Math 4318 : Real Analysis II
Mid-Term Exam 1
14 February 2013
Name:
Definitions:
True/False:
Proofs:
1.
2.
3.
4.
5.
6.
Total:
Definitions and Statements of Theorems
1. (2 points) For a function f (x) defined on (a, b) and for x0 ∈ (a, b) state the definition
of f 0 (x0 ).
Solution:
f 0 (x0 ) = lim
x→x0
f (x) − f (x0 )
.
x − x0
2. (2 points) State the Mean Value Theorem for a function that is continuous on [a, b]
and differentiable on (a, b).
Solution: Suppose that f : [a, b] → R is continuous on [a, b] and differentiable on
(a, b). Then there exists a c ∈ (a, b) such that
(b − a)f 0 (c) = f (b) − f (a)
3. (3 points) State the Second Fundamental Theorem of Calculus.
Solution: Let f be integrable on [a, b]. For x ∈ [a, b] let
Z x
f (t)dt.
F (x) =
a
Then F is continuous on [a, b]. If f is continuous at x0 in (a, b), then F is differentiable at x0 and F 0 (x0 ) = f (x0 ).
4. (3 points) State one (of the many) equivalent definitions for a function f to be integrable
on [a, b].
Solution: A function f on [a, b] is integrable if and only if for any > 0 there
exists a partition P of [a, b] such that
U (f ; P ) − L(f ; P ) < .
True or False (1 point each)
For Questions 1-5, let a, b ∈ R with a < b and let f be a function that maps the interval
[a, b] to R.
1. If f is differentiable on (a, b), then f is continuous on (a, b).
Solution: True
2. If f is continuous on [a, b], then f is integrable on [a, b].
Solution: True
3. If f is monotonic on [a, b], then f is integrable on [a, b].
Solution: True
4. If f is integrable on [a, b], and x ∈ [a, b] then F (x) =
Rx
a
f (t)dt is continuous.
Solution: True
5. A bounded function f is integrable on [a, b] if and only if there exists an > 0 such
that for all partition P of the interval [a, b]
U (f ; P ) − L(f ; P ) < .
Solution: False
6. Every continuous function on [a, b] is differentiable on (a, b).
Solution: False
7. If f and g are integrable on [a, b], then f g is integrable on [a, b].
Solution: True
8. The upper Darboux integral of f , U (f ), is always less than or equal to the lower
Darboux integral L(f ).
Solution: False
9. A function is Riemann integrable if and only if it is Darboux integrable.
Solution: True
10. If P and Q are partitions of [a, b] and P ⊂ Q then
L(f ; P ) ≤ L(f ; Q) ≤ U (f ; Q) ≤ U (f ; P ).
Solution: True
Proofs
1. (10 points) Let f be defined on R and suppose that
|f (x) − f (y)| ≤ (x − y)2
∀x, y ∈ R.
Prove that f is a constant function.
Solution: It suffices to prove that f 0 (x) = 0 for all x ∈ R. Let > 0 be given. Then we
need to show that there exists δ > 0 such that
f (x + h) − f (x)
f (x + h) − f (x) <
− 0 = h
h
when |h| < δ But, by the property of the function we have that
2
f (x + h) − f (x) ≤ h = |h| .
|h|
h
So choose δ = , and if |h| < δ we have
f (x + h) − f (x)
− 0 < ,
h
and since > 0 was arbitrary we have that f 0 (x) = 0.
2. (10 points) Suppose that f is defined and differentiable for every x > 0 and that f 0 (x) → 0
as x → ∞. Set
g(x) := f (x + 1) − f (x).
Prove that g(x) → 0 as x → ∞.
Solution: Use MVT to write
g(x) = f (x + 1) − f (x) = f 0 (tx )
where x < tx < x + 1. So as x → ∞ then tx → ∞ as well. More precisely, let > 0
be given. Then select M such that f 0 (x) < for all x > M . Then if x > M , and for
tx ∈ (x, x + 1) we have that f 0 (tx ) < as well. Thus, for any x > M we have that
g(x) = f (x + 1) − f (x) = f 0 (tx ) < and so g(x) → 0 as x → ∞.
3. (15 points) Two functions f : R → R and g : R → R are equal up to order n at the point
a if
f (a + h) − g(a + h)
lim
= 0.
h→0
hn
Suppose that f (x), f 0 (x), . . . , f (n) (x) exist and are continuous at a. Prove that the function
f and the function
n
X
f (k) (a)
(x − a)k
g(x) =
k!
k=0
are equal up to order n at the point a.
Solution: Consider
P
f (a + h) − nk=0
f (a + h) − g(a + h)
=
hn
hn
f (k) (a) k
h
k!
As h → 0 we have that this approaches 00 since in the numerator we have f (a) − g(a) =
f (a) − f (a) = 0. Since f 0 (a) exists, and g 0 (a) exists, we by L’Hopital that
P
(k)
f (a + h) − nk=0 f k!(a) hk
f (a + h) − g(a + h)
lim
= lim
h→0
h→0
hn
hn
P
(k) (a)
hk
f 0 (a + h) − nk=1 f(k−1)!
.
