Definition A map f : U ⊂ Rm → Rn is called a Lipschitz map on U if there exists a constant C ≥ 0 such that
kf (x) − f (y)k ≤ Ckx − yk for all x, y ∈ U.
If one can choose a (Lipschitz) constant C < 1 such that the above Lipschitz condition hold on U, then f is called a contraction map.
Remarks
(a) If f is Lipschitz on an open subset U of Rm, then f is continuous on U since
x→ylim |f (x) − f (y)|| ≤ lim
x→y |x − y| = 0 ∀ x, y ∈ U.
(b) For each x ∈ R, let f (x) = |x|. Since |f (x) − f (y)| = ||x| − |y|| ≤ |x − y|, f is Lipschitz with Lipschitz constant C = 1.
Definition Let U be a subset of Rm, and f be a map that maps U into U, i.e. f : U → U. A point p ∈ U is said to be a fixed point of f if f (p) = p.
Fixed Point Theorem for Contractions Let f : Rm → Rm be a contraction map. Then f has a unique fixed point.
Proof Let x1 be arbitrary point in Rm and set x2 = f (x1); inductiveley, set xn+1= f (xn), n ∈ N.
Observe that
kx3− x2k = kf (x2) − f (x1)k ≤ Ckx2− x1k and, inductively, that
kxn+1− xnk = kf (xn) − f (xn−1)k ≤ Ckxn− xn−1k ≤ Cn−1kx2− x1k.
If m ≥ n, then
kxm− xnk ≤ kxm− xm−1k + kxm−1− xm−2k + · · · + kxn+1− xnk
≤ {Cm−2 + Cm−3+ · · · + Cn−1}kx2− x1k
≤ Cn−1
1 − Ckx2− x1k → 0 as n → ∞ since 0 < C < 1.
Therefore, {xn} is a Cauchy sequence in the complete space Rm. Let
p = lim
n→∞xn. Then p is a fixed point of f since
p = lim
n→∞xn+1= lim
n→∞f (xn) = f ( lim
n→∞xn) = f (p).
In fact, there is only one fixed point for f. Suppose p, q are two distinct fixed points of f, then kp − qk = kf (p) − f (q)k ≤ Ckp − qk =⇒ p = q.
Definition Let X be a metric space and let
C(X) = {f : X → C | f is a complex-valued continuous, bounded function defined on X}.
For each f ∈ C(X), define the supremum norm k k : C(X) → R by kf k = sup
x∈X
|f (x)|.
Remarks
(a) Since f is bounded, kf k < ∞ for all f ∈ C(X).
(b) kf k ≥ 0 and kf k = 0 if and only if f (x) = 0 ∀x ∈ X;
(c) kcf k = |c|kf k ∀c ∈ C;
(d) kf + gk ≤ kf k + kgk ∀f, g ∈ C(X).
(e) Let the function d : C(X) × C(X) → R be defined by
d(f, g) = kf − gk ∀ f, g ∈ C(X).
Then C(X), k k is a metric space.
Definition Let E be a subset of a metric space X. For each n ∈ N, let f, fn : E → C be complex-valued functions deined on E. We say that
fn converges to f pointwise on E if
n→∞lim fn(x) = f (x) ∀x ∈ E.
fn converges uniformly on E to f if
∀ > 0, ∃N = N () ∈ N such that if n ≥ N then |fn(x) − f (x)| ≤ ∀ x ∈ E.
Remarks
(a) fn converges uniformly to f : E → C on E if and only if
n→∞lim kfn− f k = lim
n→∞sup
x∈E
|fn(x) − f (x)| = 0.
⇐⇒ ∀ > 0, ∃ N = N () ∈ N such that if m, n ≥ N then kfm− fnk ≤ .
(b) Theorem Let E be a subset of a metric space X and let fn : E → C be a sequence of continuous complex-valued functions defined on E. If fn converges uniformly to f : E → C on E then f is continuous.
