Real Analysis Cheat Sheet

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

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 convergent 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. P Convergence of ∞ Series ak = a → lim sn = a P Cauchy Criteria for Series ak converges iff, given ε > 0, ∃N ∈ N s.t. n > m ≥ N implies |am+1 + am+2 + . . . + an | < ε

1

Comparison Test Assume 0 ≤ ak ≤ bk . ∀k ∈ N P P • if |bk | converges, then ak converges P P • if |ak | diverges, then bk diverges P P Absolute Convergence If |ak | converges, so does 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 F C 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. T∞ Nested Compact Sets If K1 ⊃ K2 ⊃ . . . are compact sets, then n=1 6= ∅ T Def. of Open Cover Given a set A ⊂ R, and open cover is a collection {Oλ : λ ∈ Λ} of open sets s.t. A ⊂ λ∈Λ Oλ

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 f (x) = f (c) x→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 f (x) does not exist. x→c

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 2

5

The Derivative

Def. of Derivative Let g : A → R. Then g 0 (c) = lim g(x)−g(c) x−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 = g 0 (f (c))f 0 (c) Darboux’s Theorem If f is differentiable on [a, b] and f 0 (a) < α < f 0 (b), then ∃c ∈ (a, b) with f 0 (c) = α Mean Value Theorem If f : [a, b] → R is continuous and differentiable, then ∃c ∈ (a.b) s.t. f 0 (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)]g 0 (c) = [g(b) − g(a)]f 0 (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 fn (x) = n→∞

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 f 0 • 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 f 0 = 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 f 0 = g Pk Def. of Convergence for Series of Functions Let fk (x) = n=1 fn (x) P • fn (x) converges pointwise if fk (x) converges pointwise P • fn (x) converges uniformly if fk (x) converges uniformly P Theorem Let fn be continuous on A ⊂ R and assume fn converges to f on A. Then f is continuous on A. P Cauchy Criteria for Series of Functions fn converges uniformly on A iff ∀ε > 0, ∃N ∈ N s.t. n, m ≥ N and x ∈ A implies |sn (x) − sn (x)| < ε P Weierstrass M-testP Let (fn ) be a sequences of functions on ⊂ R. Let Mn > 0 satisfy |fn (x)| ≤ Mn ∀x ∈ A. If Mn converges, then fn converges uniformly. Power series

∞ X

an xn

n=0

• If a power series converges for x0 ∈ R, then it converges absolutely if |x| < |x0 | P • If an xn converges absolutely at a point x0 , then it converges uniformly on [−c, c] with c = |x0 | P∞ Abel’s Theorem Let g(x) = n=0 an xn 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. P P Differentiated Series If an xn converges ∀x ∈ (−R, R) then the differentiated series an nxn−1 converges at each x ∈ (−R, R) Taylor Series f (x) =

∞ X

an xn with an =

f (n) (0) n!

n=0

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 (N +1)! x

Def. of Equicontinuous (fn ) is equicontinous on A if ∀ε > 0 ∃δ > 0 s.t. ∀n ∈ N we have that |x − y| < δ implies |fn (x) − fn (y)| < ε 3

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 ]} Pn Pn • Then L(f, P ) = k=1 mk ∆xk and U (f, P ) = k=1 Mk ∆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 Pε s.t. U (f, Pε ) − L(f, Pε ) < ε Properties of the Integral Assume a ≤ y < x ≤ b, and c ∈ R Rx Ry Rx • a f− a f= y f Rb Rb • | a f | ≤ a |f | Rb • a c = c(b − a) Fundamental Theorem of Calculus Rb • If f : [a, b] → R is integrable and F 0 (x) = f (x) ∀x ∈ (a, b), then a f = F (b) − F (a) Rx • Let g : [a, b] → R be integrable and define G(x) = 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{ n1 : 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. Thus, 0 is greatest lower bound.

8.2

Limit of a Series

Verify that lim √ n→∞

2 = 0. n+3

Let ε > 0 2 Then | √n+3 − 0| = √

√2 n+3

ε ⇒ n+3 √2 ⇒ n + 3 > 2ε ⇒ n > ε42 − 3 Choose N = [ ε42 ] + 1 p 2 Then |2/ 4/ε2 − 0| = | 2/ε |=ε 2 So for n ≥ N , | √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). Since xn ≤ yn ≤ zn , this implies yn ∈ (l − ε, l + ε). So (yn ) → l. 4

1 n

< x by Archimedean Property.

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) = 3 x is continuous at c = 0. Let ε > 0. Choose δ = ε3 . 1 Then 0 < |x| < δ implies |x 3 | < 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 ε < + =ε 2 2 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 nx 1 + nx2 x x = → 2 as n → ∞ 1/n + x2 x

fn (x) =

So the pointwise limit of fn on (0, ∞) is x1 To show uniform convergence, we must find N ∈ N s.t. n ≥ N implies |fn (x) − f (x)| < ε nx 1 − | 1 + nx2 x nx2 − 1 − nx2 =| | x + nx3 1 = 0 be arbitrary. We must show that ∃N ∈ N s.t. n ≥ N implies |fn g − f g| < ε ∀x ∈ A. Since fn converges uniformly, we can find an N s.t. |fn − f | < ε/M ∀n ≥ N . Then, |fn g − f g| = |g||fn − f | ≤ M |fn − f | ε < M( ) = ε M 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 fn converges 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 P > 0 satisfy |fn (x)| ≤ Mn ∀x ∈ A. Assume Mn converges. Let ε > 0 We must P show that |fm+1 (x) + . . . + fn (x)| < ε Since 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 P So 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

(Mk − mk )∆x

k=1

= ∆x

n X

f (xk ) − f (xk−1 )

k=1

= ∆x(a − b) Given ε > 0, choose a partition Pε with equal subintervals with common length satisfying ∆x < U (f, Pε ) − L(f, Pε ) = ∆x(b − a) ε < (b − a) b−a =ε

6

ε b−a .

Then