= lim
h→0
nhn−1
Again, evaluating the numerator and denominator as h → 0 we find 00 since the numerator
now gives f 0 (a) − g 0 (a) = f 0 (a) − f 0 (a) = 0, and can again apply L’Hopital’s rule to see
that
P
(k)
f (a + h) − nk=0 f k!(a) hk
f (a + h) − g(a + h)
lim
= lim
h→0
h→0
hn
hn
P
(k) (a)
hk
f 0 (a + h) − nk=1 f(k−1)!
= lim
h→0
nhn−1
P
(k) (a)
f 00 (a + h) − nk=2 f(k−2)!
hk
= lim
.
h→0
n(n − 1)hn−2
By induction, this can be repeated n times since we know that f (k) (a) exists for 0 ≤ k ≤ n,
and arrive at
f (a + h) − g(a + h)
f (n) (a + h) − f (n) (a)
=
lim
= 0.
h→0
h→0
hn
n!
lim
4. (15 points) A function f on [a, b] is called a step-function if there exists a partition
P = {a = x0 < x1 < . . . < xN = b}
of [a, b] and constants cj ∈ R such that f (x) = cj for all x ∈ [xj−1 , xj ].
(a) Prove that a step function is integrable on [a, b].
Rb
(b) Compute a f (x) dx when f is a step function.
Solution: Recall that f is integrable on [a, b] if and only if for any > 0 there exists a
partition P of [a, b] such that
U (f ; P ) − L(f ; P ) < .
Let f be a step function, and let P be the partition describing the step function. Let
> 0 be given, and let P = P . Then note that
U (f ; P ) = U (f ; P ) =
N
X
sup
f (x)(xj − xj−1 ) =
j=1 x∈[xj−1 ,xj ]
N
X
cj (xj − xj−1 ).
j=1
A similar computation gives
L(f ; P ) = L(f ; P ) =
N
X
j=1
inf
x∈[xj−1 ,xj ]
f (x)(xj − xj−1 ) =
N
X
cj (xj − xj−1 ).
j=1
Thus, we have
U (f ; P ) − L(f ; P ) =
N
X
j=1
So f is integrable on [a, b].
cj (xj − xj−1 ) −
N
X
j=1
cj (xj − xj−1 ) = 0 < .
5. (15 points) Let f be a bounded function on [a, b]. Suppose that there exist sequences
{Un } and {Ln } of upper and lower Darboux sums for f such that limn→∞ (Un − Ln ) = 0.
Show that f is integrable and
Z b
f (x) dx = lim Un = lim Ln .
n→∞
a
n→∞
Solution: Let Ln be the lower Darboux sum, and suppose that it is associated to a
partition Pn , i.e, Ln = L(f ; Pn ). Similarly, let Un be the upper Darboux sum and
suppose that it associated to a partition Pn0 , i.e. Un = U (f ; Pn ). Let Qn = Pn ∪ Pn0 . Note
that we have
U (f ; Qn ) − L(f ; Qn ) ≤ U (f ; Pn0 ) − L(f ; Pn ) = Un − Ln
Thus, given > 0, choosing n sufficenly large will allow us to conclude that there exists
a partition Qn so that
U (f ; Qn ) − L(f ; Qn ) ≤ U (f ; Pn0 ) − L(f ; Pn ) = Un − Ln < .
Thus, we have that f is integrable. It suffices to show that lim Un =
then have
Rb
a
f (t)dt since we
Z
lim Ln = lim (Ln − Un + Un ) = lim (Ln − Un ) + lim Un = 0 +
n→∞
n→∞
n→∞
n→∞
f (t)dt
a
as well. However, given > 0, select N so that
Un − Ln < .
Rb
Rb
Note that Ln ≤ a f (t)dt ≤ Un , and so − a f (t)dt ≤ −Ln . Thus we have
Z
b
f (t)dt ≤ Un − Ln < Un −
a
if n ≥ N . Therefore, lim Un =
Rb
a
f (t)dt.
b
√
6. (15 points) Let r ≥ 2 be an integer. Let f (x) = r x defined on (0, ∞). Using the definition
of derivative, compute
f 0 (x0 )
P
r−1−j j
for any x0 ∈ (0, ∞). Hint: ar − br = (a − b) r−1
b.
j=0 a
Solution: Using the hint, let a =
√
√
r
x and b = r x0 , then we have
r−1
j
X
√
r−1−j
√
r
r
x − x0
x r x0r .
x − x0 =
j=0
Thus we have
√
√
r
x − r x0
=
x − x0
and so
√
r
f 0 (x0 ) = lim
x→x0
r−1
X
x
r−1−j
r
j
r
!−1
x0
j=0
√
x − r x0
= lim
x→x0
x − x0
r−1
X
x
r−1−j
r
j
r
!−1
.
x0
j=0
Now since x, x0 ∈ (0, ∞) we know that the limit above is defined and so
lim
x→x0
r−1
X
x
r−1−j
r
j
r
!−1
=
x0
However, for j = 0, . . . , r − 1, the function x
lim
x→x0
r−1
X
lim
x→x0
j=0
x
r−1−j
r
j=0
j
r
x0 =
r−1
X
r−1−j
r
r−1−j
r
x0
j
r
r−1
X
r−1
r
r−1
X
j=0
r−1
r
f (x0 ) = rx0
−1
x0
!−1
.
is continuous on (0, ∞ we have
Using this we see that
0
j
r
j=0
x0 = x0
j=0
x
r−1−j
r
1 1−r
= x0 r .
r
r−1
1 = rx0 r .