MTH265 Final
Mattia Janigro
20 December 2013
1
The Real Numbers
Axiom of Completeness Every nonempty set of R that is bounded above has a supremum. Def. of Sup/Inf
• s (i) is an upper (lower) bound • if b is any bound, then s ≤ b (i ≥ b) Lemma
• s = supA iff ∀ε > 0 ∃ a ∈ A s.t. s − ε < a • i = infA iff ∀ε > 0 ∃ a ∈ A s.t. i + ε < a Def. of Bounds
• A ⊂ R is bounded above if ∃ b ∈ R s.t. a ≤ b ∀a ∈ A • A ⊂ R is bounded below if ∃ b ∈ R s.t. b ≤ a ∀a ∈ A Archimedean Property • Given x ∈ R , ∃n ∈ N s.t. n > x • Given y ∈ R with y > 0, ∃n ∈ N s.t. 1 n < y Triangle Inequalities • |a + b| ≤ |a| + |b| • |a − b| ≤ |a| + |b| • ||a| − |b|| ≤ |a − b|
Nested Interval Property Consider intervals In= [an, bn] = {x ∈ R : an≤ x ≤ bn} where I1⊃ I2⊃ . . .
ThenT∞
n=1In6= ∅
Def. of Equality a, b ∈ R are equal iff ∀ε > 0 it follows that |a − b| < ε
2
Sequences and Series of Real Numbers
Def. of Limit (an) → a if ∀ε > 0∃N ∈ N s.t. n ≥ N implies |an− a| < ε
Monotone Convergence Theorem If a sequence is monotone and bounded, then it converges.
Bolzano-Weierstrass Theorem Every sequence that is bounded has a convergent subsequence. Subsequences of a con-vergent sequence converge to the same limit.
Cauchy Criteria for Sequences (an) is a Cauchy sequence is a Cauchy sequence if ∀ε > 0, ∃N ∈ N s.t. m, n ≥ N implies
|am− an| < ε. And, a sequence converges iff it is Cauchy.
Convergence of ∞ Series P ak= a → lim sn = a
Comparison Test Assume 0 ≤ ak≤ bk. ∀k ∈ N
• ifP |bk| converges, thenP ak converges
• ifP |ak| diverges, thenP bk diverges
Absolute Convergence IfP |ak| converges, so does P ak. If a series converges but does not converge absolutely, we say
it converges conditionally.
Alternating Series Alternating series which are decreasing to 0 converge.
3
Topology of R
Def. of Open Set A set O is open if ∀x ∈ O, ∃ε > 0 with Vε(x) ∈ O
Union and Intersection of Sets
• The union of arbitrary collection of open sets is open. • The finite intersection of open sets is open.
Def. of Limit Point x is a limit point of a set A ⊂ R if every nbhd Vε(x) contains points of A other than x.
Alternate x is a limit point of A iff ∃(an) ⊂ A with x /∈ (an) and lim an= x
Def. of Isolated Point a is an isolated point of A if a ∈ A but a is not a limit point.
Def. of Closed Set F ⊂ R is closed if it contains all of its limit points.
Alternate F is closed iff every Cauchy sequence of points in F has a limit in F .
Def. of Closure Let A ⊂ R, and let L be the set of all limit points of A. Then ¯A = A ∪ L is the closure of A. Closure Theorem If A ⊂ R, then ¯A is closed and is the smallest closet set containing A.
Complements of Sets
• O ⊂ R is open iff OC is closed.
• F ⊂ R is closed iff FC is open.
Def. of Compact Set A set K is compact iff every sequence of points in K has a convergent subsequence with a limit in K. A set is closed and bounded iff it is compact.This also implies that the set has a finite open cover.
Nested Compact Sets If K1⊃ K2⊃ . . . are compact sets, then T∞n=16= ∅
Def. of Open Cover Given a set A ⊂ R, and open cover is a collection {Oλ: λ ∈ Λ} of open sets s.t. A ⊂Tλ∈ΛOλ
4
Functional Limits and Continuity
Def. of Continuity A function f is continuous at c if ∀ε > 0, ∃δ > 0 s.t. |x − c| implies |f (x) − f (c)| < ε Alternate lim
x→cf (x) = f (c)
Alternate 2 If (xn) → c with xn∈ A for all n, then f (xn) → f (c)
Divergence Criteria Let f : A → R, and let c be a limit point of A. If ∃(xn) → c and (yn) → c but lim f (xn) 6= lim f (yn),
then lim
x→cf (x) does not exist.
