vcd γ − Vxd γ
< Vcd γ − V (γ, P 0 ) +
2
< + = .
2 2
Note that the second inequality in the steps above makes use of the bound (7), while the
third is a consequence of (5).
5
3. For f ∈ C[0, 1], does the following limit exist? If yes, evaluate it. If not, explain why not.
n
1X
log k
lim
.
f
n→∞ n
log
n
k=2
Solution using Q1. We claim that
n
1X
lim
f
n→∞ n
k=2
log k
log n
= f (1).
Now if αn is a non-decreasing step function with step size n1 at x = logn k for some positive
integer 2 ≤ k ≤ n. Then by a formula we proved in Homework 6 Q2,
Z 1
n
n
X
1X
1
log k
f (logn k) =
f dαn =
f
n
n k=2
log n
0
k=2
x
In fact, we see that one can take αn (x) = bnn c .(bnx c takes integer values at x = logn k.) Now
we calculate limn→∞ αn (x). If 0 < x < 1 then x − 1 < 0 and 0 ≤ |αn (x)| ≤ nx−1 → 0 as
n → ∞. At x = 1, αn (x) = 1 for all n. Hence αn → α pointwise where α(x) = 0, 0 < x < 1
and α(1) = 1. Hence by Q1,
Z 1
Z 1
f dα = f (1)
f dαn →
0
0
Direct Proof. Let > 0, f ∈ C[0, 1] is bounded by M > 0 and there is a δ > 0, |x − 1| <
δ =⇒ |f (x) − f (1)| < . Now observe that
log k
k > n1−δ =⇒ 0 ≤ 1 −
<δ
log n
Using this and use the bound |f (x)| ≤ M when k ≤ n1−δ . Then
X
1 n
1 X 1 X
log
k
log
k
≤
f
+
f
−
f
(1)
−
f
(1)
n
n
n
log n
log n
2≤k≤n1−δ
k=2
k>n1−δ
log
k
f
− f (1)
log n
1
1
2M n1−δ + (n − n1−δ )
n
n
which could be made less than 2 if n is sufficiently large.
≤
4. Many of the results we have studied in this course thus far focus on approximation of continuous functions by other special functions, such as polynomials or trigonometric sums or
polygonal functions, usually in the sup norm. Let us now think about approximation in the L2
norm of Riemann-integrable functions, which are not necessarily continuous, by continuous
functions. Suppose that f is a bounded, 2π-periodic function that is Riemann-integrable on
[−π, π].
(a) Show that there is a continuous function g on [−π, π] satisfying ||f − g||2 < .
(b) Can the function g above be chosen to be continuous and 2π-periodic?
(c) Can g be chosen to be a trigonometric polynomial?
6
Solution. If f ∈ R[−π, π] and is bounded by M > 0 then there is a partition P of [a, b] such
that |U (f, P ) − L(f, P )| < 8M π . Let g be a polygonal (piecewise linear) function such that
g(xi ) = f (xi ) for all xi ∈ P . Then g is continuous and for all P ∗ ⊇ P, (Let Ii = [xi−1 , xi ] be
the ith interval in P ∗ .)
Z π
X
X
|f − g|dx ≤
|f (xi ) − g(xi )|∆xi +
≤
| sup f − inf f |∆xi +
Ii
8M
8M π
Ii
−π
I
x ∈P ∗
i
i
i≥1
≤ U (f, P ∗ ) − L(f, P ∗ ) +
8M π
<
4M π
Hence since |f + g| ≤ |f | + |g|
kf −
gk22
Z
π
|f − g||f + g|dx ≤ 2π
=
−π
2M = 4M π
We can choose g to be 2π-periodic since as the g above satisfies g(π) = f (π) = f (−π) = g(−π)
and can be extended to be a continuous 2π periodic function.
Finally since g ∈ C 2π there is a trigonometric polynomial T = Sn g such that kg − T k < . Then
kf − T k2 ≤ kf − gk2 + kg − T k2 ≤ 2.