Download (a) Let f(x)

Math 335
Assignment 1 Answers
21 January 2015
√
(1) (a) Let f (x) = x. Prove
√ √that√f is uniformly continuous on the interval
√
[1, ∞). Hint: ( s − t)( s + t) = s − t
Given > 0 let δ = . If s and t ∈ [1, ∞) and |s − t| < δ then |f (s) − f (t)| =
|s−t|
|s−t|
√ √ ≤
< 2δ < as required.
2
| s+ t|
(b) What theorem implies that f is also uniformly continuous on [0, 1]?
(c) Prove that f is uniformly continuous on [0, ∞).
For any function f that is uniformly continuous on both [0, 1] and [1, ∞), given
> 0 there exist δ1 > 0 and δ2 > 0 such that
if s and t ∈ [0, 1] and |s − t| < δ1 then |f (s) − f (t)| < 2 ;
(α)
if s and t ∈ [1, ∞] and |s − t| < δ2 then |f (s) − f (t)| < 2 .
(β)
Let δ3 = min(δ1 , δ2 ), If s and t ∈ [0, ∞) and |s − t| < δ3 , consider three cases.
(i) s, t ∈ [0, 1], then by (α), |f (s) − f (t)| < .
(ii) s, t ∈ [1, ∞], you get the same conclusion using (β).
(iii) s < 1 < t, (or t < 1 < s), then |s − 1| < δ3 ≤ δ1 and |t − 1| < δ3 ≤ δ2 and so
|f (s) − f (t)| ≤ |f (s) − f (1)| + |f (1) − f (t)| < 2 + 2 = Alternatively, for any function f that is uniformly continuous on [0, 2] (instead
of [0, 1]) and also on [1, ∞) and with δ1 and δ2 similar to the above, take δ3 =
min(1, δ1 , δ2 ). Now if |s − t| < δ3 , then |s − t| < 1 and so both s and t belong
simultaneously to one of the two intervals and the proof proceeds as before with
no case (iii).
(2) Let f (x) = x + 2x2 sin x1 for x 6= 0 and f (0) = 0.
(a) Use the usual calculus formulae to calculate f 0 (a) for a 6= 0.
f 0 (a) = 1 + 4a sin a1 − 2 cos a1 .
(b) Show that f is differentiable at 0 and that f 0 (0) = 1.
(0)
| f (x)−f
− 1| = |2x sin x1 | ≤ 2|x| so the limit of the left side is 0 as x → 0
x−0
(c) Show that f 0 is not continuous at 0.
1
This follows from the fact that lim cos does not exist. It also follows from the
x→0
x
calculations in part (d) below.
(d) Although f is everywhere differentiable and its derivative at 0 is positive, show that (surprisingly) f is not monotone increasing in any
neighbourhood of 0.
1
1
f 0 ( 2nπ
) = −1 and f 0 ( (2n+1)π
) = 3. No further comments are required.
(e) Draw a rough graph of f near 0.
(3) If f is differentiable with continuous derivative in an interval (a, b) and if
f ( c) > 0 for some c ∈ (a, b), prove that f in monotone increasing in some
nbhd (c − δ, c + δ) of c.
0
By continuity, there exists a δ > 0 such that if |x − c| < δ then |f 0 (x) − f 0 (c)| < f 2(c)
and hence f 0 (x) > 0 for all such x. Now if x1 < x2 are in the interval (c − δ, c + δ),
then, by the Mean Value Theorem, there exists x3 in the same interval such that
f (x2 ) − f (x1 ) = f 0 (x3 ) (x2 − x1 ) > 0 proving strict monotinicity
(4) (29.4) Let f be defined on R and suppose that |f (x) − f (y)| ≤ (x − y)2 for all
x, y ∈ R. Prove that f is a constant function.
(x0 )
For any x0 , we get | f (x)−f
| ≤ |x − x0 . Take limits as x → x0 to get that f 0 is 0
x−x0
everywhere, so f is a constant function.
(5) (28.16) Let g be a function defined on an open interval I containing c. Prove
that g has a derivative at c if and only if there exist a real number A and a
function u on I such that
g(x) − g(c) = (x − c)(A − u(x)) for x ∈ I, u(c) = 0; lim u(x) = 0.
(1)
0
0
If g (c) exists, take A = g (c) and u(x) =
g(x)−g(c)
x−c
x→c
0
− g (c) if x ∈ I and x 6= c and
u(c) = 0. Verifying equations (1) is straightforward. Conversely, if eq (1) is satisfied,
− A = u(x) for x ∈ I, x 6= c. Taking limits proves that g 0 (c) exists
then g(x)−g(c)
x−c
and equals A. This also determines u(x) uniquely except at c. This is an error in the
statement of the question as equation (1) is valid at x = c regardless what u(c) is.
However if we insist that u is continuous at 0, then u(c) must be 0.
Note: Another useful formulation is that there exist a real number A and
a function
η(x)
η on I such that g(x) − g(c) = A(x − c) + η(x) for every x ∈ I and lim
=0
x→c
x−c
(6) Prove the Chain Rule using the results of the previous question.
Hint: Substitute x by f (t) and c by f (a) in eq (1).
Assume the usual set-up: f is defined in an open interval I including a and is differentiable at a and g is defined in an open interval J including c = f (a) and is differentiable
at c, f (I) ⊂ J. By the previous question, there exists a function u on J such that
u(c) = 0, lim u(y) = 0 and g(y) − g(c) = (y − c)(g 0 (c) − u(y)) for y ∈ J. Therefore
y→c
g(f (x)) − g(f (a)) = (f (x) − f (a)) (g 0 (c) − u(f (x))) for x ∈ I,
f (x) − f (a)
g(f (x)) − g(f (a))
=(
) (g 0 (f (a)) − u(f (x))) for x near a, x 6= a.
x−a
x−a
All that is required now is to show that lim u(f (x)) = 0 as this implies that the limit
x→a
of the right side of the last displayed equation is f 0 (a)g 0 (f (a)) proving that g ◦ f is
differentiable at a and its derivative at a is g 0 (f (a))f 0 (a) as required.
To prove this last bit, note that f is continuous at a and u is continuous at c = f (a),
so u ◦ f is continuous at a, hence lim u(f (x)) = u(f (a)) = u(c) = 0.
x→a
It may be tempting to argue that as x → a, f (x) → c and as y → c, u(y) → 0 so
u(f (x)) → 0 as x → a, but there is a subtle error in this and will be discussed in class.