Composition of Functions If f is continuous at c ∈ A and g is continuous at f (c) ∈ B, then g ◦ f (x) = g(f (x)) is continuous at x = c
Compactness of Compositions Let f : A → R be continuous on A. If K ⊂ A is compact, then f (K) is compact as well. Extreme Value Theorem If f is continuous on a compact set K, then f is bounded and attains its minimum and maximum
on K.
Def. of Uniform Continuity f : A → R is uniformly continuous if ∀ε > 0 ∃δ > 0 s.t. |x − y| < δ implies |f (x) − f (y)| < ε Continuity on Compact Sets A function f which is continuous on a compact set K is also uniformly continuous on K.
Intermediate Value Theorem Let f : [a, b] → R be continuous and let L satisfy f (a) < L < f (b). Then ∃c ∈ (a, b) with f (c) = L
5
The Derivative
Def. of Derivative Let g : A → R. Then g0(c) = lim
x→c
g(x)−g(c) x−c
If g is differentiable at c ∈ A, then g is continuous at c. The converse of this is false.
Chain Rule Let f : A → R, g : B → R where f (A) ⊂ B. If f is differentiable at c ∈ A and g is differentiable at f (c), then [g(f (c))]0= g0(f (c))f0(c)
Darboux’s Theorem If f is differentiable on [a, b] and f0(a) < α < f0(b), then ∃c ∈ (a, b) with f0(c) = α Mean Value Theorem If f : [a, b] → R is continuous and differentiable, then ∃c ∈ (a.b) s.t. f0(c) =f (b)−f (a)b−a General MVT Assume f and g satisfy the assumptions of the Mean Value Theorem. Then ∃c ∈ (a, b) s.t.
[f (b) − f (a)]g0(c) = [g(b) − g(a)]f0(c)
6
Sequences and Series of Functions
Def. of Pointwise Convergence of Functions Given (fn), where f : A → R, fn→ f pointwise if ∀x ∈ A, lim
n→∞fn(x) =
f (x)
Def. of Uniform Convergence of Functions fn → f uniformly on A iff ∀ε > 0, ∃N ∈ N s.t. if n ≥ N, then ∀x ∈ A,
|fn(x) − f (x)| < ε
Cauchy Criteria for Sequences of Functions fn converges uniformly on A iff ∀ε > 0, ∃N ∈ N s.t. n > m ≥ N and
x ∈ A implies |fn(x) − fm(x)| = |fm+1(x) + fm+2(x) + . . . + fn(x)| < ε
Theorems Regarding fn, f , and f0
• If fn → f uniformly, and each fn is continuous, then f is continuous as well.
• Let fn→ f pointwise on [a, b] and assume each fn is differentiable. If (fn0) → g uniformly, then f is differentiable
and f0 = g on [a, b]
• Let (fn) be a sequence of differentiable functions on [a, b] and assume (fn0) → g uniformly. If ∃x0 ∈ [a, b] where
fn(x0) converges, then (fn) → f uniformly, and f0= g
Def. of Convergence for Series of Functions Let fk(x) =P k
n=1fn(x)
• P fn(x) converges pointwise if fk(x) converges pointwise
• P fn(x) converges uniformly if fk(x) converges uniformly
Theorem Let fn be continuous on A ⊂ R and assumeP fn converges to f on A. Then f is continuous on A.
Cauchy Criteria for Series of Functions P fn converges uniformly on A iff ∀ε > 0, ∃N ∈ N s.t. n, m ≥ N and x ∈ A
implies |sn(x) − sn(x)| < ε
Weierstrass M-test Let (fn) be a sequences of functions on ⊂ R. Let Mn > 0 satisfy |fn(x)| ≤ Mn ∀x ∈ A. If P Mn
converges, then P fn converges uniformly.
Power series
∞
X
n=0
anxn
• If a power series converges for x0∈ R, then it converges absolutely if |x| < |x0|
• IfP anxn converges absolutely at a point x0, then it converges uniformly on [−c, c] with c = |x0|
Abel’s Theorem Let g(x) =P∞
n=0anxn converge at x = R > 0. Then the series converges uniformly on [0, R].
Corollary If a power series converges pointwise on A ⊂ R, then it converges uniformly on any compact subset of A. Differentiated Series If P anxn converges ∀x ∈ (−R, R) then the differentiated series P annxn−1 converges at each
x ∈ (−R, R) Taylor Series f (x) = ∞ X n=0 anxn with an= f(n)(0) n!
