Homework 4 Solutions - Math 321, Spring 2015
1. Weierstrass’s second theorem states that any continuous 2π-periodic function f on R is
uniformly approximable by trigonometric polynomials. The aim of this exercise is to prove
this statement.
(a) Deduce Weierstrass’s second theorem from his first in the special case when
f is even. We sketched a proof of this in class, but fill in the details.
Proof. Suppose that f is even and 2π-periodic. Define g(t) = f (arccos(t)) for t ∈
[−1, 1], where θ = arccos(t) is the unique value in [0, π] such that cos θ = t. We know
that arccos is a continuous function on [0, π], hence g ∈ C[0, 1]. By Weierstrass’s first
approximation theorem, let p be a polynomial on [−1, 1] such that
sup g(t) − p(t) < .
t∈[−1,1]
(1)
Substituting t = cos θ, we find that g(cos θ) = f (θ), and T (θ) = p(cos θ) is a trigonometric polynomial such that
sup f (θ) − T (θ) < .
θ∈[0,π]
Since f and cos θ are both even functions, the inequality above implies that
sup f (θ) − T (θ) < ,
θ∈[−π,π]
which is the conclusion of Weierstrass’s second theorem.
(b) Explain why a verbatim adaptation of the proof does not work if f is odd.
Proof. If we repeat the same argument as above, we reach the following inequality
analogous to (1):
sup f (θ) − T (θ) < ,
θ∈[− π2 , π2 ]
where T (θ) = p(sin θ) for some polynomial p. Since both f and T are now odd,
we have no knowledge of f on [ π2 , π] based on its values on [0, π2 ] and hence cannot
upgrade this inequality as before to the desired one, namely,
sup f (θ) − T (θ) < .
θ∈[−π,π]
(c) Given a general function f and a number > 0, invoke part (a) to find
trigonometric polynomials P and Q that approximate the even functions
f (x) + f (−x)
(f (x) − f (−x)) sin x
and
to within 10 . Now use P and Q to find a trigonometric polynomial R that
uniformly approximates f with error at most , thereby proving Weierstrass’s second theorem.
1
2
Proof. We write
(2)
(3)
f (x) + f (−x) = P (x) + E1 (x),
f (x) − f (−x) sin x = Q(x) + E2 (x),
where ||Ei ||∞ < 10 for i = 1, 2, and ||·||∞ denotes the sup norm on [0, 2π]. Multiplying
both sides of (2) and (3) by sin2 x and sin x respectively and adding the two, we obtain
f (x) sin2 x = R1 (x) + F1 (x),
(4)
(5)
where R1 (x) = P (x) sin2 x + Q(x) sin x is a trigonometric polynomial and the error
term F1 (x) = E1 (x) sin2 x + E2 (x) sin x is at most /5 in sup norm. Since (4) holds
for an arbitrary continuous 2π-periodic f , we may replace f (x) by f (x − π2 ), thereby
obtaining a trigonometric polynomial R̃1 with the property
π
f (x − ) sin2 x = R̃1 (x) + F̃1 (x), x ∈ [0, 2π]
2
where ||F̃1 ||∞ < /5. Setting y = x −
(6)
π
2
and using the 2π-periodicity of f gives
f (y) cos2 y = R2 (y) + F2 (y),
y ∈ [0, 2π]
where R2 (y) = R̃1 (y + π2 ) is yet another trigonometric polynomial and the error term
F2 (y) = F1 (y + π2 ) satisfies ||F2 ||∞ < 5 . Setting y = x in (6) and adding to (4) we
obtain
f (x) = R(x) + F (x),
with R(x) = R1 (x) + R2 (x), F (x) = F1 (x) + F2 (x),
so that R is a trigonometric polynomial and ||F ||∞ < 5 + 5 < . This is the conclusion
of Weierstrass’s second approximation theorem.
2. Let A be a normed algebra. If B is a subalgebra of A, conclude that B is
a subalgebra of A. Clarification: Recall that for us, an algebra is a vector
space equipped with a vector multiplication that is associative, left and right
distributive and compatible with scalars.
Proof. If a, b ∈ B then there are sequences {an }, {bn } in B converging to a, b respectively.
Now to show that it is a subalgebra, we have verify the following: If a, b ∈ B then
ab, a + b, αa ∈ B for any scalar α. This can be done e.g. showing that the sequence {an bn }
in B converges to ab. Other properties are automatically inherited as it is a subset of a
bigger algebra.