(c) Theorem Let X be a subset of a metric space and let C(X), k k be the space of complex- valued continuous, bounded functions defined on X equipped with the supremum norm.
Then the metric space C(X), k k is a complete metric space, i.e. every Cauchy sequence fn in C(X) converges to f ∈ C(X) since one can show that fn converges uniformly to f on X.
(d) Theorem Let fn : [a, b] → C be a sequence of differentiable functions defined on [a, b]. If fn(x0) converges for some x0 ∈ [a, b] and if fn0 : [a, b] → C converges uniformly on [a, b], then fn converges uniformly on [a, b] to a function f and
f0(x) = lim
n→∞fn0(x) ∀ x ∈ [a, b].
Examples
(a) For each n ∈ N, let fn : R → R be defined by fn(x) = x
n for each x ∈ R. Show that fn
converges to f ≡ 0 pointwise (not uniformly) on R.
(b) For each n ∈ N, let fn : R → R be defined by fn(x) = x2+ nx
n for each x ∈ R. Show that fn converges to f (x) = x pointwise (not uniformly) on R.
(c) For each n ∈ N, let fn : R → R be defined by fn(x) = 1
nsin(nx + n) for each x ∈ R. Show that fn converges uniformly to f ≡ 0 on R.
(d) Let fn(x) = 1nsin(n2x). Notice that this converges pointwise to zero.
(e) For each n ∈ N, let fn : I = [0, 1] ⊂ R → R be defined by fn(x) = xn for each x ∈ I.
Show that fn converges to f, defined by f (x) = 0 for 0 ≤ x < 1 and f (1) = 1, pointwise (not uniformly) on I. Notice this example shows that pointwise convergence does not imply continuity and note that area and derivative may not be preserved by pointwise convergence.
(f) Let fm(x) = lim
n→∞(cos(m!πx))2n. If m!x ∈ Z, then fm(X) = 1. If x ∈ Q, then x = p q =⇒
m!x ∈ Z for m ≥ q. Thus for x ∈ Q, then lim
m→∞fm(x) = 1. For x 6∈ Q, fm(x) = 0. So:
fm(x) → f (x) =
(1 x ∈ Q 0 x 6∈ Q (g) f (x) = lim
n→∞fn(x). Recall that f is continuous at x if and only if lim
t→xf (t) = f (x). Thus if f (x) is continuous:
limt→x lim
n→∞fn(t) = lim
n→∞lim
t→xfn(t).
(h) Note that normally cannot reorder limits. For example, if we let Sm,n = m
m + n Fix n, then
m→∞lim Sm,n = 1 and therefore lim
n→∞ lim
m→∞Sm,n = 1. But if we fix m, limn→∞Sm,n = 0 and therefore lim
m→∞ lim
n→∞Sm,n = 0.
Definitions Let E be a subset of a metric space X and let fn : E → C be a sequence of functions defined on E. We say that
(a) {fn} is pointwise bounded on E if {fn(x) | n ∈ N} is bounded for every x ∈ E, i.e. there exists a function φ : E → R such that |fn(x)| ≤ φ(x) for all x ∈ E and for all n ∈ N.
(b) {fn} is bounded (or uniformly bounded) on E if there exists a constant M such that kf k = sup{f (x) | x ∈ E} ≤ M, for all n ∈ N.
Definition Let E be a subset of a metric space X and let F = {f : E → C} be a set of complex-valued functions defined on E. Then F is said to be uniformly equicontinuous on E if, for each real number > 0 there is a number δ = δ() > 0 such that if x, y belong to E and kx − yk < δ and f is a function in F , then kf(x) − f(y)k <
Theorem (Arzela-Ascoli) Let K be a compact subset of Rp and letF be a collection of func- tions which are continuous on K and have values in Rq. The following properties are equivalent:
(a) The family F is bounded and uniformly equicontinuous on K.
(b) Every sequence from F has a subsequence which is uniformly convergent on K.
Proof
(b) ⇒ (a) Claim: F is bounded.
