Tyler Clark Spring 2012 MATH 532 - Real Analysis
1
Ackn Ackno owled wledgm gmen ents ts
I would like to thank my professor, Dr. Ferhan Atici for teaching this course. The notes to follow are based upon the course she taught. Texts
• Bartle, R. G., & Sherbert, D. R. “Introduction to Real Analysis, 4th ed.” John Wiley & Sons Inc: New York, NY. 2011.
• Bartle, R. G. “The Elements of Integration and Lebesgue Measure.” John Wiley & Sons, Inc: New York, NY. 1966.
Typed notes prepared in 2012 by Tyler Clark 1
2 2.1
Course Notes Metric Spaces
Definition 1. The metric space M is a complete metric space if every Cauchy sequence is
convergent in M . Example 1.
R is
a complete metric space
Example 2. C [0, 1] is a complete metric space with d(f, g ) = sup | f (t) − g (t) | , t ∈ [0 , 1]. Theorem 1. Any closed subset A of a complete metric space M is complete.
Proof. Let A ⊂ M be closed and < xn > be any Cauchy sequence in A. Since M is a complete metric space, < xn > is convergent in M , so xn → x ∈ M . Since A is closed, x ∈ A . Thus, A is complete. Definition 2. Let < M,ρ > be a metric space. If T : M → M , we say that T is a
contraction on M if there exists α ∈ (0 , 1) such that ρ (tx, ty) ≤ αρ (x, y), x, y ∈ M . Theorem 2 (Contraction Mapping Theorem) . Let < M,ρ > be a complete metric space.
If T is a contraction on M , then there exists one and only one x ∈ M such that T x = x . Proof. Note that ρ(T 2 x, T 2 y ) = ρ (T (T (x)), T (T (y )))
≤ αρ (T x , T y ) ≤ α 2 ρ(x, y). One can show that ρ(T n x, T n y ) ≤ α n ρ(x, y) for every n. Let x0 ∈ M and define < xn > in M as follows. x1 = T x0 , x2 = T 2 x0 = T x1 , . . . , xn = T n x0 = T n−1 x1 , . . . Claim 1. < xn > is a Cauchy sequence in M .
2
Proof. Let ε > 0 be given. Let us consider ρ (xn , xm ). For n < m there esists p such that ρ(xn , xm ) = ρ (xn , xn+ p )
≤ ρ (xn , xn+1 ) + ρ(xn+1 , xn+2 ) + . . . + ρ(xn+ p−1 , xn+ p ) = ρ (T n x0 , T n x1 ) + ρ(T n+1 x0 , T n+1 x1 ) + . . . + ρ(T n+ p−1 x0 , T n+ p−1 x1 )
≤ α n ρ(x0 , x1 ) + αn+1 ρ(x0 , x1 ) + . . . + αn+ p−1 ρ(x0 , x1 ) = ρ (x0 , x1 )αn [1 + α + α2 + . . . + α p−1 ]
≤ ρ (x0 , x1 )αn [1 + α + α2 + . . . + α p−1 + α p + . . .] = ρ (x0 , x1 )α = ρ (x0 , x1 )
n
1 1−α
αn
1−α
(since lim αn = 0)
<ε
n
→∞
Thus, < xn > is Cauchy. M is complete implies < xn > → x ∈ M . Since T is a continuous function, we have that
T lim xn = T x =⇒ lim T xn = T x . n
→∞
→∞
n
(1)
Furthermore, consider < T xn > such that < T xn > is a subsequence of < xn > . Since xn → x 1 , we have
limn→∞ T xn = x.
Equations 1 and 2 imply that T x = x showing existence of solution. To show uniqueness, assume there are two fixed points, y , z ∈ M such that y = z . Thus, T y = y and T z = z . Thus we have that
0 = ρ (y, z ) = ρ (T y , T z ) ≤ αρ (y, z ) = 0. This implies that 1 ≤ α which is a contradiction. Example 3. Consider T (x) = x 2 for 0 < x ≤ 13 . Prove that T is a contraction on (0 , 13 ],
but T has no fixed point. Let ρ(x, y ) =| x − y | and ρ(T x , T y) = ρ (x2 , y 2 ) =| x 2 − y 2 |. We need to show that ρ(T x , T y) ≤ α | x − y | . That is,
2
2
|x −y | ≤ α . |x−y| 3
(2)
Consider T (x) = x 2 on(0, 13 ] with x, y ∈ (0 , 13 ] on [x, y]. Then we have that T is continuous on [x, y] and differentiable on ( x, y). By the Mean Value Theorem, there exists c ∈ ( x, y) such that
( ) =
T c
T (y ) − T (x) 2 =⇒ | T (c)| = 2c ≤ . y−x 3
Thus, we have that α ∈ (0 , 1) and T (x) = x 2 is a contraction for 0 < x ≤ 13 . For T to have a fixed point, we need x2 = x . This implies that x2 − x = 0 =⇒ x (x − 1) = 0 =⇒ x = 0, 1 ∈ (0 , 13 ]. Thus, T does not have a fixed point. Theorem 3. Let f (t, y ) be continuous and satisfy a Lipschitz condition with Lipschitz
constant K on Ω = {(t, y ) : |t − t0 | ≤ a, |y − y0 | ≤ b }. Let M be a number such that b , a}. Then there exists a unique |f (t, y)| ≤ M for (t, y) ∈ Ω. Choose 0 < α < min{ k1 , M
solution of,
Exercise 1. Define
y = f (t, y ), |t − t0 | ≤ α y (t0 ) = y 0 t
T (φ)(t) := y 0 +
f (s, φ(s) ds.
