Karlsruhe Institute of Technology
Institute for Analysis
Prof. Dr. D. Hundertmark
M. Sc. S. Avramska-Lukarska
WS 2012/2013
02.11.2012
Functional Analysis
Exercise Sheet 2
Solutions
Exercise 2
(a)
(i)
• d1 (x, y) = min{d(x, y), 1} = 0 ⇔ d(x, y) = 0 ⇔ x = y.
• symmetry is clear.
• triangle inequality:
d1 (x, y) = min{d(x, y), 1} ≤ min{d(x, z) + d(z, y), 1}
≤ min{d(x, z), 1} + min{d(z, y), 1}
= d1 (x, z) + d1 (z, y),
so d1 is a metric on M .
(ii)
d(x,y)
• d2 (x, y) = d(x,y)+1
= 0 ⇔ d(x, y) = 0 ⇔ x = y.
• symmetry is clear.
• triangle inequality:
1
d(x, y)
=1−
d(x, y) + 1
d(x, y) + 1
1
≤1−
d(x, z) + d(z, y) + 1
d(z, y)
d(x, z)
+
=
d(x, z) + d(z, y) + 1 d(x, z) + d(z, y) + 1
d(x, z)
d(z, y)
≤
+
d(x, z) + 1 d(z, y) + 1
= d2 (x, z) + d2 (z, y),
d2 (x, y) =
so d2 is also a metric on M .
(b) First we show that d is a metric.
• obviously d(x, y) ≥ 0 for all x, y ∈ R and
d(x, y) = 0 ⇔ arctan(x) = arctan(y) ⇔ x = 0
since arctan is injective.
• symmetry is clear.
• triangle inequality:
d(x, y) = |arctan(x) − arctan(y)| = |arctan(x) − arctan(z) + arctan(z) − arctan(y)|
≤ |arctan(x) − arctan(z)| + |arctan(z) − arctan(y)| = d(x, z) + d(z, y).
Next we show that d and | · | are equivalent on R. So let (xn ) ⊂ R be such that xn → x ∈ R w.r.t
d.
− π2 , π2 , we
Then we have arctan(xn ) → arctan(x) w.r.t. | · |. Since y 7→ tan(y) is continuous on
also have
xn = tan(arctan(xn )) → tan(arctan(x)) = x
w.r.t. | · |.
On the other hand, let (xn ) ⊂ R be such that xn → x w.r.t. | · |. Then the continuity of y 7→ arctan(y)
implies
d(xn , x) = |arctan(xn ) − arctan(x)| → 0.
(R, d) is not complete: consider (xn ) = (n) ⊂ R. Since arctan(xn ) = arctan(n) → π2 , it follows that
(xn ) is a Cauchy sequence w.r.t. d Assume that there exists some x ∈ R such that xn → x in (R, d).
Then, due to the considerations above, we also have xn → x in (R, | · |). But xn = n → ∞ ∈
/ R, a
contradiction.
Exercise 3
(a) A is closed: Let (fn ) ⊂ A with fn → f w.r.t. k · k∞ -norm.
Let a ∈ [0, 1]. First we show that there exists x ∈ [0, 1] with f (x) = a. (1)
Since fn ∈ A, for every n ∈ N there exists xn ∈ [0, 1] with fn (xn ) = a.
Furthermore, since (xn ) ⊂ [0, 1] and [0, 1] is compact, there exists a subsequence (xnk ) with xnk → x ∈ [0, 1].
Therefore, we have
f (x) = lim fnk (xnk ) = lim a = a.
k→∞
k→∞
Let b ∈ [0, 1]c . We will show that b ∈
/ f ([0, 1]). (2)
Suppose that there exists x ∈ [0, 1] with f (x) = b. Since [0, 1]c is open in R, there exists ε > 0 with
Bε (b) ⊂ [0, 1]c . Furthermore, since fn → f uniformly, there exists n ∈ N:
|fn (x) − f (x)| < ε,
i.e., fn (x) ∈ Bε (b) and thus fn (x) ∈
/ [0, 1], which is a contradiction!
By (1) it follows that [0, 1] ⊂ f ([0, 1]) and by (2) it follows that f ([0, 1]) ⊂ [0, 1]. Thus f ∈ A.
(b) B is not closed: Choose fn (x) =
1
n x, x
∈ [0, 1]. Then we have fn ∈ B, kfn − 0k∞ =
1
n
→ 0 but f ≡ 0 ∈
/ B.
(c) B is not open: Let f ∈ B and assume that B is open, i.e., there exists ε > 0 such that Bε (f ) ⊂ B. We will
show that there exists a sequence (fn ) ⊂ X \ B with fn → f w.r.t. k · k∞ -norm. (3)
This will give us a contradiction, since, by (3) there exists n ∈ N with kfn − f k∞ < ε, i.e., fn ∈ Bε (f ) ⊂ B,
but we know that fn ∈ X \ B.
To show (3) define
(
f (x), x ≥ n1 ,
fn (x) =
f ( n1 ), x ∈ [0, n1 ).
It holds: fn ∈ C([0, 1]), fn is not injective and
1
kfn − f k∞ = sup |f ( ) − f (x)|.
n
x∈[0, 1 )
n
Let ε > 0. Since f is uniformly continuous, there exists δ > 0 with
|f (x) − f (y)| < ε whenever |x − y| < δ.
Choose n0 ∈ N so that
1
n0
< δ. Then it holds kfn − f k < ε for n ≥ n0 .
(d) D is closed: Let (fn ) ⊂ D with fn → f in X. Then we have
1 1 1 − fn
f
= f
≤ kf − fn k∞ → 0,
2
2
2
i.e., f ( 21 ) = 0.