MEASURE THEORY AND L^{p} SPACES WOMP Program September 16, 2004

1. Various types of convergence for sequences of measurable functions

1.1 ”Almost everywhere” convergence

Definition Let (X, A, μ) be a measure space and f_{n} : X → R̄ a sequence of functions. The sequence of functions f_{n} converges μ-a.e (i.e. almost everywhere) (or, when no confusion is possible, just a.e.) on X to the function f if the set

A = {x; f_{n}(x) ↛ f(x)}

has μ−measure 0.

Example: The sequence f_{n}(x) = (−x)^{n} defined for all x ∈ [0, 1] is a sequence of functions convergent μ-a.e. with respect to the Lebesgue measure μ, since for x ∈ [0, 1) we have lim_{n→∞} f_{n}(x) = 0 and for x = 1 the limit of f_{n}(x) doesn’t exist and the set A = {1} has μ(A) = 0.

Theorem Let f_{n} : X → R̄ be a sequence of measurable functions on X. If f_{n} converges μ− a.e. to f then f is measurable.

Proof: Let A = {x; f_{n}(x) ↛ f(x)}. From hypothesis we have that μ(A) = 0. Let’s associate to the sequence f_{n}, the sequence f̄_{n} : X → R̄ defined by:

f̄_{n}(x) = \{ \begin{matrix}f_{n}(x) & if x ∈ X − A \\ 0 & if x ∈ A \end{matrix}

It clear that f̄_{n} are measurable functions on X and

lim_{n→∞} f̄_{n}(x) = f̄(x) = \{ \begin{matrix}f(x) & if x ∈ X − A \\ 0 & if x ∈ A \end{matrix}

Since the limit of a sequence of measurable functions is also a measurable function and since f̄(x) = f(x)(μ − a.e.) we have that f is measurable as well. √

Theorem If the sequence of measurable functions f_{n} : X → R̄ converges a.e. to two functions f and g then f = g-a.e.

Proof: Let A = {x; f_{n}(x) ↛ f(x)}and B == {x; f_{n}(x) ↛ g(x)} with μ(A) = μ(B) = 0 and f̄, ḡ be the associated functions as in the previous proof. Then we have that:

f̄(x) = f(x), ∀x ∈ X − A, ḡ(x) = g(x), ∀x ∈ X − B

hence, since for x ∈ X − (A ∪ B), f̄(x) = ḡ(x), we get that for all x ∈ X − (A ∪ B), f(x) = g(x), thus f = g μ-a.e. since 0 ≤ μ(A ∪ B) ≤ μ(A) + μ(B) = 0.

Exercise: It’s very easy to prove analogously the following

Theorem A sequence of measurable and a.e. finite functions f_{n} : X → R̄ is convergent a.e. on X if and only if the sequence is Cauchy-a.e.

1.2 ”Almost uniform” convergence

Definition Let (X, A, μ) be a measure space and f_{n} : X → R̄ a sequence of functions finite a.e. The sequence is said to converge almost uniformly to the function f (for short a.u.) if , for any ϵ > 0 there exists a set E_{ϵ} ∈ A with μ(E_{ϵ}) < ϵ so that the sequence converges uniformly on X − E_{ϵ} to f.

It’s natural to ask whether there exists any relation between a.e. convergence and a.u. convergence. A first result in this direction is given by the following:

Theorem Let f_{n} : X → R̄(n ∈ N). If the sequence f_{n} converges a.u. to f then it converges a.e. on X to f.

Proof: Let’s take ϵ = \frac{1}{m}(m ∈ N) in the previous definition; then there exists the set E_{m} ∈ A so that μ(E_{m}) < \frac{1}{m} and f_{n} converges uniformly on X − E_{m} to f.

Let’s denote by E = ∩_{m=1}^{∞}E_{m}. From this we have:

μ(E) ≤ μ(E_{m}) < \frac{1}{m}

for any m ∈ N. hence μ(E) = 0.