t0
Show that T is a contraction. Definition 3. A subset K of a metric space M is said to be compact if every open cover of
K contains a finite subcover.
Recall exercise 1.
t
T (φ(t)) = y 0 +
f (x, φ(s)) ds
t0
T : C [0, 1] → C [0, 1] φ → T φ : [0, 1] →
R t
t → T φ(t) = y 0 +
f (s, φ(s)) ds
t0
T (φ) = ρ (T (φ(t)), T (ψ (t)) ≤ αρ (φ(t), ψ (t))
=⇒ |T (φ(t)) − T (ψ(t))| ≤ α |φ(t) − ψ(t)| 4
Note 1. The sup-norm is defined as d(f, g ) = f − g = sup |f (t) − g (t)|, t ∈ [0 , 1]. Note 2. T is an operator
Recall open covers. Theorem 4. A closed subset of a compact set in a metric space M is compact.
Proof. Let F ⊂ K such that K is compact and F is closed and {Gα } be an open cover for the set F . Therefore, F ⊂
Gα . Furthermore, K = F ∪ F c ⊂
α
Gα ∪ F c where F is
α
closed and F c is open. Since K is compact, it has a finite open subcover. Then n
F ∩ K
⊂ (
n c
Gα ∪ F ) ∩ F =⇒ F i
i=1
Theorem 5.
R is
⊂ (
n
) ⊂
Gα ∩ F i
i=1
Gα . i
i=1
not a compact metric space.
Proof. Let Gn = (−n, n), n ∈
N,
R
⊂
Gn .
∈
n N
Let x ∈
R.
There exists n ∈
N such
that n · 1 > ( x). Therefore, − n < x < n , x ∈ G n . Now N
we assume that
Gn has a finite subcover so that
⊂
R
Gn . Choose x ∈ i
R such
that
i=1
∈
n N N
x > max {n1 , n2 , . . . , nN }. This implies that x
∈
Gn . Thus we get a contradiction. i
i=1
Theorem 6. If K is compact, then K is bounded. Theorem 7. If K is compact, then K is closed. Theorem 8 (Heine-Borel Theorem) . A subset K of R is compact iff K is closed and
bounded. Example 4. [a, b] is compact in
R.
Corollary 1. If F is closed and K is compact, then F ∩ K is compact (intersection of
closed sets is closed). Theorem 9. Let < M,ρ > be a compact metric space. If f is a continuous function from
M into a metric space M ∗ , then f is uniformly continuous on M .
5
Theorem 10. Let f : [a, b] →
R be
continuous. Then f is uniformly continuous.
Theorem 11. Let f be a continuous function from the compact metric space M to M 1 .
Then the range f (M ) is also compact in M 1 . Theorem 12 (****). Let f : [a, b] →
R be
continuous. Then f achieves its maximum and
minimum values.
2.2
Sequences of Functions
Let < xn > be a sequence of real numbers We have x n → x for lim xn = x ∈
→∞
n
Example 5. lim
1
→∞ 2
n
R.
n
=0
Let < f n > be a sequence of functions defined such that f 1 : R → R x → f 1 (x) f 2 : R → R x → f 2 (x)
.. . Note 3. If we let f n be a constant for all n , we can get such that f 1 (n) = 12 , f 2 (n) =
f 3 (n) =
1 , . 23
1 , 22
. . which generates example refex2. Thus, a sequence of numbers is just the
special case of a sequence of functions. Example 6. f 1 (x) = x , f 2 (x) = x 2 , f 3 (x) = x 3 , . . . =⇒ < xn >.
For < f n >→ f , we can have this converging pointwise or uniform. Definition 4. Pointwise Convergence: Let < f n > be a sequence of functions. f n → f
pointwise on E ⊂ R: f 1 : E → R, f 2 : E → there exists an n ∈
N such
R, .
. .. Let x ∈ E and ε > 0 be given. Then
that |f n (x) − f (x)| < ε whenever n ≥ N .