Proof Suppose that F is not bounded, then
∃ a sequence {fn}n∈N⊂F such that kfnkK ≥ n, i.e. ∃ xn ∈ K such that kfn(xn)k ≥ n.
This implies that fn cannot have any subsequence that converges uniformly on K.
Otherwise if fn had a subsequence, still denoted by fn, that converges uniformly on K to a uniformly continuous function f, then
∞ = lim
n→∞kfnkK ≤ lim
n→∞kfn− f kK+ lim
n→∞kf kK < ∞ which is a contradiction.
Hence, F is bounded.
(b) ⇒ (a) Claim: F is uniform equicontinuous.
Proof Suppose that F is not uniformly equicontinuous on K, then
∃ 0 > 0 such that ∀ n ∈ N, ∃ fn ∈F and xn, yn ∈ K such that
kxn− ynk < 1
n and kfn(xn) − fn(yn)k ≥ 0.
This implies that fn cannot have any subsequence that converges uniformly on K.
Otherwise if fn had a subsequence, still denoted by fn, that converges uniformly on K to a uniformly continuous function f, then
0 < 0 ≤ lim
n→∞kfn(xn) − fn(yn)k
= lim
n→∞kfn(xn) − f (xn) + f (xn) − f (yn) + f (yn) − fn(yn)k
≤ lim
n→∞kfn(xn) − f (xn)k + lim
n→∞kf (xn) − f (yn)k + lim
n→∞kfn(yn) − f (yn)k
≤ lim
n→∞kfn− f kK+ lim
n→∞kf (xn) − f (yn)k + lim
n→∞kfn− f kK
= 0 which is a contradiction.
Note that in the last equality
n→∞lim kfn− f kK = 0 since fnconverges to f uniformly on K,
and
n→∞lim kf (xn) − f (yn)k = 0 since f is uniformly continuous on K, and kxn− ynk < 1 n. Hence, F is uniform equicontinuous.
(a) ⇒ (b) Use the diagonal process to find a convergent subsequence.
Let S = K ∩ Qp = {xi}i∈N.
Then S is a countable dense subset of K such that S = K, i.e.
∀ x ∈ K and ∀ δ > 0, ∃ xi ∈ S such that xi ∈ B(δ, x) = {z ∈ Rp | kz − xk < δ}.
Let {fn} be a sequence inF . Then
{fn(x1)} is bounded in Rq ⇒ {fn} has a subsequence {fn1} converges at x1. {fn1(x2)} is bounded in Rq ⇒ {fn1} has a subsequence {fn2} converges at x2, x1.
· · · ·
{fnk−1(xk)} is bounded in Rq ⇒ {fnk−1} has a subsequence {fnk} converges at xk, xk−1, . . . , x1.
· · · · i.e.
{ f1 f2 · · · fk · · · }
∪
{ f11 f21 · · · fk1 · · · }(x1) → f (x1)
∪
{ f12 f22 · · · fk2 · · · }(xi) → f (xi) for 1 ≤ i ≤ 2
∪
{ f13 f23 · · · fk3 · · · }(xi) → f (xi) for 1 ≤ i ≤ 3
∪ ...
∪
{ f1k f2k · · · fkk · · · }(xi) → f (xi) for 1 ≤ i ≤ k
∪ ...
By setting gn= fnn, we obtain a subsequence of {fn}.
For each 1 ≤ j ≤ k ∈ N, since
gn= fnn∈ {fmj}m∈N ∀ n ≥ j =⇒ lim
n→∞gn(xj) = f (xj) ∀ 1 ≤ j ≤ k ∈ N.
Hence,
n→∞lim gn(xj) = f (xj) ∀ xj ∈ S.
(a) ⇒ (b) Claim: The subsequence gn (of fn) converges uniformly to f on K.
Proof For each > 0, since F is uniformly equicontinuous on K, there exists a δ = δ() > 0 such that
if x, y ∈ K satisfying that kx − yk < δ, then kh(x) − h(y)k < ∀ h ∈F .