If x ∈ X − E = X − ∩_{m=1}^{∞}E_{m} = ∪_{m=1}^{∞}(x − E_{m}) then there exists an m_{0} so that x ∈ X − E_{m_{0}} so the sequence f_{n} converges at this point.

Let’s denote by

A_{n} = {x; x ∈ X, |f_{n}(x)| = +∞}

and by A = ∪_{n=1}^{∞}A_{n}; it’s clear then that A ⊂ E and since μ(E) = 0 the sequence f_{n} converges a.e. on X to the function f finite a.e. √.

One could ask if the converse is true and the answer is NO, in general. However, if the space X is of finite measure then the converse holds. This is known as:

Egorov’s Theorem Let (X, A, μ) be a measure space, with finite measure i.e. μ(X) < ∞. If the sequence f_{n} : X → R̄ of measurable, finite a.e. functions on X converges a.e. to the function f then the sequence converges a.u. to f, as well.

The proof is a bit intricate and the reader is referred to [1].

Observation: If the space is not of finite measure the the result of Egorov’s theorem doesn’t hold in general, as one can see taking X = N and A = P(X) (i.e. the set of subsets of X) and for any finite set A ⊂ N we take μ(A) to be the number of elements of A and for A infinite μ(A) = ∞.

Let χ_{n} be the characteristic function of the set {1, 2, …, n}. It’s clear that lim_{n} χ_{n}(x) = 1 for any point x ∈ X but the sequence χ_{n} is not convergent a.u. to 1 because it’s not Cauchy a.u. (the sequence f_{n} is Cauchy a.u. if for any ϵ > 0 there exists a set E_{ϵ} ∈ A with μ(E_{ϵ}) < ϵ so that f_{n} − f_{m} → 0 uniformly on X − E_{ϵ} when n, m → ∞.)

1.3 Convergence ”in measure”

The concept of convergence in measure is of particular importance in probability theory where it’s known as convergence in probability. Hence of particular interest for the study of the convergence in measure will be the case when μ(X) < ∞.

Definition Let (X, A, μ) be a measure space, and f, f_{1}, f_{2}, … real, measurable functions, finite a.e. on X. We’re saying that the sequence f_{n} converges in measure to f if, for any ϵ > 0,

lim_{n→∞} μ(x; |f_{n}(x) − f(x)| ≥ ϵ) = 0.

One part of the relation between a.u. convergence and the convergence in measure is given by the following:

Theorem If f_{n} : X → R̄ is a sequence of measurable functions converging a.e. to f then f_{n} converges in measure to f.

Proof: Let ϵ > 0 and A ∈ A so that μ(A) < ϵ and f_{n} converges in measure to f on X − A. If δ is an arbitrary positive number then there exists an n_{0} ∈ N so that

|f_{n}(x) − f(x)| < δ

for n ≥ n_{0} and ∀x ∈ X − A. However

x; |f(x) − f_{n}(x)| ≥ δ ⊂ A

so that

μ(x; |f_{n}(x) − f(x)| ≥ δ) ≤ μ(A) < ϵ

for n ≥ n_{0} which shows that f_{n} converges in measure to f. √ Combining the above theorem with Egorov’s theorem we get: Theorem Let (X, A, μ) be a finite measure space. Any sequence (f_{n}) of measurable and a.e. finite

functions convergent a.e. to the function f finite a.e. is convergent in measure to the same function f.

Remark: If the measure of the space X is not finite, then the result of the last theorem doesn’t hold, in general, as one can see by takin X = R and μ to be the Lebesgue measure, with A = M the class of Lebesgue measurable sets. If f_{n} is the characteristic function of the interval [n, n + 1] then

lim_{n} f_{n}(x) = 0, ∀x ∈ R

but μ(x ∈ R; |f(x) − f_{n}(x)| ≥ 1) = μ([n, n + 1]) = 1 ≠ 0; so the sequence (f_{n}) doesn’t converges in measure to the function f = 0.