6
Example 7. Let < f n (x) = (1 − |x|)n > on (0, 1). For x ∈ (0 , 1), then
(1 − |x|)n = (1 − x)n , lim (1 − x)n = 0 since (1 − x)n < 1.
→∞
n
Claim 2. (1 − |x|)n → 0 pointwise on (0, 1) \ {0}. Note 4. |(1 − |x|)n − 0| = | (1 − |x|)n | ≤ (1 − |x|)N = ε for n ≥ N .
ln(1 − |x|)N = ln (ε) =⇒ N =
ln(ε) ln (1 − |x|)
ln(ε) . Then we have ln(1 − |x|) |(1 − |x|)n − 0| = | (1 − |x|)n | < (1 − |x|)N = ε whenever n ≥ N .
Proof. Let ε > 0 and x ∈ (0 , 1) be given. Choose N >
xn+1 = L exists. n→∞ xn
Theorem 13. Assume all sn = 0 and lim
• If L < 1, lim S n = 0.
→∞
n
• If L > 1, lim S n = +∞.
→∞
n
Example 8. Let < f n (x) = nx n > for x ∈ (0 , 1). Since lim
(n + 1)xn+1
→∞
n
have lim nxn = 0.
nxn
= x < 1 , we
→∞
n
Claim 3. < nxn >→ 0 pointwise on (0, 1).
Let ε > 0 and x ∈ (0 , 1) be given. Then there exists r ∈
(n+1)xn+1 nxn
→ , choose − . x
(n+1)xn+1 nxn
R such
that x < r < 1. Since
ε = r − x. Then there exists N ∈ N such that
x <ε
Let k = N + 1. Then for all n > k we have n − 1 > N , so that
(n+1)xn+1 nxn
< x + ε = r .
It follows that for all n > k , 0 < nxn < ∗ n − 1)xn−1 r < ( n − 2)xn−2 r2 < .. . < kxk rn−k . Let M =
kx k rk
. Then we obtain 0 < nxn < M rn for all n > k . Since rn → 0, we can say
that there exists N ∗ ∈
N such
ε that | nxn − 0| < M r n < M M = ε for n ≥ N ∗ .
7
Example 9. Let f n (x) = n1 sin(nx), x ∈
lim
1
→∞ n
n
R.
By the squeeze theorem, we know that
sin(nx) = 0.
Claim 4. Note 5.
Let x ∈
1 n
sin(nx) → 0 pointwise on R.
1 We have sin(
R and ε
n
nx) − 0| =
1 n
sin(
1
nx)| <
n
≤
1
< ε whenever n ≥ N .
N
> 0 be given. Choose N ε > 1.
Definition 5. Uniform Convergence: Let < f n > be a sequence of functions. We say
f n → f uniformly on E ⊂ R if given ε > 0, there exists N ∈ N such that | f n (x) − f (x)| < ε
whenever n ≥ N and x ∈ E . Example 10. Recall Example 7. Does < f n (x) > → 0 uniformly?
Let ε =
1 2
and n > N =⇒ (1 − x)n < 12 . Consider (1 − x)N =
1 2
=⇒ 1 − x
1 = 2 1
=⇒ x = 1 − Thus, we should choose x = 1 −
1 2
1 N +1
1 N +1
2
1 N +1
. Then we have that
(1 − x)N +1 = [1 − (1 − 2−
1 N +1
)]N +1 =
1 1 < . 2 2
Thus, < f n (x) > does not converge uniformly to 0. Example 11. Recall Example 8. Does < f n (x) > → 0 uniformly?
Let ε = 1 and n > N =⇒ (N + 1)xN +1 < 1. Consider (N + 1)xN +1 = 1 =⇒ x N +1 = =⇒ x = 8
1 N + 1
1 N + 1
1 N +1
Thus, we should choose x =
1 N +1
1 N +1
. Then we have that
(N + 1)xN +1 = (N + 1)
1 N + 1
N +1 N +1
= 1 < 1 .
Thus, < f n (x) > does not converge uniformly to 0. Theorem 14 (Cauchy Criterion for Uniform Convergence) . The sequence of functions
< f n > defined on E converges uniformly on E if and only if for ε > 0, there exists N ∈ N
such that |f n (x) − f m (x)| < ε whenever n, m ≥ N and x ∈ E . Theorem 15 (**). Suppose lim f n (x) = f (x) for x ∈ E . Let M n = sup |f n (x) − f (x)|.
→∞
n
∈
x E
Then f n → f uniformly if and only if M n → 0 as n → ∞. Example 12. Let f n (x) =
x for x ∈ 1 + nx2
R.
Prove that < f n > is uniformly convergent.
