Advanced Calculus Handout 3 March 28, 2011
Example (More details for Remark 2 in Handout 1): Let 0 < a < 1, define f : R → R by f (x) =
ax + x2sin1
x if x 6= 0
0 if x = 0
. Then, f0(x) =
a + 2x sin 1
x − cos 1
x if x 6= 0
0 if x = 0
, where we use
the definition of the derivative to compute f0(0), and f00(x) =
(2 − 1
x) sin 1 x − 2
xcos1
x if x 6= 0 does not exist if x = 0 . Claim: There is no solution to the following system of equations
(1)
(f0(x) = 0
f00(x) = 0 ⇐⇒
2x −1
2 − 1 x2 −2
x
!
sin1
x cos1 x
=−a 0
We prove this claim by contradiction. Suppose that system (1) has a solution for some x, then observe that, for this x in the system (1), the following system
(2)
2x −1
2 − 1 x2 −2
x
! x1 x2
=−a 0
has a solution x1 = sin 1
x, x2 = cos1
x. (3)
However, if we solve the system (2) by using Cramer’s rule, we shall get
x1 =
−a −1 0 −2 x
2x −1
2 − 1 x2 −2
x
= 2a/x
−4 + (2 − 1 x2)
; x2 =
2x −a
2 − 1 x2 0
2x −1
2 − 1 x2 −2
x
= 2a − a/x2
−4 + (2 − 1 x2)
which implies that x21+ x22 = a2[4x2+ (2x2− 1)2]
(1 + 2x2)2 = a2[4x4+ 1]
1 + 4x2+ 4x4 ≤ a2 < 1 (4) This is a contradiction since, by (3), x1 = sin 1
x, x2 = cos1
x is a solution of the system (2), and satisfies that x21 + x22 = sin2 1
x + cos2 1
x = 1 > a2 = x21+ x22 by (4). Therefore, the system (1) does have a solution. This implies that when f0(x) = 0, f00(x) 6= 0, therefore, f is not 1 − 1 around x.
Note that the Inverse Function Theorem does not apply here since f0 is not continuous at x = 0.
Examples of curves in parametric form:
(1) Let f : R → R2 be defined by f (t) = (t, |t|). Note that the f is not differentiable at t = 0, and f (R) = the graph of y = |x| over R.
(2) Let f : R → R2 be defined by f (t) = (t3, t2). Note that the f is everywhere differentiable, but f0(0) = (0, 0), and f (R) = the graph of y3 = x2 over R with a vertical cusp at (0, 0), i.e.
lim
x→0±
dy
dx = ±∞.
(3) Let f : R → R2 be defined by f (t) = (t3−4t, t2−4). Note that f is differentiable and f0(t) 6= (0, 0) for all t ∈ R. But, f (2) = f (−2) = (0, 0) i.e. f is not a globally one-to-one function, and the curve f (R) is not differentiable at (0, 0).
Example 4(b) in Handout 1: Let F = {fn(x) = xn | n ∈ N x ∈ I = [0, 1]}. Note that for each n ∈ N, and any x, y ∈ I, there exists a cn lying between x and y such that
(∗) |fn(x) − fn(y)| = n cn−1n |x − y| (by the Mean Value Theorem).
Advanced Calculus Handout 3 (Continued) March 28, 2011 For each n, choosing x = 1 and 0 ≤ y < 1 in (∗), we have
1 − yn == n cn−1n (1 − y)
Note that the left-hand-side approaches to 1 as n goes to ∞, i.e. n cn−1n on the right-hand-side ap- proaches to 1
1 − y as n goes to ∞. This implies that if 1 > > 0, then |fn(1) − fn(y)| ≥ for all 0 ≤ y < 1, and for all sufficiently large n. Hence,F is not uniformly equicontinuous.
Some notions you need from before:
Exercise: (1a) Let S ⊂ Rp. When is S said to be open in Rp ? closed in Rp ? bounded? compact?
What is the limiting (or accumulation, or cluster) points set of S?
Exercise (1b) Let S ⊂ R, f : S → Rq. When is f said to be continuous on S? Let W ⊂ Rq, what is f−1(W )?What is f (S)? What is the graph of f over S? If S is compact, what can you say about f (S), and the continuity of f on S?
(2a) The Completeness Axiom (Sec. 1.5 in the book): Let ∅ 6= S ⊂ R, and let R be equipped with the Euclidean distance kx − yk for any x, y ∈ R (note that kx − yk = |x − y| in R).If S has an upper bound, then, by “bisecting” the set S, the least upper bound (or the supremum) of S, denoted sup S = sup{x | x ∈ S},exists in R, If S has a lower bound, then the greatest lower bound (or the infimum) of S, denoted inf S = inf{x | x ∈ S}, exists in R. Using this completeness property of R (or by “bisecting” S), one can show that
(2b) Theorem 1.18: Every bounded sequence {xn} in R (=⇒ sup{xn| n ∈ N}, inf{xn | n ∈ N}
exist) has a convergent subsequence ({xnk} converging to sup{xn} ∈ R or inf{xn} ∈ R.) By “bisecting” S, one can be easily generalize Theorem 1.18 to higher dimension Rp,
(2c) Theorem 1.19: Every bounded sequence {xn} ⊂ Rp has a convergent subsequence (that converges to a point in Rp.)
(2d) Theorem 1.20: {xn} is a Cauchy sequence in Rp iff {xn} is convergent in Rp. [Note:
You should check that {xn} is a Cauchy sequence in Rp =⇒ {xn} is bounded in Rp]
Exercise: explain how (2a)−(2d) indicate that Rp, equipped with the Euclidean distance (metric), has the completeness property if and only if every Cauchy sequence {xn} is convergent in Rp. [e.g. Let xndenote the n digit decimal representation of √
2, x1 = 1, x2 = 1.4, x3 = 1.41, · · · . Note that xn is a Cauchy sequence in Q with lim
n→∞xn = √
2 /∈ Q, i.e. The Cauchy sequence {xn} converges but it does not converge in Q, so Q does not have completeness property. ]
(3) Let S be a compact subset of Rp, f : S → Rq be a continuous map. Then f is uniformly contin- uous on S, i.e. ∀ > 0, ∃ δ = δ() > 0, such that for any x, y ∈ S with kx − yk < δ we have kf (x) − f (y)k < . Compare this with f is continuous on S, i.e. f is continuous at each x ∈ S, ∀ >
0, and for each x ∈ S, ∃ δ = δ(, x) > 0, such that for any y ∈ S with ky − xk < δ we have kf (y) − f (x)k < . [e.g. Let S = (0, 1), f (x) = 1
x. Then for each > 0, x ∈ (0, 1) =⇒ x
2 > 0, since
|f (y) − f (x)| = |y − x|
yx < 2|y − x|
x2 , we may choose δ = min{x2
2 , 1}, such that if |y − x| < δ then
|f (x) − f (y)| < 2|y − x|
x2 < 2δ
x2 ≤ . Note that δ here varies according to x, the continuity of f is Not uniform on (the noncompact set) (0, 1).]
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