This implies that for each each x ∈ K and for each gn ∈F , since S = K ⊂
∞
[
i=1
B(δ, xi),
(∗) ∃ xi ∈ S such that if kx − xik < δ =⇒ kgn(x) − gn(xi)k <
3 ∀ n ∈ N.
Also, for the given > 0, since lim
n→∞gn(xi) = f (xi), there exists L ∈ N such that (†) ∃ L ∈ N such that if m, n ≥ L =⇒ kgm(xi) − gn(xi)k <
3. Thus for each x ∈ K and for each m, n ≥ L, combining (∗) and (†), we have
kgn(x) − gm(x)k ≤ kgn(x) − gn(xi)k + kgn(xi) − gm(xi)k + kgm(xi) − gm(x)k< 3 +
3 + 3 = .
This implies that
for any m, n ≥ L, we have kgn− gmkK < , i.e. gn converges uniformly on K to a uniformly continuous function f .
Examples Justify that the Arzela-Ascoli Theorem is not applicable in the following sequence of functions.
(a) fn(x) = x + n for x ∈ [0, 1]
(b) fn(x) = xn for x ∈ [0, 1]
(c) fn(x) = 1
1 + (x − n)2 for x ∈ [0, ∞)
Examples Let fn : [0, 1] → R be continuous and be such that |fn(x)| ≤ 100 for every n and for all x ∈ [0, 1] and the derivatives fn0(x) exist and are uniformly bounded on (0, 1).
(a) Show that there is a constant M such that
| fn(x) − fn(y) | ≤ M | x − y | ∀ x, y ∈ [0, 1] and ∀ n ∈ N.
Solution Let M be a constant such that |fn0(x)| ≤ M for all x ∈ (0, 1). By the mean value theorem, we get
| fn(x) − fn(y) | ≤ M |x − y| ∀x, y ∈ [0, 1] and ∀ n ∈ N.
(b) Prove that fn has a uniformly convergent subsequence.
Solution We apply the Arzela-Ascoli Theorem by verifying that {fn} is equicontinuous and bounded. Given , we can choose δ = /M, independent of x, y, and n. Thus {fn} is equicontinuous. It is bounded because kfnk = sup
x∈[0,1]
|fn(x)| ≤ 100.
Example Let the functions fn : [a, b] → R be uniformly bounded continuous functions. Set Fn(x) =
Z x a
fn(t) dt for x ∈ [a, b].
Prove that Fn has a uniformly convergent subsequence.
Solution Since kFnk ≤ kfnk(b − a), Fn is uniformly bounded. Also, since |Fn0(x)| ≤ kfnk, Fn is equicontinuous by the preceding result. Hence, Fn has a uniformly convergent subsequence by Arzela-Ascoli Theorem.
Examples
(a) For each n ∈ N, let fn be defined for x > 0 by fn(x) = 1
nx. For what values of f does
n→∞lim fn(x) exist? Is the convergence uniform on the entire (0, ∞)? Is the convergence uniform for x ≥ 1?
(b) For each n ∈ N, let fn be defined on R by fn(x) = nx
1 + n2x2. Show that lim
n→∞fn(x) exists for all x ∈ R. Is this convergence uniform on R?
(c) Show that lim
n→∞(cos πx)2n exists for all x ∈ R. Is the limiting function contiuous? Is this convergence uniform?
(d) Let fn be defined on the interval I = [0, 1] by the formula
fn(x) =
(1 − nx, if 0 ≤ x ≤ 1/n 0 if 1/n < x ≤ 1.
Show that lim
n→∞fn(x) exists on I. Is this convergence uniform on I?
(e) Let fn be defined on the interval I = [0, 1] by the formula
fn(x) =
(nx, if 0 ≤ x ≤ 1/n
n(1 − x)/(n − 1) if 1/n < x ≤ 1.
Show that lim
n→∞fn(x) exists on I. Is this convergence uniform on I? Is the convergence uniform on [c, 1] for c > 0?
