4) Stone-Weierstrass theorem

Question 1

intro stone-weierstrass thm

Answer 1

In the original and
most known form says that any continuous function on a compact interval can be uniformly
approximated by a sequence of polynomials. Polynomials have many nice properties which
make this dense subset particularly useful: easy computation, derivation and integration, etc.
Yet, we will prove here a more general version of the Stone–Weierstrass theorem that applies
to general compact metric spaces

This is an interestingly sounding theorem: it states that a subset of C(X, R) which is closed under algebraic operations and separates points automatically has topological property—it is dense. Its consequences are striking. Before we prove this theorem let’s look at some of them

Question 2

Thm 4.1 (Stone–Weierstrass).

Answer 2

Suppose that X is a compact metric space and let C(X, R) be the Banach space of real valued continuous functions on X with norm ∥·∥∞.
Suppose that A ⊂ C(X, R) is a unital subalgebra of C(X, R), i.e.
* A is a linear subspace,
* 1 ∈ A,
* A · A ⊂ A, or in other words f, g ∈ A implies that also f · g ∈ A.
Suppose furthermore that A separates points, i.e. for any two x, y ∈ X with x ≠ y there
exists a function f ∈ A such that f(x)≠ f(y). Then, A is dense in C(X, R). (If A is dense then for any nhd of arbitrary size we can approx with arbirtrary precision)

Question 3

stone weirstrass more info

Answer 3

two seemingly unrelated properties?
We are saying that A is algebra-property closed under multiplication + (algebraic identities)

We are saying about the approx: elements are becoming closer (inequalities)

continuous and compactness help this bridge

Question 4

separates points means

Answer 4

we can distinguish any two points on the interval

so for any two points in x,y X that arent equal there is a funcition st f(x) is noit f(y)

Question 5

Corollary 4.2. [Weierstrass approximation theorem]

Answer 5

The space of polynomials R[x] is
dense in C([a, b], R) for any compact interval [a, b] in the in the ∥ · ∥∞ norm.

In other words, any continuous function can be approximated with arbitrary accuracy by a
polynomial in the supremum norm

e.g. consider any complicated continuous function on [a,b] an approximation p(x) = Σa_k x^k
gives a function that is within a distance Ɛ
)

Question 6

why we aproximate Weirstrass

Answer 6

(we are considering polynomials to be integrated, if we wish to integrate a continuous function we approximate by polynomials

Question 7

polynomials form a unital algebra

Answer 7

sum of polynomials polynomial
multiply two polynomials gives polynomals

Question 8

Corollary 4.3. The space of polynomials is dense in…
extended to compact set of R^n

Answer 8

The space of polynomials R[x1, . . . , xn] is dense in C(K, R) for any
compact subset K of R^n
in the ∥ · ∥∞ norm.

we need domain to be compact

(This is the higher dimensional version of the above theorem and states that a continuous
functions of n-variables can be approximated by polynomials in n variables.)

Question 9

Corollary 4.4. finite linear span related to trigonometric polynomials stone-weir

Answer 9

Let C(S^1, R) be the space of continuous functions on the unit circle, or, equivalently, the space of 2π-periodic real valued functions on R. Then the finite linear
span of the set

∪_m∈N {1,sin(mx), cos(mx)} is dense in C(S^1, R).

(we have collection as vector space function = 1 in our set and is closed under multiplication)

Question 10

Why are trig functions forming linear space

Answer 10

cos(nx)*sin(mx)
= 0.5(sin(n+m)x)-sin(n-m)

sin(A+B) = sinA *cosB +cosAsinB

Question 11

consider a continuous function defined on C[-pi,pi]

Answer 11

we can approx using trig functions

diagram: oscillating trig functions

Question 12

f ∧ g

Answer 12

f ∧ g = min{f, g},
f ∨ g = max{f, g}
Note that of f and g are continuous, then so are f ∧ g and f ∨ g (demonstrate this!)

diagram:
consider 2 functions, max takes all the max parts from the two
a function itself, continuous and will still be a function from our set (stones)

Question 13

Theorem 4.5 (Stone’s Theorem

Answer 13

. Let X be a compact metric space and suppose that
there is a subset A of C(X, R) such that

1) A is closed under the operations ∧ and ∨, this means f, g ∈ A implies f ∧ g ∈ A and f ∨ g ∈ A.

2) for any pair of points x ≠ y and numbers a, b ∈ R there is a function f ∈ A such that f(x) = a and f(y) = b.

Then, A is dense in C(X, R) in the topology induced by the norm ∥ · ∥∞ (the uniform topology)

Question 14

Corollary 4.6. [Stone–Weierstrass (complex version)]

Answer 14

Suppose that X is a compact
metric space and let C(X, C) be the complex Banach space of complex valued continuous functions on X with norm ∥ · ∥∞. Suppose that A ⊂ C(X, C) is a unital ∗-subalgebra
of C(X, C), i.e.
* A is a linear subspace,
* 1 ∈ A,
* A · A ⊂ A, or in other words f, g ∈ A implies that also f · g ∈ A.
* if f ∈ A the also overline(f) ∈ A.