One can ask as well if the converses of the last two theorems are valid. In general the answer is NO, the convergence in measure doesn’t imply a.e. convergence, nor a.u. convergence. This can be seen from the following:

Example: Let X = [0, 1], μ = the Lebesgue measure and A = M the class of Lebesgue measurable sets. If, for any k ∈ N and any i = 1, 2, …, k we denote by

E_{k}^{i} = [\frac{i − 1}{k}, \frac{i}{k}]

and by χ_{k}^{i} the characteristic function of the set E_{k}^{i} then the sequence ψ_{n} made by {χ_{1}^{1}, χ_{2}^{1}, χ_{2}^{2}, χ_{3}^{1}, χ_{3}^{2}, χ_{3}^{3}, …, }, i.e. ψ_{n} = χ_{k}^{i} where \frac{k(k−1)}{2} < n ≤ \frac{k(k+1)}{2}, one notice that it is convergent in measure to 0, since for any 0 < ϵ ≤ 1, A_{n}(ϵ) = \{x; |ψ_{n}(x)| ≥ ϵ\} = [\frac{i−1}{k}, \frac{i}{k}] so that μ(A_{n}) = \frac{1}{k} and for ϵ > 1, A_{n} = Φ. Thus, for any ϵ > 0, μ(A_{n}) → 0 when n → ∞. One the other hand one can see that the sequence ψ_{n}(x) is not converging in any point x ∈ (0, 1], since the sequence ψ_{n}(x) contains an infinity of terms equal to 0 and an infinity of terms equal to 1.

2. Pointwise convergence and the convergence of the integrals

The following situation appears naturally and very often in various instances: given a sequence of functions f_{n} which converges pointwise (i.e. almost everywhere) to a function f what can we say about the relation between ∫_{X} f_{n} and ∫_{X} f? One should be cautious in asserting that the convergence of the functions implies the convergence of the integrals. Indeed, if one thinks of the case of the Riemann integrals one remembers that ONLY the UNIFORM CONVERGENCE of the sequence of the functions implies the convergence of the corresponding integrals.

One of the most remarkable features of the Lebesgue integral is that the above question can be answered positively under much weaker conditions than in the case of the Riemann integral.

2.1 The case of positive functions

One of the most used theorems, in the case that the functions are POSITIVE is:

Monotone Convergence Theorem (Lebesgue-Beppo-Levi) Let f_{n} : X → [0, ∞] be a sequence of measurable functions so that:

(∗)  0 ≤ f_{1}(x) ≤ f_{2}(x) ≤ … ≤ f_{n}(x) ≤ f_{n+1}(x) ≤ …, ∀x ∈ X.

If

f(x) = lim_{n→∞} f_{n}(x), ∀x ∈ X,

then

∫_{X} fdμ = lim_{n→∞} ∫_{X} f_{n}dμ.

Proof: From the monotony of the integral we get that ∫_{X} f_{n}dμ ≤ ∫_{X} f_{n+1}du, ∀n ∈ N, which implies the existence of the limit lim_{n→∞} ∫_{X} f_{n}dμ.

On the other hand f, being the limit of a sequence of measurable functions, is a measurable function as well, hence, from (∗) we have that f_{n}(x) ≤ f(x), for ∀x ∈ X and ∀n ∈ N hence:

∫_{X} f_{n}dμ ≤ ∫_{X} fdμ

for ∀n ∈ N. Passing to the limit in n we get:

lim_{n→∞} ∫_{X} f_{n}dμ ≤ ∫_{X} fdμ

Let’s show that the opposite inequality takes place and to this extent take s to be a simple function so that 0 ≤ s ≤ f and take an ϵ ∈ (0, 1). Consider the sets:

E_{n} = {x ∈ X; f_{n}(x) ≥ ϵs(x)}, ∀n ∈ N

Since the functions f_{n} and s are measurable , one has that the elements of the increasing sequence of sets E_{n} are measurable as well. Also, one can see that X = ∪_{n=1}^{∞}E_{n}. Indeed, if x ∈ X, and f(x) = 0 then s(x) = 0, which means that x ∈ E_{1}. If f(x) > 0 then, since ϵ < 1 we have ϵs(x) < f(x), which means that there exists a set E_{n} in the sequence so that x ∈ E_{n}.

