Some informal notes about Lp-spaces and convergence
Version September 17, 2003
Harald E. Krogstad, IMF
It is said that David Hilbert, at one of the early meetings on quantum mechanics, whispered
to his neighbor: ”What is really this Hilbert space they are talking about”?
The story is hardly true, but Hilbert space has been one of the most loved topics in physics
and applied mathematics ever since the theory was formalized in the first part of the previous
century.
In this note I summarize some (actually simple) points we are using all the time, and which
you need to think over once (but only once) in your lifetime.
The note is likely to expand later.
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L1 and L2
Let us first recall that the Lp -spaces consists of functions with finite L p -norm ( p < ∞ here):
Z
1/p
p
|f (x)|
< ∞.
kf kp =
T
In general, a norm, k·k, is supposed to fulfil
(i)
kf k = 0 ⇔ f = 0,
(ii)
kαf k = |α| kf k ,
(iii) kf + gk ≤ kf k + kgk .
The Lebesgue-integral is not able to deal with (i): The function’s values on sets of measure
0 do not affect the value of the integral. In the L p -theory one has decided to weaken (i)
and simply say that f = 0 if f (x) = 0 a.e.
That is the same as saying that a function f ∈ L p represents a whole bunch of functions
(a class), namely all the functions g such that g (x) = f (x) a.e.
This may sometimes be confusing, e.g. when we say that f ∈ L p is continuous: What we
then mean is that the class of f contains a continuous function (only one!).
The most important Lp -spaces for us are L1 and L2 .
The latter is a Hilbert space, which is characterized by having a scalar product,
Z
hf, gi =
f (x) g (x) dx.
T
If the functions take complex values, the definition contains a complex conjugate:
Z
hf, gi =
f (x) g (x)dx.
T
(Different books have different ways of writing this).
The norm in a Hilbert space is defined by means of the scalar product, and is equal to the
regular L2 -norm:
Z
1/2
1/2
kf k2 = hf, f i =
f (x) f (x)dx
.
T
1
The scalar product is involved in the most important inequality in Hilbert space, namely
the Schwarz’ inequality:
|hf, gi| ≤ kf k2 kgk2 .
To prove Schwarz’ inequality, consider the polynomial p (λ) = kf + λgk 22 and observe that
0 ≤ p (λ) (Fill in details!).
It is natural to ask whether one of the spaces above is contained in the other, and this can
be resolved as follows:
For a measure space T with finite measure (m (T ) < ∞), L 2 (T ) is contained in L1 (T ):
Consider a function f ∈ L2 (T ) and the set A where |f (x)| > 1. There |f (x)| < |f (x)| 2 ,
and hence
Z
Z
Z
|f (x)| dx
|f (x)| dx +
|f (x)| dx =
T \A
A
T
Z
Z
2
1 · dx
|f (x)| dx +
≤
T \A
A
Z
|f (x)|2 dx + m (T ) < ∞.
≤
T
A more direct proof using the Schwarz’ inequality goes as follows:
Z
Z
|f (x)| dx =
1 · |f (x)| dx ≤ k1k2 kf k2 < ∞,
T
since
R
T
T
12 dt = m (T ) < ∞. Thus, L2 (T ) ⊂ L1 (T ), and
kf k1 ≤ m (T )1/2 kf k2 .
The opposite inclusion is not true: Let T = [0, 1] and consider f (x) = x −1/2 which is in L1 ,
but not in L2 .
For a measure space with infinite measure, m (T ) = ∞, none of the spaces are included in
each other!
Consider T = [0, ∞] and the function f (x) = x −1/2 , set to 0 for x > 1. Then consider
0, x < 1,
g (x) =
1/x x > 1.
NB! A probability space Ω has finite measure, P (Ω) = 1. Thus, L 2 (Ω) ⊂ L1 (Ω), and
r E (|X|) ≤ E |X|2 .
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Convergence
Mathematical arguments always talk about convergence, convergent series, sequences etc.
There exists a lot of different types of convergence, and that may be confusing.