Example Let f : [a, b] → R be differentiable satisfying
0 < m ≤ f0(x) ≤ M for every x ∈ [a, b]
and let
f (a) < 0 < f (b).
Given x1 ∈ [a, b], define the sequence {xn} by xn+1 = xn− 1
Mf (xn) for each n ∈ N.
Prove that this sequence is well-defined and converges to the unique root ¯x of the equation f (¯x) = 0 in [a, b].
Hint Consider φ : [a, b] → R defined by φ(x) = x − f (x)
M . Note that the φ map [a, b] into [a, b] = dom(f ).
Solution For any x ∈ [a, b], since 0 < f0(t) ≤ M for all t ∈ [a, b], we have f (x) − f (a) =
Z x a
f0(t)dt ≤ M Z x
a
dt = M (x − a) which implies that
a = x − f (x)
M +f (a)
M ≤ x − f (x) M ,
where we have used the assumption that f (a) < 0 in the last inequality.
Similarly,
f (b) − f (x) = Z b
x
f0(t)dt ≤ M Z b
x
dt = M (b − x) which implies that
b = f (b)
M + x −f (x)
M + ≥ x − f (x) M , where we have used the assumption that f (b) > 0 in the last inequality.
Hence, the function
a ≤ φ(x) = x − f (x) M ≤ b maps [a, b] into [a, b], and the sequence
xn+1= xn− 1
Mf (xn) = φ(xn), for n ∈ N, is well defined if x1 ∈ [a, b].
For any x, y ∈ [a, b], since
|φ(x) − φ(y)| = |x − y −f0(c)
M (x − y)|, for some c lying between x and y.
Hence,
|φ(x) − φ(y)| ≤ (1 − m
M)|x − y|
and φ is Lipschitz with Lipschitz constant 0 ≤ 1 − m M < 1.
Hence, φ is a contraction and has a unique fixed point ¯x ∈ [a, b].
Since
|xn+1− ¯x| = |φ(xn) − φ(¯x)| ≤ (1 − m
M)|xn− ¯x| ≤ (1 − m
M)n|x1− ¯x| ≤ (1 − m
M)n(b − a), and
¯
x = lim
n→∞xn+1 = lim
n→∞xn− lim
n→∞
f (xn)
M = ¯x − f (¯x) M . Hence, f (¯x) = 0.
Examples For each k ∈ N, let fk : R → R be differentiable, satisfying |fk0(x)| ≤ 1 for all x ∈ R and fk(0) = 0.
(a) For each x ∈ R, prove that the set {fk(x)} is bounded.
Solution For each x ∈ R, since |fk(x)| = |fk(x) − fk(0)| ≤ |fk0(ck) (x − 0)| ≤ |x|, where ck
lies between x and 0, the set {fk(x)}∞k=1 is bounded.
(b) Use Cantor’s diagonal method to show that there is an increasing sequence n1 < n2 < n3 <
· · · of positive integers such that, for every x ∈ Q, we have {fnk(x)} is a convergent sequence of real numbers.
Solution Let Q = {x1, x2, · · · }. Since {fk(x1)} is bounded, we can extract a convergent subsequence, denoted {fk1(x1)}, out of {fk(x1)}. Next, the boundedness of {fk1(x2)} implies that we can extract a convergent subsequence, denoted by {fk2(x2)}, out of {fk1(x2)}. Con- tinuing this way, the boundedness of {fkj(xj+1)} implies that we can extract a convergent subsequence {fkj+1(xj+1)}, out of {fkj(xj+1)} for each j ≥ 1. Let fnk = fkk for each k ≥ 1.
Then {fnk} is a subsequence of {fn} and {fnk} converges at each xj ∈ Q.
Example LetF be a bounded and uniformly equicontinuous collection of functions on D ⊂ Rp to R and let f∗ be defined on D → R by
f∗(x) = sup{f (x) | f ∈F , x ∈ D}
Show that f∗ is continuous on D to R. Show that the conclusion may fail if F is not uniformly equicontinuous.