Taking into account the monotony of the Lebesgue integral with respect to the domain and with respect to the function, we get that:

∫_{X} f_{n}dμ ≥ ∫_{E_{n}} f_{n}dμ ≥ ϵ ∫_{E_{n}} s dμ

which implies that

lim_{n→∞} ∫_{X} f_{n}dμ ≥ ϵ lim_{n→∞} ∫_{E_{n}} s dμ

It’s useful to recall now the fact the for a simple function s one can easily prove that ν(E) = ∫_{E} sdμ defines a measure. Remembering the continuity of a measure on a sequence of increasing sets we have that the last inequality becomes:

lim_{n→∞} ∫_{X} f_{n}dμ ≥ ϵ ∫_{X} s dμ.

Since the last relation is valid for any ϵ ∈ (0, 1) we get:

lim_{n→∞} ∫_{X} f_{n}dμ ≥ ∫_{X} sdμ

for any simple function s which satisfies 0 ≤ s ≤ f, thus taking into account the definition of the integral this implies that:

lim_{n→∞} f_{n}dμ ≥ ∫_{X} f dμ

which finishes the proof taking into account that the reverse inequality was already proved. √ A very easy consequence is the following: Theorem If f, g : X → [0, ∞] are measurable functions then

∫_{X} f + g dμ = ∫_{X} f dμ + ∫_{X} g dμ

Proof: By the theorem of structure of measurable functions one has that there exists the sequences of simple functions f_{n}, g_{n} so that f_{n} ↑ f and g_{n} ↑ g. Let’s notice that h_{n} = f_{n} + g_{n} is an increasing sequence of simple functions converging pointwise to f_{g}, and applying the previous theorem to the sequence h_{n} one gets:

lim_{n→∞} ∫_{X} h_{n}dμ = ∫_{X}(f + g)dμ

One the other hand, from the linearity of the Lebesgue integral defined on simple functions we have:

∫_{X} h_{n}dμ = ∫_{X} f_{n}dμ + ∫_{X} g_{n}dμ, ∀n ∈ N

so, applying again the monotone convergence theorem to the sequences f_{n} and g_{n} we get the desired conclusion. √

It’s a very easy and straightforward exercise to prove the following: Theorem If f_{n} : X → [0, ∞] is a sequence of measurable functions and

f(x) = Σ_{n=1}^{∞}f_{n}(x), ∀x ∈ X,

then

∫_{X} f dμ = Σ_{n=1}^{∞} ∫_{X} f_{n}dμ.

The next theorem, widely used, shows that assuming basically nothing, one can still obtain some information about the convergence of the integrals:

Fatou’s Lemma Let f_{n} : X → [0, ∞] a sequence of measurable functions. Then:

∫_{X}(lim inf_{n} f_{n})dμ ≤ lim inf_{n} ∫_{X} f_{n}dμ.

Proof: Let

g_{n}(x) = inf_{k≥n} f_{k}(x), ∀n ∈ N, x ∈ X

and

f(x) = lim inf_{n} f_{n}(x), ∀x ∈ X.

As (g_{n}(x)) is an increasing sequence for any x ∈ X, one has that

f = sup_{n} g_{n} = lim_{n→∞} g_{n}

Since g_{n} and f are inf and respectively limit of measurable functions, they are measurable as well, so the integrals ∫_{X} g_{n}dμ and ∫_{X} f dμ are defined.

Applying now the monotone convergence theorem to the sequence g_{n} we have (taking int account the last equality) that:

lim_{n→∞} ∫_{X} g_{n}dμ = ∫_{X} f dμ.