Suppose furthermore that A separates points, i.e. for any two x, y ∈ X with x ≠ y there
exists a function f ∈ A such that f(x) ≠ f(y). Then, A is dense in C(X, C).

(complex conjugation: for complex numbers |z|= srt(Z*overline(Z))

constructed as collection of polynomials if you take bounded functions not continuous then not seperable so useful if continuous functions we can approximate all sets

Question 15

corollary 4.7 using complex version and e Euler identity e^iϕ = cos ϕ + isin ϕ

Answer 15

The linear span of the set {e
imφ | m ∈ Z} is dense in C(S^1, C)

Note that we need both positive and negative values of m in e^imφ, the set {e^imφ | m ∈ N0} is
not dense in C(S^1, C).

Question 16

Definition 4.8 (Separable Metric Space).

Answer 16

A metric space X is called separable if there exists a countable dense subset of X

Question 17

Corollary 4.9

(Suppose K ⊂ R^ n
is compact. Then the space of polynomials with real coefficients and n variables is dense in the space of continuous functions C(K). Of course every polynomial with
real coefficients may be approximated by one with rational coefficients. Thus the set of rational
polynomials Q[x1, . . . , xn] is dense in C(K). However, the space of rational polynomials is a
countable set. In this way one obtains…)

Answer 17

Let K be a compact subset of R^n
then the Banach space C(K) is
separable.

The following statement shows that continuous functions make only a tiny fraction of all
bounded functions:

Question 18

Exercise 4.10. Let X be an infinite set, show that the space B(X) of bounded functions
on X is not separable. (Hint: present a set of disjoint balls of radius 1/2
parametrised by all
real numbers.)

Answer 18

Exercise 4.10. Let X be an infinite set, show that the space B(X) of bounded functions
on X is not separable. (Hint: present a set of disjoint balls of radius 1
2
parametrised by all
real numbers.)

Question 19

Proof discussion: thm 4.5 stones theorem shown over 2 lectures 11/7/23

Answer 19

in notes but:
we show any function g can be approx by elements in A (ie for Ɛ>0 there is an f s.t. for all x in X /g(x)-f(x)/ < Ɛ
so g(x)- Ɛ < f(x) <g(x) + Ɛ
ie. diagram a function that is between a channel 2 Ɛ wide around g(x)

as its continuous the function doesnt immediately jump for nhds, for each two points we choose a function f_x,y, it is good for certain nhds but not elsewhere maybe (ie between the channel for a certain domain, nhd of x and nhd of y) O_x,y = {z in X : f_x,y(z) < g(z) + Ɛ}
this set is open as f_xy(z)-g(z) < Ɛ and the diff of two continuous sets is continuous. The set gives the preimage of this which by thm is also open

we repeatedly use by induction for finite functs min and max. The nhds in the domain form a cover of X even if we fix x.
(by construction f_xy = g)
open covering of a compact set:
We then find a finite subcover for each fixed x.. There are finitely many points y_1,…,yn s.t. O_x,y is an open cover
union k=1,..n of O{xy_k} of X

f_{xy}(x) < g(z) + Ɛ
take f_x = f_xy1∧f_xy2∧…f_xyz
functions are a finite number of patches which cover the entire space, upper bound is not exceeded. Then we define O_x for f_x(z)> g(z) + Ɛ which gives finite subcovering and we can show inequality required….

Question 20

define f ∧ g by f and g alternate

Answer 20

used in Stone Weierstrass theorem 4.1.
f ∧ g = (f+g)/2 - |f − g|/2

f ∨ g = (f+g)/2 + |f − g|/2

these help to describe how max and min of functs are expressed

Question 21

Sketch of proof of the Stone Weierstrass theorem 4.1

Answer 21

First we observe that if B is the
closure in C(X, R) of A from Thm. 4.1, then B will also be a unital point separating subalgebra of C(X, R) (exercise!).
Step 1: If f is non-negative and in B, so is √f.
Using the defn for min,max we need to show |a| = sqrt(a^2) To get sqrt we will use Taylor series

**Step 2: show B is closed under the operations ∧ and ∨.

**Step 3: need to show that for a pair of points x ̸= y and a,b in R there is a function in A st f(x)=a and f(x)=b

Final Step: As we can see all the conditions of Stone’s theorem are satisfied and therefore
B is dense in C(X, R). Since B is closed in C(X, R) this means that B = C(X, R). Thus,
A is dense in C(X, R).

(if closed contains all limiting points, if dense means we can always build sequence approaching, A produced B as its closure and it will be dense

Question 22

Sketch of proof of the Stone Weierstrass theorem 4.1
Step 1: If f is non-negative and in B, so is √f.

Answer 22

If f∈B show √f∈B
To see this note that it is enough to show this for 0 ≤ f < 1
(we will aprox √x by polynomials, addition and multiplication by coefficients and multiplications of elements are in our algebra )
because in case f ≠ 0 we can compute
√f =
√2 √(∥f∥_∞) [√(1/(2∥f∥∞)f)].