Stone-Weierstrass Theorem Let f : [a, b] → C be a continuous complexed-valued function.
Then there exists a sequence of polynomials Pn such that
n→∞lim Pn(x) = f (x) uniformly on [a, b].
If f is a real-valued function, then Pn may be taken as real-valued.
Proof We may assume, without loss of generality, that [a, b] = [0, 1].
For if For if the theorem is proved for this case, consider the composition of f : [a, b] → C and x = a + t(b − a) : [0, 1] → [a, b].
We may also assume that
f (0) = f (1) = 0.
For if the theorem is proved for this case, consider
g(x) = f (x) − f (0) − x[f (1) − f (0)] for x ∈ [0, 1]
Here g(0) = g(1) = 0, and if g can be obtained as the limit of a uniformly convergent sequence of polynomials, it is clear that the same is true for f, since f (x) − g(x) = f (0) + x[f (1) − f (0)]
is a polynomial.
Extend f : [0, 1] → C to f : R → C by defining
f (x) =
(f (x) x ∈ [0, 1]
0 x ∈ R \ [0, 1].
Then f : R → C is uniformly continuous on R.
For each n ∈ N, let Qn(x) = cn(1 − x2)n, where cn is chosen such that Z 1
−1
Qn(x) dx = 1 i.e. cn Z 1
−1
(1 − x2)n = 1.
Since
Z 1
−1
(1 − x2)ndx = 2 Z 1
0
(1 − x2)ndx
≥ 2 Z 1/√
n 0
(1 − x2)ndx
≥ 2 Z 1/√
n 0
(1 − n x2) dx
≥ 4
3√ n
> 1
√n. Hence we have
cn <√
n ∀ n ∈ N.
For any δ > 0, this implies that Qn(x) = cn(1 − x2)n ≤√
n (1 − δ2)n ∀ δ ≤ |x| ≤ 1 =⇒ Qn → 0 uniformly in δ ≤ |x| ≤ 1.
Now set
Pn(x) = Z 1
−1
f (x + t) Qn(t) dt for each x ∈ [0, 1].
Since f (x) =
(f (x) x ∈ [0, 1]
0 x ∈ R \ [0, 1]. =⇒ f (x + t) = 0 if x + t /∈ [0, 1] (i.e. t /∈ [−x, 1 − x]), we have
Pn(x) =
Z 1−x
−x
f (x + t) Qn(t) dt
= Z 1
0
f (s) Qn(s − t) ds by setting s = x + t, ds = dt
= Z 1
0
f (s) cn1 − (s − x)2n
ds
= a polynomial in x.
This shows that {Pn} is a sequence of polynomials.
Given > 0, we choose δ > 0 such that
if |y − x| < δ =⇒ |f (y) − f (x)| < 2. Let M = sup
x∈[0,1]
|f (x)|. Since
Z 1
−1
Qn(t) dt = 1 and Qn(t) ≥ 0, we see that for 0 ≤ x ≤ 1,
|Pn(x) − f (x)| =
Z 1
−1
f (x + t) Qn(t) dt − f (x) Z 1
−1
Qn(t) dt
=
Z 1
−1
f (x + t) − f (x) Qn(t) dt
≤ Z 1
−1
|f (x + t) − f (x)| Qn(t) dt
≤ 2M Z −δ
−1
Qn(t) dt + 2
Z δ
−δ
Qn(t) dt + 2M Z 1
δ
Qn(t) dt
≤ 4M√
n 1 − δ2n
+ 2
< if n is sufficiently large and this proves that Pn converges to f uniformly on [0, 1].
Corollary For each interval [−a, a] ⊂ R, there exists a sequence of polynomials {Pn| Pn : [−a, a] → R, n ∈ N}
such that
( Pn(0) = 0
limn→∞Pn(x) = |x| uniformly on [−a, a].
