Define a unbounded functional on C00 ⊂ l 2 . Justify your answer.
Answers
Basic properties of Banach and metric spaces 5
We also obtain
kαfk1 =
Z 1
0
|α| |f(t)| dt = |α| kfk1,
kf + gk1 =
Z 1
0
|f(t) + g(t)| dt ≤
Z 1
0
(|f(t)| + |g(t)|) dt = kfk1 + kgk1.
As a result, k · k1 is a norm on X. Next, consider the functions given by
fn(t) =
1,
1
2 ≤ t ≤ 1,
nt −
n
2 + 1,
1
2 −
1
n ≤ t ≤
1
2
,
0, 0 ≤ t ≤
1
2 −
1
n
,
for n ∈ N with n ≥ 3. For m ≥ n ≥ 3 we have
kfn − fmk1 =
Z 1
2
1
2 − 1
n
|fn(t) − fm(t)| dt ≤
1
n
−→ 0
as n → ∞, so that (fn) is a Cauchy sequence with respect to k · k1. We suppose
that (fn) would converge to some f ∈ X with respect to k · k1. Fix any a ∈ (0,
1
2
)
and take n ∈ N with 1
2 −
1
n ≥ a. We then obtain that
0 ≤
Z a
0
|f(t)| dt =
Z a
0
|f(t) − fn(t)| dt ≤ kfn − fk1 −→ 0
as n → ∞, i.e., R a
0
|f(t)| dt = 0. The continuity of f implies as above that f = 0 on
[0, a] for every a < 1
2
, and hence f(1/2) = 0. On the other hand,
0 ≤
Z 1
1
2
|f(t) − 1| dt =
Z 1
1
2
|f(t) − fn(t)| dt ≤ kf − fnk1 −→ 0
as n → ∞. As a consequence, f is equal to 1 on [ 1
2
, 1], which is a contradiction.
d) Let X = C(R) and a < b in R. We see as in b) that p(f) = supt∈[a,b]
|f(t)|
defines a seminorm on X. Moreover, if p(f) = 0, then f = 0 on [a, b], but of course
f does have to be the 0 function. ♦
Remark. The vector space X = C([0, 1]) is infinite dimensional since the functions
pn ∈ X, n ∈ N, given by pn(t) = t
n are linearly independent. If one interprets a
function 0 ≤ u ∈ X as a mass density, then kuk1 is the total mass and kuk∞ is the
maximal density. ♦
Before discussing further examples, we want to investigate various ‘topological’
concepts in a more general framework without vector space structure.
Definition 1.5. A distance d on a set M 6= ∅ is a map d : M ×M → R+ satisfying
a) d(x, y) = 0 ⇐⇒ x = y (definiteness),
b) d(x, y) = d(y, x) (symmetry),
c) d(x, y) ≤ d(x, z) + d(z, y) (triangle inequality)
for all x, y, z ∈ M. The pair (M, d) (or just M) is called a metric space. A sequence
(xn) ⊆ M converges to a limit x ∈ M if
∀ ε > 0 ∃ Nε ∈ N ∀ n ≥ Nε : d(x, xn) ≤ ε, (1.1)
and it is a Cauchy sequence if
∀ ε > 0 ∃ Nε ∈ N ∀ n, m ≥ Nε : d(xm, xn) ≤ ε.
(M, d) or d is called complete if each Cauchy sequence converges in (M, d).