Problem. Suppose f : R → R is a function such that |f (x)−f (y)| ≤ |x−y|2
for all x and y. Show that f (x) = C for some constant C.
Solution. We claim that f is differentiable and has derivative 0 everywhere.
To show this, we observe that if x 6= c, then
f (x) − f (c) |f (x) − f (c)|
≤ |x − c|.
x−c =
|x − c|
So,
f (x) − f (c)
≤ |x − c|
x−c
−|x − c| ≤
for all x ∈ R \ {c}. Since limx→c (−|x − c|) = 0 and limx→c |x − c| = 0, the
squeeze lemma (Corollary 3.1.11) shows that
f (x) − f (c)
= 0.
x−c
lim
x→c
Hence f 0 (c) = 0.
By Proposition 4.2.5, f is constant.
Problem. Let c ∈ (a, b) and let d ∈ R. Define f : [a, b] → R as
(
d if x = c,
f (x) :=
0 if x 6= c.
Rb
Prove that f ∈ R[a, b] and compute a f using the definition of the integral.
Solution. Let P = {a, c − 2d
, c + 2d
, b} for > 0 sufficiently small that c − 2d
and c + 2d are contained in [a, b]. Compute
L(P, f ) = (c − − a)(0) + (0) + (b − c − )0 = 0,
d
U (P, f ) = (c − − a)(0) + d + (b − c − )0 = .
d
Rb
Rb
Since a f is the supremum of L(P, f ) over all partitions P , and a f is the
infimum of all U (P, f ) over all partitions P , this shows that
Z b
f ≥ 0,
a
Z
b
f ≤
a
for any > 0. This proves that
Z
b
Z
f−
b
f ≤
a
a
1
2
for any > 0. Hence
proven that
Rb
af
=
Rb
a f,
showing that f ∈ R[a, b]. Also, we’ve
Z
0≤
b
f ≤
a
for any > 0, proving that
Rb
a
f = 0.