Taking a general viewpoint, a sequence of numbers, functions etc., {x n }∞
n=1 is said to converge to x when n → ∞ if the distance between x n and x tends to 0 when n → ∞. What
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distance we talk about depends on the circumstances: If the x-es are numbers, the distance
is simply |xn − x|, whereas if we are dealing with functions, the distance may be a norm,
for example an Lp -norm.
The concept of a Cauchy sequence is important when we deal with convergence. A Cauchy
sequence {xn }∞
n=1 has the property that the distance between its members tends to 0 when
n → ∞. This may be expressed as
sup |xm − xn | −→ 0.
n→∞
m>n
Obviously, a convergent sequence (convergent to x) is a Cauchy sequence, since
sup |xm − xn | = sup |xm − x + x − xn | ≤ sup |xm − x| + |xn − x| ,
m>n
m>n
m>n
and both terms on the RHS tend to 0. However, the definition of a Cauchy sequence as
such does not mention convergence.
Nevertheless, for the real or complex numbers, all Cauchy sequences converge. This is a very
fundamental property of the numbers. For function spaces is not so obvious: The collection
of Riemann-integrable functions with finite L p -norm does not this property. But the big
triumph of the Lebesgue integral is that it makes the L p -spaces complete:
If we use the Lebesgue integral, all Cauchy sequences in the L p - spaces converge to functions
in Lp ! Spaces with this property are called complete spaces.
A complete, normed space is called a Banach space, and if it, in addition, has a scalar
product defining the norm, it is a Hilbert space.
Collection of functions in a function space with this property is called closed (a generalization from closed sets on the real line).
Let us now turn to some questions about different types of convergence.
There is no simple connection between pointwise convergence a.e., that is
fn (x) −→ f (x) a.e.
(1)
kfn − f k −→ 0.
(2)
n→∞
and convergence in norm,
n→∞
In general, the first does not imply the second, or the opposite.
In some cases it is, however, possible to say more:
If the conditions of monotone or dominated convergence apply, then (1) implies (2) (In
L1 ).
The following proposition is also quite useful:
Consider a series
∞
X
fk (x)
k=1
and assume that
∞
X
kfk k1 < ∞.
k=1
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Then
P∞
n=1 fn (x)
converges absolutely a.e. and f (x) =
P∞
n=1 fk
(x) is in L1 .
(This proposition is proved in the note about the Lebesgue-integral, but try to derive
yourself by considering the sequence {s n } defined as
sn (x) =
n
X
|fk (x)| ,
k=1
and apply the Monotone Convergence Theorem).
Even if kfn − f kp → 0 does not imply that
fn (x) −→ f (x) a.e.
n→∞
in general, it is possible to find a sub-sequence {f nk }∞
k=1 such that
fnk (x) −→ f (x) a.e.
k→∞
This is easy to see in L1 , and a similar argument is true in all L p -spaces ( p < ∞): Since
kfn − f k1 → 0 , it is possible to find a subsequence {n k }∞
k=1 such that
kfnk − f k1 ≤
and then
1
2k
∞
X
kfnk − f k1 < ∞.
∞
X
(fnk (x) − f (x))
k=1
By the proposition above,
k=1
converges absolutely a.s., and therefore the k-th term in this convergent series tends to 0,
that is,
fnk (x) − f (x) −→ 0 a.e.
k→∞
Advanced probability theory contains numerous results about (numerous types of) convergence. One theorem says that a convergent sequence of independent random variables in
L2 (Ω) converges pointwise a.s., but here independence is essential.
You should consult Exercise Set 1 for an example of a sequence of functions which converges
pointwise to 0 for all x without necessarily converging in norm.
Finally, let
fn (ω) = χAn (ω) , n = 1, 2, · · ·
where {An } is a sequence of independent sets such that P (A n ) −→ 0 and
n→∞
X
P (An ) = ∞.
n=1
Then kfn kL1 (Ω) = P (An ) −→ 0. What does the Borel-Cantelli Lemma say about lim n→∞ fn (ω)?
n→∞
(It is possible to make up simpler cases using functions like the ones in Exercise Set 1).
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