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Real Analysis
Math 125A, Fall 2012
Sample Final Questions
- Define f : R → R by
f (x) =
x
3
1 + x^2
Show that f is continuous on R. Is f uniformly continuous on R?
Solution.
- To simplify the inequalities a bit, we write
x
3
1 + x
2
= x −
x
1 + x
2
For x, y ∈ R, we have
|f (x) − f (y)| =
x − y −
x
1 + x
2
y
1 + y
2
≤ |x − y| +
x
1 + x^2
y
1 + y^2
- Using the inequality 2|xy| ≤ x
2
2 , we get
∣ ∣ ∣ ∣
x
1 + x
2
y
1 + y
2
x − y + xy
2 − x
2 y
(1 + x
2 )(1 + y
2 )
[
1 + |xy|
(1 + x^2 )(1 + y^2 )
]
|x − y|
[
1 + x
2
2
(1 + x^2 )(1 + y^2 )
]
|x − y|
1 + y
2
1 + x
2
|x − y|
≤ |x − y|
|f (x) − f (y)| ≤ 2 |x − y| for all x, y ∈ R.
Therefore f is Lipschitz continuous on R, which implies that it is uni-
formly continuous (take δ = ǫ/2).
- Does there exist a differentiable function f : R → R such that f
′ (0) = 0
but f
′ (x) ≥ 1 for all x 6 = 0?
Solution.
- No such function exists.
- We have
f
′ (0) = lim x→ 0
[
f (x) − f (0)
x
]
The mean value theorem implies that for for every x 6 = 0, there is some
ξ strictly between 0 and x (so ξ 6 = 0) such that
f (x) − f (0)
x
= f
′ (ξ) ≥ 1.
- Since limits preserve inequalities, it follows that
lim x→ 0
[
f (x) − f (0)
x
]
so we cannot have f
′ (0) = 0.
- (a) Suppose fn : A → R is uniformly continuous on A for every n ∈ N
and fn → f uniformly on A. Prove that f is uniformly continuous on A.
(b) Does the result in (a) remain true if fn → f pointwise instead of uni-
formly?
Solution.
- (a) Let ǫ > 0. Since fn → f converges uniformly on A there exists
N ∈ N such that
|fn(x) − f (x)| <
ǫ
for all x ∈ A and n > N.
Choose some n > N. Since fn is uniformly continuous, there exists
δ > 0 such that
|fn(x) − fn(y)| <
ǫ
for all x, y ∈ A with |x − y| < δ.
Then, for all x, y ∈ A with |x − y| < δ, we have
|f (x) − f (y)| ≤ |f (x) − fn(x)| + |fn(x) − fn(y)| + |fn(y) − f (x)| < ǫ,
which implies that f is uniformly continuous on A.
- (b) The result does not remain true if fn → f pointwise. For example,
consider fn : [0, 1] → R defined by fn(x) = x
n
. Then fn is uniformly
continuous on [0, 1] because it is a continuous function on a compact
interval, but fn → f pointwise where
f (x) =
0 if 0 ≤ x < 1,
1 if x = 1.
The limit f is not even continuous on [0, 1].
- Define fn : [0, ∞) → R by
fn(x) =
sin(nx)
1 + nx
(a) Show that fn converges pointwise on [0, ∞) and find the pointwise limit
f.
(b) Show that fn → f uniformly on [a, ∞) for every a > 0.
(c) Show that fn does not converge uniformly to f on [0, ∞).
Solution.
|fn(x)| ≤
1 + nx
→ 0 as n → ∞
so fn(x) → 0. Also, fn(0) = 0 for every n, so fn(0) → 0. Thus, fn → 0
pointwise on [0, ∞).
|fn(x)| ≤
1 + na
na
for all a ≤ x < ∞,
so given ǫ > 0 take N = 1/a and then |fn(x)| < ǫ for all n > N,
meaning that fn → 0 uniformly on [a, ∞).
- (c) If (fn) converges uniformly on [0, ∞), then it must converge to the
pointwise-limit 0. Let xn = π/(2n). Then
fn(xn) =
1 + π/ 2
Therefore, if 0 < ǫ 0 ≤ 1 /(1 + π/2), there exists x ∈ [0, ∞) such that
fn(x) ≥ ǫ 0 ,
which means that fn does not converge uniformly to 0 on [0, ∞).
- Suppose that
f (x) =
∑^ ∞
n=
sin nx
n
3
, g(x) =
∑^ ∞
n=
cos nx
n
2
(a) Prove that f, g : R → R are continuous.
(b) Prove that f : R → R is differentiable and f
′ = g.
Solution.
sin nx
n^3
n^3
∞ ∑
n=
n^3
cos nx
n^2
n^2
∑^ ∞
n=
n^2
the Weierstrass M-test implies that both series converge uniformly (and
absolutely) on R.
- Each term in the series is continuous, and the uniform limit of contin-
uous functions is continuous, so f , g are continuous on R.
- (b) The series for g is the term-by-term derivative of the series for f.
Since the series for g converges uniformly, the theorem for the differen-
tiation of sequences implies that f is differentiable and f
′ = g.
- Let P = { 2 , 3 , 5 , 7 , 11 ,... } be the set of prime numbers.
(a) Find the radius of convergence R of the power series
f (x) =
p∈P
x
p = x
2
3
5
7
11 +...
(b) Show that
0 ≤ f (x) ≤
x
2
1 − x
for all 0 ≤ x < 1.
Solution.
- (a) We write the series as
f (x) =
∞ ∑
n=
anx
n
where
an =
1 if n is prime,
0 if n isn’t prime.
|anx
n | ≤ |x|
n for every n = 2, 3 , 4 ,....
Therefore, if |x| < 1 the series converges by comparison with the con-
vergent geometric series
|x|
n
. Furthermore, if |x| > 1, the terms in
the series do not approach 0. So the radius of convergence of the series
is R = 1.
- (b) As in (a), and using the sum of the geometric series, we have for
0 ≤ x < 1 that
p∈P
x
p ≤
∑^ ∞
n=
x
n = x
2
∑^ ∞
n=
x
n
x
2
1 − x
which proves the result.