On the other hand we have that g_{n} ≤ f_{n} for ∀n ∈ N, which implies that

∫_{X} g_{n}dμ ≤ ∫_{X} f_{n}dμ, ∀n ∈ N

The last two inequalities imply:

∫_{X} f dμ = lim_{n→∞} ∫_{X} g_{n}dμ ≤ lim inf_{n} ∫_{X} f_{n}dμ

which finishes the proof.√

Remark: In the statement of Fatou’s lemma we can even have strict inequality. Indeed, let:

f_{n}(x) = \{ \begin{matrix}n & for x ∈ (0, \frac{1}{n}] \\ 0 & for x ∈ (\frac{1}{n}, 1] \end{matrix}

Let’s notice that lim inf_{n} f_{n} = 0 so that ∫_{0}^{1}(lim inf_{n} f_{n})dμ = 0 and, on the other hand lim inf_{n} ∫_{0}^{1} f_{n}dμ = 1 hence:

∫_{0}^{1}(lim inf_{n} f_{n})dμ < lim inf_{n} ∫_{0}^{1} f_{n}dμ

2.2 The case of general functions

Let’s recall that for an arbitrary function f : X → R̄ we define the positive and the negative part of f(which are both positive functions) as being:

f^{+} = sup{f, 0}, f^{−} = − inf{f, 0}

and taking into account the properties of measurable functions it’s easy to show that f is measurable if and only if f^{+} and f^{−} are measurable. Then we are calling f integrable if the integrals ∫_{X} f^{+} dμ, ∫_{X} f^{−} dμ are finite.

It is then very easy to prove the:

Theorem (The absolute convergence of the Lebesgue integral) The measurable function f : X → R̄ is integrable on X if and only if |f| is integrable on X.

We can now state and prove the most largely used criterion which insures that the integrals of pointwise convergent sequences of functions have convergent integrals, namely:

The dominated convergence theorem Let f_{n} : X → R̄ be a sequence of measurable functions. If g : X → [0, ∞] is integrable on X so that

(a)  |f_{n}| ≤ g, ∀n ∈ N

and the sequence f_{n} converges a.e. to the function f, then f is integrable and

lim_{n→∞} ∫_{X} f_{n}dμ = ∫_{X} f dμ

Proof: Relation (a) implies, taking into account the absolute convergence of the Lebesgue integral, that f_{n} is integrable for ∀n ∈ N. Also, passing to the limit in n in the relation (a) we get:

|f| ≤ g(a.e.)

hence, using again the absolute convergence of the Lebesgue integral, we get that f is integrable.

In order to prove the last part of the conclusion we need to consider first the sequence of functions g +f_{n} and to observe that the functions of this sequence are measurable and nonnegative , thus, by Fatou’s lemma we have:

∫_{X} lim inf_{n}(g + f_{n})dμ ≤ lim inf_{n} ∫_{X}(g + f_{n})dμ

hence

∫_{X} g dμ + ∫_{X}(lim inf_{n} f_{n})dμ ≤ ∫_{X} g dμ + lim inf_{n} ∫_{X} f_{n} dμ

and since f is integrable we get:

∫_{X}(lim inf_{n} f_{n})dμ ≤ lim inf_{n} ∫_{X} f_{n}dμ.

Recall now that the sequence f_{n} converges a.e. to the function f and then lim inf_{n} f_{n} = f (a.e.), which implies that the last inequality can be rewritten as:

(∗)  ∫_{X} f dμ ≤ lim inf_{n} ∫_{X} f_{n} dμ

On the other hand the sequence of functions (g − f_{n})_{n∈N} satisfies the assumptions of Fatou’s Lemma, as well, and this implies that:

∫_{X} lim inf_{n}(g − f_{n}) dμ ≤ lim inf_{n} ∫_{X}(g − f_{n}) dμ

which implies

∫_{X} g dμ − ∫_{X}(lim sup_{n} f_{n})dμ ≤ ∫_{X} g dμ − lim sup_{n} ∫_{X} f_{n} dμ.

We get from this that:

(∗∗)  ∫_{X} f dμ ≥ lim sup_{n} ∫_{X} f_{n} dμ.

