Analysis Homework #5
Solutions
1. Suppose that f is differentiable with
f ′ (x) = x2 f (x),
f (0) = 1.
Show that f (x)f (−x) = 1 for all x ∈ R and that f (x) > 0 for all x ∈ R.
• Setting g(x) = f (x)f (−x) for convenience, we easily get
g ′ (x) = f ′ (x)f (−x) + f ′ (−x)(−x)′ f (x)
= x2 f (x)f (−x) + (−x)2 f (−x)(−1)f (x) = 0,
hence g(x) is constant and
f (x)f (−x) = g(x) = g(0) = f (0)f (0) = 1.
• To show that f (x) is always positive, suppose f (x0 ) ≤ 0 at some point x0 . Then
f (x0 ) ≤ 0,
f (0) > 0
and f is continuous, so we can use Bolzano’s theorem to conclude that f (x∗ ) = 0 at
some point. This actually gives the contradiction 0 = f (x∗ )f (−x∗ ) = 1.
2. Suppose that f is differentiable with f ′ (x) = 2f (x) for all x. Show that there exists
some constant C such that f (x) = Ce2x for all x.
• Setting g(x) =
f (x)
e2x
for convenience, we easily get
g ′ (x) =
f ′ (x) · e2x − 2e2x · f (x)
f ′ (x) − 2f (x)
=
= 0.
e2x e2x
e2x
This shows that g(x) is actually constant, say g(x) = C, so
g(x) = C
=⇒
f (x)
=C
e2x
=⇒
f (x) = Ce2x .
3. Show that ex ≥ x + 1 for all x ∈ R.
• We compute the minimum value of f (x) = ex − x − 1. Since the derivative
f ′ (x) = ex − 1
is negative when x < 0 and positive when x > 0, this function is decreasing for the
former values of x and increasing for the latter. In particular,
f (0) = e0 − 0 − 1 = 0
is the minimum value attained by the function, so f (x) ≥ f (0) = 0 for all x ∈ R.
4. Suppose that f is differentiable with
|f (x) − f (y)| ≤ |x − y|2
for all x, y ∈ R.
Show that f is actually constant. Hint: you need to show that f ′ (y) = 0 for all y.
• Recall the definition of the derivative, according to which
f (x) − f (y)
.
x→y
x−y
f ′ (y) = lim
Taking the absolute value of both sides, we then easily find that
f (x) − f (y) |f (x) − f (y)|
′
|f (y)| = lim
= lim
.
x→y
x→y
x−y
|x − y|
On the other hand, the given inequality implies
0≤
|f (x) − f (y)|
≤ |x − y|.
|x − y|
Here, both the left hand side and the right hand side go to zero as x → y, so
|f (x) − f (y)|
=0
x→y
|x − y|
lim
by the Squeeze Law. In particular, |f ′ (y)| = 0 for all y and f is constant, indeed.