Lagrange Remainder Theorem Givenx 6= 0 and x ∈ (−R, R), ∃c with |c| < |x| s.t. ∀n ∈ N, En(x) = f
N +1(c)
(N +1)!x N +1
Def. of Equicontinuous (fn) is equicontinous on A if ∀ε > 0 ∃δ > 0 s.t. ∀n ∈ N we have that |x − y| < δ implies
7
The Riemann Integral
Def. of Upper and Lower Sums Consider an interval [a, b] with partition P = {x0, x1, . . . , xn}.
• For each subinterval [xk−1, xk], let mk = inf{f (x) : x ∈ [xk−1, xk]} and Mk = sup{f (x) : x ∈ [xk−1, xk]}
• Then L(f, P ) =Pn
k=1mk∆xk and U (f, P ) =P n
k=1Mk∆xk
• U (f ) = inf{U (f, P ) : P ∈ P} and L(f ) = sup{L(f, P ) : P ∈ P} Def. of Refinement Q is a refinement of P if P ⊂ Q.
Lemma If P1, P2 are partitions of [a, b], then L(f, P1) ≤ U (f, P2)
Def. of Integrable f is Riemann integrable on [a, b] if U (f ) and L(f ) exist and are equal.
Alternate A bounded function f on [a, b] is integrable iff ∀ε > 0, there exists a partition Pεs.t. U (f, Pε) − L(f, Pε) < ε
Properties of the Integral Assume a ≤ y < x ≤ b, and c ∈ R • Rx a f − Ry a f = Rx y f • |Rb af | ≤ Rb a |f | • Rb a c = c(b − a)
Fundamental Theorem of Calculus
• If f : [a, b] → R is integrable and F0(x) = f (x) ∀x ∈ (a, b), thenRb
a f = F (b) − F (a)
• Let g : [a, b] → R be integrable and define G(x) =Rx
a g for x ∈ [a, b]. Then G is continuous and if g is continuous
at c ∈ [a, b], then G is differentiable at c with G0(c) = g(c) Integrable Functions f is integrable if one of the following holds:
• f is bounded and has only a finite number of discontinuities • f is continuous
• f is monotone
8
Exercises
8.1
Infimum of a Set
Show that i = inf{1
n : n ∈ N} = 0
Clearly 0 is a lower bound because 1/n is positive.
If x ∈ R is a lower bound for {n1 : n ∈ N}, then we have x ≤ 0 because ∀x > 0 ∃n ∈ N s.t. 1
n < x by Archimedean Property.
Thus, 0 is greatest lower bound.
8.2
Limit of a Series
Verify that lim
n→∞ 2 √ n + 3 = 0. Let ε > 0 Then |√2 n+3− 0| = 2 √ n+3< ε ⇒ √ n+3 2 > ε ⇒√n + 3 > 2ε ⇒ n > 4 ε2 − 3 Choose N = [ε42] + 1 Then |2/p4/ε2− 0| = | 2 2/ε| = ε So for n ≥ N , |√2 n+3− 0| < ε
8.3
Squeeze Theorem
Show that if xn≤ yn ≤ zn∀n ∈ N, and lim xn= lim zn= l, then lim yn = l.
We have that ∃N ∈ N s.t. n ≥ N implies xn, zn ∈ (l − ε, l + ε).
8.4
Isolated Point
Let a ∈ A. Prove that a is an isolated point of A iff there exists a ε-nbhd Vε(a) s.t. Vε(a) ∩ A = {a}
⇒ Assume a ∈ A is an isolated point of A. Then a is not a limit point. So ∃Vε(a) s.t. Vε(a) ∩ A = ∅ or Vε(a) ∩ A = {a}.
Since a ∈ A, it must be that Vε(a) ∩ A = {a}.
⇐ Assume ∃Vε(a) s.t. Vε(a) ∩ A = {a}.
Then a is not a limit point. Therefore a is an isolated point by definition.
8.5
Continuity of a Function
Show that g(x) = √3x is continuous at c = 0.
Let ε > 0. Choose δ = ε3.
Then 0 < |x| < δ implies |x13| < ε.
8.6
Cauchy Criterion for Uniform Convergence
Prove the Cauchy Criterion for Uniform Convergence using the Cauchy Criterion for real numbers.
The Cauchy Criterion for real numbers states that (an) is Cauchy (and thus converges) if ∀ε > 0 ∃N ∈ N s.t. m, n ≥ N
implies |am− an| < ε.
⇒ Assume (fn) converges uniformly on A. Let f (x) = lim fn(x).