Combining the relations (∗) and (∗∗) we have:

∫_{X} f dμ = lim sup_{n} ∫_{X} f_{n} dμ ≤ lim sup_{n} ∫_{X} f_{n} dμ ≤ ∫_{X} f dμ

which implies the conclusion. √

3. Comparison between Riemann and Lebesgue integrals

Let’s observe first that the Lebesgue integral is more general than the Riemann integral, and this is the result of the following:

Theorem If f : [a, b] → R is bounded and Riemann integrable on [a, b], then f is Lebesgue integrable on [a, b] and

(R) ∫_{a}^{b} f(x)dx = (L) ∫_{a}^{b} f dμ.

The proof of this theorem can be found for instance in [2].

The converse of this statement is not true in general. Indeed, the function

f(x) = \{ \begin{matrix}1 & for x rational in [a, b] \\ 0 & otherwise \end{matrix}

is Lebegue integrable and (L) ∫_{a}^{b} f dμ = 0 since f is 0 almost everywhere.

On the other hand the function is not Riemann integrable and this is proved for instance in Spivak’s book ” Calculus”. This shows that the class of functions which are Lebesgue integrable is strictly larger then that of the functions which are Riemann integrable.

In order to continue our comparison between the two integrals, we’ll denote for short the Lebesgue integral by (L) and the Riemann integral by (L).

A brief analysis of the definitions of the two types of integrals show that (L) is built on the class of measurable sets, while (R) is defined on intervals only (compact or in the case of generalized Riemann integrals even noncompact intervals)., which is strictly included in the class of measurable sets.

On the other hand in the process of (L) integration the sets of measure 0 don’t influence the integrability of the function, nor the value of the integral (the function could be even undefined on such sets), in the case of (R) one cannot neglect this.

It is natural to ask ourselves if the above mentioned theorem holds in the case of generalized Riemann integrals, and to this extent let’s start by recalling their definition.

Definition Let f : R → R. We are calling f Riemann integrable (in the generalized sense)on (−∞, ∞) if for any a, b ∈ R the function f is bounded and Riemann integrable on (a, b) and

lim_{b→∞,a→−∞} ∫_{a}^{b} f dx

exists and is finite.

The value of the above limit, denoted by (R) ∫_{−∞}^{+∞} is called the generalized Riemann integral.

We’ll show by the next example that the answer the previous question is negative, in general, namely that there exists functions Riemann integrable in the generalized sense which are not Lebesgue integrable.

To this extent let’s consider the function:

f(x) = \{ \begin{matrix}\frac{sin x}{x}, & if x ∈ (0, ∞) \\ 1 & if x = 0 \end{matrix}

which is known to be Riemann integrable in the generalized sense (again, Spivak’s Calculus would do it), on [0, ∞) but for which there is no ∫_{0}^{∞} |f|dx. On the other hand taking into account the absolute continuity of Lebesgue integral we have that f ∈ L([0, ∞)) iff |f| ∈ L([0, ∞)), and since [0, ∞) = ∪_{n=1}^{∞}[0, n) and since ν(A) = ∫_{A} |f|dμ for all sets A is a measure on the set of Lebesgue measurable sets, we get:

(L) ∫_{0}^{∞} |f|dμ = lim_{n→∞}(L) ∫_{0}^{n} |f|dμ = lim_{n}(R) ∫_{0}^{n} |f|dx = (R) ∫_{0}^{∞} |f| = +∞,

so f ∉ L([0, ∞)).

Another remarkable feature of (L) which makes it superior to (R) is the countable additivity with respect to the function or to the domain of integration which is clearly more general then the corresponding property for the (R) integral which allows only for finite additivity.

The superiority of (L) integral is also clear when it comes to passing to the limit under the integral, since for the (L) integral it suffices to have the terms of the sequence bounded by an integrable function (dominated convergence theorem) thus obtaining that the limit commutes with the integral sign (lim_{n} ∫_{X} f_{n} = ∫_{X} lim_{n} f_{n}). In the case of the(R) integral, for this we need the much more restrictive requirement that the convergence of the sequence f_{n} to be uniform.

