MATH 3150 FINAL EXAM PRACTICE PROBLEMS – FALL 2014
Problem 1. Suppose (sn ) is a sequence in R, and for each n, let σn = n1 (s1 + · · · + sn ).
(a) Show that, if (sn ) is convergent, then (σn ) is convergent and lim σn = lim sn .
(b) Find an example where (σn ) converges but (sn ) does not.
Solution.
(a) Suppose sn → s; we want to show that σn → s as well. The key estimate is
s1 + · · · + sn
s1 + · · · + sn − ns |s1 − s|
|sn − s|
≤
(1)
|σn − s| = − s = + ··· +
,
n
n
n
n
where we use the triangle inequality (n times).
Let ε > 0 be given. Since sn → s, there exists N0 ∈ N such that n ≥ N0 implies
|sn − s| < ε/2. Also, since (sn ) converges, it is bounded; in particular there exists
M > 0 such that |sn − s| ≤ M for all n. Using the latter to estimate the first N0 terms
in (1) and the former to estimate the rest, we have, for n > N0 ,
N0 M
(n − N0 )ε
N0 M
nε
N0 M
ε
+
≤
+
=
+ .
n
2n
n
2n
n
2
Now, there exists N1 such that for n ≥ N1 , (N0 M )/n < ε/2, and then
ε ε
n ≥ max {N0 , N1 } =⇒ |σn − s| < + = ε.
2 2
(b) The alternating sequence (sn ) = (1, 0, 1, 0, 1, . . .) does not converge, but
|σn − s| <
(σn ) = (1, 12 , 23 , 21 , 35 , 12 , . . .)
converges to 1/2.
Problem 2. Show that f (x) = x2 is uniformly continuous on the open interval (−1, 2).
Solution. f (x) is continuous, and restricted to a closed and bounded (hence compact) interval, say [−1, 2], it is uniformly continuous. But then it is uniformly continuous on any subset
thereof, such as (−1, 2).
Alternatively, you can give a direct ε–δ proof.
Problem 3. Define f : R −→ R by
(
x sin
f (x) =
0
1
x
x 6= 0
x = 0.
(a) Show that f is continuous, and uniformly continuous on [−1, 1].
(b) Show that f is not differentiable at x = 0.
Solution.
1
(a) f is continuous at any x 6= 0 since there it is the product of the continuous function x
and the composition of the continuous functions sin(x) and 1/x.
At x = 0 we must show that
lim f (x) = f (0) = 0.
x→0
We will show that limx→0 |f (x)| = 0 which implies the above.
lim |f (x)| = lim |x| sin x1 ≤ lim |x| = 0
x→0
x→0
x→0
since sine is bounded by 1 in absolute value.
Then f is uniformly continuous on any compact set, such as [−1, 1] since this is true
for any continuous function.
(b) The limit of the difference quotient
x sin x1
f (x) − f (0)
lim
= lim
= lim sin x1
x→0
x→0
x→0
x−0
x
does not exist. Therefore f is not differentiable at x = 0.
Problem 4. Let f (x) =
R x2
0
√
e
t
dt for x ∈ [0, +∞).
(a) Compute f (0).
(b) Show that f is differentiable on (0, +∞) and compute f 0 (x).
Solution.
R0 √
(a) f (0) = 0 e t dt = 0 since
the interval of integration has width 0.
√
t
(b) The integrand, t 7−→ e , is a continuous function on [0, +∞) and therefore by the
fundamental theorem of calculus,
Z y √
F (y) =
e t dt
0
0
√
y
is differentiable with derivative F (y) = e . Now f = F ◦ g where g(x) = x2 , so using
the chain rule,
√
f 0 (x) = e g(x) g 0 (x) = 2xex .
Problem 5. Define f : [0, 1] −→ R by
(
2 x 6= 12
f (x) =
0 x = 12 .
R1
Show that f is integrable and compute 0 f (x) dx.
Solution. It suffices to show that for any ε > 0, there exists a partition P of [0, 1] such that
U (f, [0, 1], P ) − L(f, [0, 1], P ) < ε
since this implies that supP {L(f, [0, 1], P )} and inf P {U (f, [0, 1], P )} — the lower and upper
integrals, respectively — are equal.
2
For any partition P = {0 = x0 < x1 < · · · < xN = 1}, the upper integral is
U (f, [0, 1], P ) =
N
X
N
X
sup f (x)(xi − xi−1 ) = 2
(xi − xi−1 ) = 2,
i=1 [xi−1 ,xi ]
i=1
and the lower integral is
L(f, [0, 1], P ) =
N
X
i=1
X
inf f (x)(xi − xi−1 ) = 2
[xi−1 ,xi ]
(xi − xi−1 ) + 0 · (xj − xj−1 )
1/2∈[x
/ i−1 ,xi ]
where [xj−1 , xj ] is the interval of P which contains the point x = 12 . Thus
L(f, [0, 1], P ) = 2(1 − (xj − xj−1 )).
Thus given any ε > 0, we may choose a partition such that the interval [xj−1 , xj ] containing
x = 1/2 has width (xj − xj−1 ) < ε/2. For such a partition P ,
U (f, [0, 1], P ) − L(f, [0, 1], P ) < 2 − 2(1 − ε/2) = ε.
Thus f is integrable. Then
1
Z
f (x) dx = U (f ) = 2
0
since the upper sums U (f, [0, 1], P ) are all equal to 2 so their infimum is 2.
Problem 6. Suppose f : R −→ R satisfies
|f (x) − f (y)| ≤ C |x − y|2 ,
∀x, y ∈ R
for some C ≥ 0. Show that f must be constant. [Hint: show that it is differentiable first.]
Solution. By assumption, the limit
|f (x) − f (y)|
≤ lim C |x − y| = 0
x→y
x→y
|x − y|
lim
which implies that
f (x) − f (y)
=0
x→y
x−y
and therefore f is differentiable at all y ∈ R with derivative f 0 (y) = 0. Since the derivative
vanishes identically, f must be constant.
lim
Problem 7. Suppose f : [0, +∞) −→ R is continuous and differentiable on (0, +∞), and
suppose that
f (x) + x f 0 (x) ≥ 0, ∀x > 0.
Show that f (x) ≥ 0 for all x ≥ 0. [Hint: consider the function g(x) = xf (x).]
Solution. If we define g(x) = x f (x), then by the product rule,
g 0 (x) = f (x) + x f 0 (x) ≥ 0,
Thus g is increasing on (0, +∞). Furthermore,
g(0) = 0 · f (0) = 0
so it follows that g(x) ≥ 0 for x > 0.
3
∀x > 0.
Now, f (x) = g(x)/x implies that f (x) ≥ 0 for x > 0 since 1/x is positive there, and finally
lim f (x) ≥ 0
x→0
since the limit of a non-negative function is nonnegative. Thus f (x) ≥ 0 for all x ∈ [0, ∞).
Problem 8. Let fn : A ⊂ R −→ R be a sequence of functions (not necessarily continuous),
converging uniformly to a function f : A ⊂ R −→ R. Show that, if each fn is bounded, then
f is bounded.
Solution. Since each fn is bounded, there exists Mn > 0 such that |fn (x)| ≤ Mn for all x.
Then, since fn → f uniformly, given ε = 1, there exists N such that
|f (x)| − |fN (x)| ≤ |fN (x) − f (x)| < ε = 1,
∀ x ∈ A.
Combining this with |fN (x)| ≤ MN and rearranging, we have
|f (x)| < 1 + MN ,
4
∀ x ∈ A.