3. Given a metric space (X, d), recall that B(X) is the space of all bounded realvalued functions on X. Let A be a vector subspace of of B(X). Show that A is
a sublattice of B(X) if and only if |f | ∈ A.
Proof. (⇒) |f | = max{f, 0} − min{f, 0} ∈ A.
|
, f − := min{f, 0} =
(⇐) f + := max{f, 0} = f +|f
2
f − g ∈ A so h+ , h− ∈ A so
f −|f |
.
2
Now let f, g ∈ A then h :=
max{f, g} = h+ + g, min{f, g} = h− + g
3
are in A.
4. For the examples below, explain whether the class of functions A is dense in
C(X).
(a) X = U × V , where U and V are compact metric spaces; A = the class of
functions of the form f (u, v) = g(u)h(v) where g ∈ C(U ), h ∈ C(V ).
(7)
Proof. Take U = V = [0, 1]. We can’t apply Stone-Weirestass’s Theorem as A is
not a sub-algebra (it is not closed under addition; the simplest example would be
u, v ∈ A but f (u, v) = u + v ∈
/ A). Indeed taking f (u, v) = u + v we hope that it
is not in the closure of A. Supppose on the contrary that it does then there exists
fn (u, v) = gn (u, v)hn (u, v) ∈ A such that
1
sup |fn − f | <
8
U ×V
Now using equation (7) (here we write g, h for gn , hn , since f (0, 14 ) = 14 so g(0)h( 14 ) ∈
( 81 , 38 ) and since f (0, − 14 ) = − 14 so g(0)h(− 14 ) ∈ (− 83 , − 18 ). i.e. g(0)h( 14 ) > 0 and
g(0)h(− 41 ) < 0. By intermediate value theorem, there is an v0 ∈ (− 41 , 41 ) such that
g(0)h(v0 ) = 0 since g(0)h( 14 ) > 0 we have g(0) 6= 0 so h(v0 ) = 0. Then by equation
(7),
3
3
f (1, v0 ) = 1 + v0 > , |f (1, v0 ) − g(1)h(v0 )| = |f (1, v0 )| >
4
4
so this is contradiction to equation 7 so it is not dense.
Remark: Note however that the subspace generated by the class of functions of the
form f (u, v) = g(u)h(v) is in fact a subalgebra that separates points (consider the
distance functions dU and dV ) and vanishes nowhere (the constant 1 function), hence
is dense by Stone-Weierstrass.
(b) X = a compact set in Rn ; A = the class of all polynomials in n-variables.
Proof. It is easy to verify that A is a subalgebra of C(X). f (x1 , ..., xn ) ≡ 1 ∈ A
makes A nowhere vanishing. f (x1 , ..., xn ) = xi for 1 ≤ i ≤ n will seperate the points.
And we can apply Stone-Weierstass theorem so it is dense.
5. Let X = {z : |z| = 1} be the unit circle in the complex plane.
(a) Verify that the space of functions
N
n
X
iθ
A = f : f (e ) =
cn einθ ,
θ ∈ [0, 2π),
cn ∈ R
o
n=0
is an algebra.
Proof. This is easy to verify since A consists of polynomials in eiθ .
(b) Show that A separates points in X and vanishes at no point of X.
Proof. The function eiθ separates points. The constant function 1 vanishes nowhere.
Both functions lie in A.
4
(c) Show that there exist continuous functions on X that cannot be in the
uniform closure of A.
Proof. We argue by contradiction. Assume if possible that f (eiθ ) = e−iθ (which is
continuous on X) lies in the uniform closure of A, and let {Pn (eiθ ) : n ≥ 1} ⊆ A be
a sequence such that Pn → f uniformly on X. Then,
Z 2π
Z 2π
iθ −iθ
2π =
e e dθ = lim
eiθ Pn (eiθ ) dθ = 0,
n→∞
0
0
a contradicton. The last equality follows from the fact that
Z 2π
ei(j+1)θ dθ = 0 for all j = 0, 1, 2, · · · ,
0
Z 2π
and hence
eiθ P (eiθ ) dθ = 0 for all P ∈ A.
0
(d) Explain why the statements above do not contradict the Stone-Weierstrass
theorem.
Proof. The functions in A are complex-valued. The complex version of the StoneWeierstrass theorem states that an algebra of continuous complex-valued functions
is dense if it is self-adjoint, separates points and vanishes nowhere. The algebra A is
not self-adjoint.