Since fn converges uniformly, we can find N ∈ N s.t. for ∀m, n ≥ N and x ∈ A we have
|fn(x) − fm(x)| = |fn(x) − f (x) − (fm(x) − f (x))| ≤ |fn(x) − f (x)| + |fm(x) − f (x)| < ε 2+ ε 2 = ε
So (fn) satisfies Cauchy criterion for sequences of functions.
⇐ Assume ∀ε > 0 ∃N ∈ N s.t. m, n ≥ N implies |fn(x) − fm(x)| < ε.
For a fixed x ∈ A, fn(x) is a Cauchy sequence of real numbers.
So fn(x) converges, and we can say that f (s) = lim fn(x).
For m, n ≥ N we have |fn(x) − fm(x)|ε
By Algebraic Limit Theorem, we can say that as m → ∞, |fn(x) − fm(x)| → |fn(x) − f (x)|
So then for n ≥ N , |fn(x) − f (x)| < ε
Thus (fn) converges uniformly.
8.7
Pointwise and Uniform Convergence of a Function
fn(x) = nx 1 + nx2 = x 1/n + x2 → x x2 as n → ∞
So the pointwise limit of fn on (0, ∞) is 1x
To show uniform convergence, we must find N ∈ N s.t. n ≥ N implies |fn(x) − f (x)| < ε
fn(x) − f (x)| = | nx 1 + nx2− 1 x| = |nx 2− 1 − nx2 x + nx3 | = 1 x + nx3 < ε
So N = (1ε− x)x−3. This depends on x and tends toward ∞ as x → 0, so f
8.8
Conjectures on Sequences of Functions
i) If fn → f pointwise on a compact set K, then it is not necessarily true that fn→ f uniformly on K.
Consider fn(x) = xn, which converges pointwise on [0, 1] but not uniformly.
ii) If fn → f uniformly on A and g is a bounded function on A, then fng → f g uniformly on A.
Let |g(x)| ≤ M and ε > 0 be arbitrary. We must show that ∃N ∈ N s.t. n ≥ N implies |fng − f g| < ε ∀x ∈ A.
Since fn converges uniformly, we can find an N s.t. |fn− f | < ε/M ∀n ≥ N . Then,
|fng − f g| = |g||fn− f |
≤ M |fn− f |
< M ( ε M) = ε
iii) If fn → f uniformly on A, and if each fn is bounded on A, then f must also be bounded.
Let |fn| ≤ M and ε > 0 be arbitrary. Since fnconverges uniformly, we have that for some N , n ≥ N implies |fn− f | < ε.
Then |f (x)| ≤ M + ε on A, and hence f is bounded.
iv) If fn → f uniformly on a set A, and if fn→ f uniformly on a set B, then fn→ f uniformly on A ∪ B.
Let ε > 0 be arbitrary. Because fn→ f uniformly on A we can find N1 s.t. n ≥ N1 implies |fn− f | < ε ∀x ∈ A.
Similarly, we can find N2 s.t. n ≥ N2 implies |fn− f | < ε ∀x ∈ b.
Then, if we take N = max{N1, N2}, it follows that n ≥ N implies |fn− f | < ε ∀x ∈ A ∪ B.
v) If fn → f uniformly on an interval I, and if each fn is increasing, then f is also increasing.
Consider x, y ∈ I. Since fn is increasing, we have that x < y implies fn(x) < fn(y).
Since fn(x) → f (x) and fn(y) → f (y), by order limit theorem it follows that x < y implies f (x) < f (y). Thus f is
increasing.
8.9
Proof of Weierstrass M-test
Let Mn > 0 satisfy |fn(x)| ≤ Mn∀x ∈ A.
AssumeP Mn converges. Let ε > 0
We must show that |fm+1(x) + . . . + fn(x)| < ε
SinceP Mn converges, ∃N s.t. n > m ≥ N implies that |Mm+1+ . . . + Mn| < ε by Cauchy criterion for real numbers.
Then we have |fm+1(x) + . . . + fn(x)| ≤ |Mm+1+ . . . + Mn| < ε
SoP fn converges uniformly.
8.10
Increasing Functions are Integrable
Let P be a partition where all the subintervals have equal length ∆x = xk− xk−1.
Since f is increasing, then Mk= f (xk) and mk = f (xk−1). Thus
U (f, P ) − L(f, P ) = n X k=1 (Mk− mk)∆x = ∆x n X k=1 f (xk) − f (xk−1) = ∆x(a − b)
Given ε > 0, choose a partition Pεwith equal subintervals with common length satisfying ∆x < b−aε . Then
U (f, Pε) − L(f, Pε) = ∆x(b − a)
< ε
b − a(b − a) = ε