Taylor series Σ_[k=0 to ∞]
a_nx^k
for
√[1 − x] =1 −(1/2)x−(1/8)x^2 − (1/16)x^3 −(5/128)x^4 −(7/256)x^5 − . . .
and its partial sums converge in ∥∥_∞ to √[1 − x] converges absolutely and uniformly on any interval [0, 1 − δ).

for f(x)+δ:
Therefore, the series
Σ_[k=0 to ∞] a_k(1 − f − δ)^k
converges in the Banach space B because all the partial sums are actually in B as B is a
subalgebra.
The limit of this sequence is, of course, √[f + δ.] or √[f + (1/n).] converges to √[f(x))

the polynomials √[f +(1/n)] are in B as B is a unital algebra thus as B is a closure the limiting points are also in B. so √f(x) is in B by closedness.

If we let δ go to zero, we can see that also √f ∈ B.
This works because
|f(x) + δ − f(x)| = δ( f(x) + δ + f(x))−1 ≤√δ so that the approximation is uniform.

Question 23

Sketch of proof of the Stone Weierstrass theorem 4.1
Step 2: show B is closed under ^ and v

Answer 23

**Step 2: For f,g∈B
f+g ∈B linear space
(f+g)/2 ∈B
(f-g)/2 ∈B
|f| ∈ B ( Since |f| =√[f^2] )

Thus:
f ∧ g =(f + g)/2−|f − g|/2
f ∨ g =(f + g)/2 + |f − g|/ 2
we conclude from this that B is closed under the operations ∧ and ∨.

Question 24

Sketch of proof of the Stone Weierstrass theorem 4.1
Step 3: show for x =y and a,b in R we have a function in A f(x)=a and f(y)=b

Answer 24

Assume x=y are points in X and assume that a, b are real numbers. Then, by assumption there is an element f in B such that
f(x) ̸= f(y).**
Since B is a subspace that contains the constant functions, the function
g(z) =
a + (b − a)[(f(x) − f(z))/(f(x) − f(y))]
=[(bf(x) − af(y))/(f(x) − f(y))]−(b − a)/(f(x) − f(y))*f(z)

is also in B and it satisfies g(x) = a and g(y) = b.

Question 25

max(a,b)
min(a,b)

Answer 25

max(a,b) = (a+b)/2 +|a-b|/2
min(a,b) = (a+b)/2 +|a-b|/2

adding half the size of interval to midpoint or subtracting

we do this for functions, absolute value for functions?

Question 26

express |f|
needs to be in set A for |f| in A

Answer 26

|a| = sqrt(a^2)

To get sqrt we will use Taylor series:

Question 27

A-is a unital algebra separating points, instead we use use B = overline(A)

Answer 27

Using a set that is a linear algebra
A-unital algebra separating points, instead use B = overline(A)
B- is the unital algebra separating points also and is a closure of A

B includes A, A seperates points so B includes these values for functs so also seperates points

unital algebra: We build closure by adding limiting points. for sequences in A: Say for x_n converging to x in B. y_n converging to y in B

Then Z_n converges to xy
Z_n = x_n y_n is in A as A is closed under multiplication and algebra of limits means it converges to xy so is an algebra as xy in B. (Contains required unit values also)

Question 28

exercise: Let X be an infinite set, show that the space B(X) of bounded functions on X is not separable. (Hint: present a set of disjoint balls of radius 1/2
parametrised by all
real numbers.)

Answer 28

not continuous, not going to be separable so boundedness is not sufficient to say its separable

Take x, infinite set and let N⊂X be a countable infinite subset.

(We want to demonstrate that the space is not separable, we want to show the space cannot be dense- that means if every ball contains an element of this dense set)

Consider countably many balls which are disjoint, if the number of balls is uncountable it won’t be sense.

∥f∥∞ = sup{x ∈X} |f(x)|
Within this norm, distance between two functions is not less than distance at any point
∥f-g∥∞ ≥ ∥f(x)-g(x)∥∞

We want to have an uncountable number of disjoint balls, these balls need to be centred in well separated points. For bounded functs we can use their correspondence with reals.

For uncountable model set is R or R∩[a,b], use the fact that R is uncountable and so if we take a bijection between R and
For r∈[0,1] in decimal expansion r=0.d_1d_2,….
for digits d_k = 0,1,…9
Consider a function f_r(x)=
{d_k if x=x_k in N
{0 otherwise
∥f_r∥∞ ≤ 9
take r_1 not equal to r_2
then 1≤ ∥f_r_1-f_r_2∥∞
(at least one pair of different digits)
Then B_0.5(f_{r_1} disjoint with B_0.5(f_{r_2}

If they did intersect at Z then by triangle inequality d(x_1,z)+d(x_2,z) ≥ d(x_1,x_2) we know this is ≥ 1 but
in balls:
d(x_2,z) <0.5 , d(x_1,z) <0.5

So we have uncountable collection of disjoint balls so a dense subset shall be uncountable. So the space of bounded functions cannot be separable