Homework 3 Solutions - Math 321,Spring 2015
(1) 1a. We want to find a sequence {pn } of polynomial such that pn → f uniformly
on any bounded subset of R. By The First Weierstrass Approximation Theorem,for
each positive integer N , we can find a sequence of polynomial {pn } such that pn → f
uniformly on [−N, N ]. Hence for each positive integer N we can find a polynomial
qN (picked one from the sequence) such that |f (x) − qN (x)| ≤ N1 for all x ∈ [−N, N ].
Define our (new) sequence of polynomials {pn } by pn := qn . Now if E is a bounded
subset of R then there is an M > 0 a positive integer such that E ⊆ [−M, M ]. Now
given > 0 and N such that N1 < , if n > max{N, M } then
|pn (x) − f (x)| <
1
<
n
for all x ∈ [−M, M ] hence all x ∈ E.
1b. If P (x) is a non-constant polynomial then limx→∞ |P (x)| = ∞. Since sin(x)
is bounded by 1, given M > 0 we can find x0 such that |P (x0 )| > M then
| sin(x0 ) − P (x0 )| ≥ |P (x0 )| − | sin(x0 )| > M − 1
Since M could be arbitrarily large. We have k sin(x) − P (x)k∞ = ∞ for any nonconstant polynomial P .
Hence if pn → sin uniformly on R. Then pn (x) ≡ Cn are constant polynomials. Then
for all sufficiently large n, | sin(x) − Cn | < for all x ∈ R (with out loss of generality
< 1). So | sin(0) − pn (0)| = |Cn | < .
• Case 1: C ≥ 0 then | sin( −π
) − C| = C + 1 > , which is a contradiction.
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• Case 2: C < 0 then | sin( π2 ) − C| = 1 − C > , which is a contradiction.
Hence no such sequence of polynomials could exist.
(2) Let p(x) be a polynomial such that |p(x) − f (x)| < for all x ∈ [a, b]. Now since
´b
f (x)xn dx = 0 for all n, by linearity of integral, we have
a
ˆ b
p(x)f (x)dx = 0
a
for any polynomial p. Now f is continuous on [a, b] hence bounded by M ,say. So
ˆ b
ˆ b
ˆ b
2
f (x) dx = f (x)(f (x) − p(x)) + f (x)p(x)dx = f (x)(f (x) − p(x))dx
a
a
a
≤ M (b − a).
Since M, b − a are fixed constants and > 0 could be arbitrarily small. Hence
ˆ b
f (x)2 dx = 0
a
1
2
Since f is continuous we have that f ≡ 0 on [a, b]. (For example, if f is continuous
at x0 and f (x0 ) > 0 then f (x0 ) > 0 on a small neighborhood around x0 which could
be shown by definition
´ of continuity, then would contribute positive quantity to the
integral). In general, |f (x)|2 dx = 0 does not implies f ≡ 0 though f is nonzero on
a very small set.
(3) It would be easy easier to start from derivative and do integration (Integration makes
f behaves better and easier to deal with). Since f 0 is continuous on [a, b] there is a
polynomial q such that
kf 0 − qk∞ <
<
b−a+1
(here sup norm is over [a, b].) Define a polynomial p by
ˆ x
p(x) := f (a) +
q(t)dt
a
0
Then p = q and
ˆ x
ˆ x
0
0
|f (x) − p(x)| = (f (a) +
f (t)dt) − (f (a) +
p (t)dt)
a
a
ˆ x
|f 0 (t) − p0 (t)|dt
≤
a
< (b − a)
< .
b−a+1
Next we show separability by finding a countable dense subset which will be the
set of polynomials with rational coefficients. Let Sn be the set of polynomials with
rational coefficients with degree n hence |SnS| ≤ |Qn | is countable. Then the set of
all polynomials with rational coefficients is n Sn is countable. Since Q is dense in
R we will show that any polynomial could be approximated with the norm k · kC 1 [a,b]
by a polynomial of rational coefficients. Let p be a polynomial of degree N (so N is
fixed here), let
M := sup max{|x|, |x|2 , · · · , |x|N }
x∈[a,b]
P
be a constant. Give a polynomial p such that p(x) = N
ak xk and for each ak
k=0 P
N
k
choose a rational bk such that |bk − ak | < (N +1)
2 M . Let q(x) =
k=0 bk x be another
polynomial then
|p(x) − q(x)| ≤ , |p0 (x) − q 0 (x)| ≤ so C 1 [a, b] is separable.
3
(4) 4a. Give an example of a non-polynomial continuous function with reasons why she
is not a polynomial.
4b. Suppose such a sequence of polynomials converging to an f whose degrees
bounded by N exists. Since the set of polynomials of degree ≤ N forms a vector
space, we could find {φ}N
i=0 , an orthonormal basis. For any polynomial pn of degree
≤ N then
ˆ b
n
X
pn =
pn · φi dx
an,i φi , an,i = hpn , φi i =
i=0
a
Let > 0 then for sufficiently large n, kpn − f k∞ < so kpn k∞ < kf k∞ + . Also by
Cauchy-Schwartz’s inequality,
|hpn , φi i| ≤ kpn k2 kφi k2 = kpn k2
So for each i, {an,i := hpn , φi i}∞
n=1 is a bounded sequence hence has a convergent
sequence converging to ai ∈ R.
Now start with a subsequence A1 of {an,1 } converging to a1 . Consider {an,2 }n∈A1
is also a bounded sequence and hence has a convergent subsequence converging to
a2 . Continue this way until ak we have a sequence of polynomial {pn } (renumbered)
which is a subsequence of the origin sequence such that
kpn − qk∞ → 0
where q(x) = i=0 ai x is a polynomial. So if the original converges, it must converge
to a polynomial which is a contradiction.
PN
i