We have that lim f n (x) = 0. Then
→∞
n
( )| = sup ∈ 1+
sup |f n (x) − f x
∈
x R
Consider g (x) =
x 1+nx2
x R
, then g (x) =
x x . 0 = sup − 2 nx2 x∈R 1 + nx
(1+nx2 ) xn2x (1+nx2 )2
−
= 0. This implies that x = ± √ 1n . We have
1 1 that f ( √ 1n ) = 2√ amd f (− √ 1n ) = − 2√ . n n 1 Thus we have that M n = supx∈R | 1+xnx2 | = 2√ . We have that M n → 0 implies that n
< f n > → 0 uniformly. Example 13. Let f n (x) = n 2 xn (1 − x) for x ∈ [0 , 1].We have that lim f n (x) = 0. Thus,
→∞
n
f n (x) → 0 pointwise. Let
M n = sup |n2 xn (1 − x) − 0| = sup |n2 xn (1 − x)| = max n2 xn (1 − x). x [0,1]
x [0,1]
∈
x [0,1]
∈
∈
Note 6. Since [0, 1] is compact, we can look at the max instead of the sup .
We then have g (x) = n 2 xn (1 − x) =⇒ g (x) = n 3 xn−1 (1 − x) − n2 xn . Thus we have x = 0, (n − nx) =⇒ x = 0,
maxx∈[0,1] n2 xn (1 − x) = n 2 Now, lim M n = lim
→∞
n
n
n2
n
. Thus, 1− 1+
1 + n
n
n
n
n
→∞ n + 1 n + 1
n n + 1
= M n .
n
= ∞ =⇒< f n > is not uniformly convergent. 9
Theorem 16. Supponse < f n > is a sequence of functions defined on E and that
unifromly on E .
Proof. Suppose
=
|f n (x)| ≤ M n for x ∈ E and n = 1, 2, . . .. If
M n converges, then
f n converges
m
M n converges, then < S m
M n > converges. Then < S m > is a
n=1
Cauchy sequence.
Let ε > 0 be given. There exists a natural number N ∈ N such that | S n − S m | < ε whenever n > m ≥ N . Consider
− = n
m
n
≤ | n
n
|≤ converges uniformly. . Thus, Theorem 14 implies that f k
f k
k=1
k=1
f k
k =m
f k
M k < ε,
k=m
whenever n > m ≥ N
k =m
f n
Theorem 17 (****). If < f n > is a sequence of continuous functions on E and if f n → f
uniformly on E , then f is continuous on E . Example 14. Let f (x) = x n for x ∈ [0 , 1]. Then,
lim xn
→∞
n
0 = 1
0 ( )= 1
, x ∈ [0 , 1)
=⇒ f x
, x = 1
, x ∈ [0 , 1) , x = 1.
Thus, f is not continuous on E and f n → f is not uniform on E . Theorem 18. Suppose K is compact and
1. < f n > is a sequence of continuous functions on K , 2. < f n > converges to a continuous function f on K pointwise, and 3. f n (x) ≥ f n+1 (x) for all x ∈ K and n = 1, 2, . . .. Then < f n > → f uniformly on K . Note 7. R[a, b] denotes the set of Riemann Integrable functions on [ a, b]. Theorem 19. Supponse f n ∈ R[a, b] for n = 1, 2, . . . and f n → f uniformly on [a, b]. Then
f ∈ R[a, b] and
b
lim
→∞
n
a
b
f n (x) dx =
b
lim f n (x) dx =
→∞
a n
10
a
f (x) dx.
Proof. Let εn = sup |f n (x) − f (x)|. Then |f (n(x) − f (x)| ≤ sup |f n (x) − f (x)| = ε n . x [a,b]
x [a,b]
∈
Then f n (x) − εn ≤ f (x) ≤ f n (x) + εn for x ∈ [ a, b]. Consider, b
b
(f n − εn ) dx ≤
a
Thus,
b
(f n − εn ) dx ≤
a
b
f (x) dx −
a
b
(f n − εn ) dx ≤
a
b
∈
(f n + εn ) dx.
a
b
f (x) dx ≤ 2
a
εn dx = 2εn (b − a).
a
Thus, f n → f uniformly on [a, b] if and only if εn → 0. Then b
b
f (x) dx
a
( ) − ( ) −
f x dx < ε =⇒ f ∈ R[a, b]. Consider
a
b
b
f n (x) dx
a
b
b
≤ | ( ) ≤
b
a
that ( ) − Since → 0, there exists f n x dx
a
a
ε n b
lim
→∞
n
b
f n (x) dx =
a
f n (x) − f (x)| dx ≤ ε n
f x dx
a
b
f x dx
dx = ε n (b − a). This implies
a
ε whenever n ≥ N .
N ∈ N such that εn <
ε b a
− whenever n ≥ N . Thus we have that
f (x) dx.
a
b
Note 8 (Riemann-Stieltjes Integral) .
f (x) d[α(x)] for α(x) a monotone increasing
a
function. α(x) = x gives the Riemann Integral.
11
