Homework 2 Solutions - Math 321,Spring 2015
(1) For each a ∈ [0, 1] consider fa ∈ B[0, 1] i.e. fa : [0, 1] → [0, 1] defined by
(
1 if t = a
fa (t) =
0 if t =
6 a
There are uncountably many such fa as [0, 1] is uncountable. Also if a 6= b then
kfa − fb k∞ = 1 (if it is clear from context that we are using sup norm you may omit
the ∞ symbol in the norm if you like.) Now consider all the balls B(fa , 12 ), a ∈ [0, 1]
so if fb 6= fa then fb ∈
/ B(fa , 21 ).
Now if C ⊆ B[0, 1] is dense then for each fa there is a ga ∈ C such that kga −fa k∞ < 12
and if a 6= b then
1
>0
2
Hence for a 6= b, ga 6= gb so the dense subset C cannot be countable.
kfb − ga k ≥ kfb − fa k − kfa − ga k ≥ 1 −
(2) Claim If fn is uniformly cauchy in R (or any metric space) then it is uniformly
convergent in R : Proof, for each x, {fn (x)}n≥1 is cauchy. So f (x) := limn→∞ fn (x)
exists for every x. Now let > 0 we can find N, n, m > N implies |fm (x) − fn (x)| < for all x. Now with n.m > N we have that for all x, we have
|fn (x) − f (x)| ≤ |fn (x) − fm (x)| + |fm (x) − f (x)| ≤ + |f (x) − fm (x)|
Now let m goes to ∞ we have that if n > N then |fn (x) − f (x)| < 2 for all x.(We
only take pointwise limit in m, we choose m large enough so that |f (x) − fm (x)| < .
, this m may be different for different x. Hence we have uniform convergence.
Now we will show that fn is uniformly cauchy on R. First since fn , fm are continuous,
we can find δ1 , δ2 such that for any x ∈ R,
|y − x| < δ1 =⇒ |fn (y) − fn (x)| < , |y − x| < δ2 =⇒ |fm (y) − fm (x)| < .
Since Q is dense in R. Let δ = min δ1 , δ2 , choose xr ∈ Q (which may depend on x)
such that |xr − x| < δ Also since fn is uniformly convergence on Q, there is N such
that if n.m > N then
|fn (x) − fm (x)| <
3
for all x ∈ Q. Combine these information, we have that for all n, m > N ,
|fn (x) − fm (x)| ≤ |fm (x) − fn (xr )| + |fn (xr ) − fm (xr )| + |fm (xr ) − fm (xr )| < 3
Note You may argue that the limiting f is continuous on R and use it to do this
. This is true but not completely obvious. You may have to argue why it is (e.g.
1
2
problem 7.11 in textbook) Note Important: sup norm of f : X → R (or C) is defined
as
kf k∞ = sup |f (x)|
x∈X
(sometimes it is ∞ so is not defined). We should observe that this quantity depends
only on f (and domain) , it does not depend on x so it is not necessary to write
kf (x)k∞ . The statement like kfn (x) − fn (y)k∞ < or “kfn (x) − fm (x)k∞ < for
all x” is confusing and sometimes is a misuse of notation. If you don’t like the
notation you can also write in words e.g. you can either write kfn − fm k∞ < or
|fn (x) − fn (x)| < for all x ∈ X.
(3) 3a.1) (fn )n∈N is cauchy in B(X) means it is uniformly cauchy (since we are using
sup norm) Hence as in the proof of the claim in Q2, f converges uniformly to f .
Now we have to check that f ∈ B(X). To see this, fN is bounded means there is an
M, |fN (x)| ≤ M for all x. Also there is and N such that n > N =⇒ kfN − f k < .
Hence for all x,
|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| < M + so f ∈ B(X).
3a.2) Now we prove that limn→∞ kfn k∞ = kf k∞ This follows from Triangle inequality:
|kfn k∞ − kf k∞ | ≤ kf − fn k∞ → 0(n → ∞.)
And we have the result (or one can argue that kk∞ is continuous.
3a.3) To show that supn kfn k∞ < ∞.( The face that each fn ∈ B(X) only tells us
that kfn k∞ exists as a finite number). First, there is N, ∀n ≥ N, kfn − f k∞ < 1
i.e. kfn k∞ ≤ kf k∞ + 1 for all n > N . Let M = max{kf k∞ + 1, kf1 k∞ , ..., , kfN k∞ }.
Then kfn k∞ ≤ M for all n hence supn kfn k∞ ≤ M < ∞.
PN
3b) Since TN =
n=1 kgn k∞ is cauchy. Let > 0 there is an N such that for
n, m > N , for all x ∈ X,
m
m
X
X
gk (x) ≤
kgk k∞ < k=n+1
k=n+1
P
This implies SN = N
n=1 is a cauchy sequence in B(X) (sup norm). Using Q3a) we
conclude that SN converges in B(X). The triangle inequalities is true for finite sum
but by
PQ 3a3) we can take
P N → ∞ to get the required
P triangle inequality.
(4) 4a) | n≤N an sin nx| ≤ P
|a
|
for
all
x
∈
R
and
n
n≤N
n≤N |an | < ∞. By Weierstrass
M-test,
we conclude that n≤N an sin nx converges uniformly on R. Same is true for
P
a
n≤N n cos nx.
3
4b)If |x| > 0 then 1 + x2 > 1. Hence for x 6= 0 we can apply geometric series:
∞
∞
X
x2
1
x2 X
x2
1
=1
=
=
1
2
n
2
2
n
2
(1
+
x
)
1
+
x
(1
+
x
)
1
+
x
1
−
2
1+x
n=1
n=0
While at x = 0 the series obviously converges to 0. The limiting function is not
continuous at x = 0 so the convergence cannot be uniform.
Direct proof: limx→0+ (1 + x2 )N = 1 so we can find x ∈ (0, 1] such that (1 + x2 )N ≤ 2
so with this x,
X
X
X
x2
x2
x2
1
1
=
=
−
≥
2
n
2
n
2
n
2
N
(1 + x )
(1 + x )
(1 + x )
(1 + x )
2
n≥1
n≥N +1
n≤N
Hence the convergence can’t be uniform.
(5) Write x ∈ [0, 1] in base 3 expansion x = 0.a1 a2 ... then we know that ai = 0, 2. Then it
is not show that f (3n t) = a2n . Now pick any (x, y) ∈ [0, 1]2 whose base 2 expansion are
x = 0.a0 a1 a2 ..., y = 0.b0 b1 b2 ... Let t ∈ [0, 1] has base 3 expansion 0.(2a0 )(2b0 )(2a1 )...
then we can check that t ∈ ∆ and s(t) = (x, y). This shows [0, 1]2 ⊆ s(∆). On the
other hand,
∞
∞
X
X
−k−1
2k
2−k−1 = 1
2
f (3 t) ≤
0≤
k=0
k=0
so 0 ≤ x(t) ≤ 1 and the same is true for y(t) hence s(∆) ⊆ [0, 1]2 .