4. L^{p} Spaces

4.1 The concept of normed spaces. Completeness. Banach and Hilbert Spaces

Some thought will show that concepts of real analysis, complex analysis and several variables functions theory are deeply related, and that it might be useful to express all these concepts in an unitary manner, in an abstract setting, obtaining results in abstract spaces. Then, by choosing these abstract spaces to be either R, C, R^{n}, C^{n} (and of course there are much more examples) one gets a lot of important theorems for free, and one is able to understand what are the peculiarities of various spaces and what are the common features. This is basically the main idea behind creating functional analysis (and topology, differential geometry, abstract algebra, and so on).

It turns out that a common framework for R, C and R^{n} for instance is the concept of normed linear space, which is just a linear space endowed with a norm.

Definition A function ||.|| : X → R+ is called a norm on the linear space X if it satisfies the following properties:

(N_{1})  ||x|| = 0 ⇔ x = θ

where θ is the origin of the linear space X.

(N_{2})  ||λx|| = |λ|||x||, ∀λ ∈ R, x ∈ X

(N_{3})  ||x + y|| ≤ ||x|| + ||y||, ∀x, y ∈ X

One can define a topology on these normed spaces by declaring open sets the union of open balls, where an open ball of center x and radius r is defined to be:

B(x, r) = {x ∈ X; ||x|| < r}

Having a topology one can talk of limits and convergence and one can decide if a sequence has a limit with respect to this topology or not. It’s useful at this point to recall that in R, R^{n}, or C there was an equivalent way of deciding wheather a sequence has a limit or not, even without knowing or suspecting what that limit might be, namely by checking if that sequence if Cauchy (recall that a sequence x_{n} is Cauchy iff ∀ϵ > 0, ∃n_{0}, s.t. ∀n, m ≥ n_{0} |x_{n} − x_{m}| ≤ ϵ where |.| stands for the norm).

While in all the spaces known so far the concept of Cauchy sequence and sequence which has a limit were equivalent, this is no longer true in the general case of normed spaces. The normed spaces for which this two concepts are equivalent, namely a sequence has a limit iff it’s Cauchy, are called complete (for the very good reason that if one has a space where this equivalence is not true one can embed it in a large one by ”completing” it, i.e. by adding the missing elements which should be limits for some Cauchy sequences but which are not in the space).

A normed linear spaces which is complete with respect to the topology induced by the norm is called a Banach space.

While Banach spaces resemble a lot the spaces we were used to so far, namely R, R^{n} or C, there is a feature which is still missing and this is the geometrical flavor of these spaces. This is captured by the concept of Hilbert spaces.

A Hilbert space is a particular type of Banach space, where the norm is of a very particular type namely it’s obtained from a scalar product which is an application < ., . >: X × X → R satisfying the following properties:

(SP_{1})  < x, y >=< y, x >, ∀x, y ∈ X

(SP_{2})  < x, y_{1} + y_{2} >=< x, y_{1} > + < x, y_{2} >, ∀x, y_{1}, y_{2} ∈ X

(SP_{3})  < λx, y >= λ < x, y >, ∀x, y ∈ X, λ ∈ R

(SP_{4})  < x, x >≥ 0; < x, x >= 0 ⇔ x = 0

4.2 The Spaces L^{p} and L^{p} for 1 ≤ p < ∞ Let (X, A, μ) be an arbitrary measure space, and

1 ≤ p < ∞.

Definition The application f : X → R̄ is called p^{th} power integrable if : I. f is measurable II. ∫_{X} |f|^{p} dμ < ∞

We’ll denote by L^{p}(X, A, μ) or L^{p}(X) or L^{p} the set of all functions which are p^{th} power integrable on X.

On L^{p} we’ll introduce the equivalence relation ρ given by

fρg ⇔ f(x) = g(x)(a.e.)

We’ll denote by L^{p}(X, A, μ) or L^{p}(X) or L^{p} the quotient space L^{p}/ρ.

In the following we’ll show that L^{p} is a normed linear space. For the sake of simplicity we’ll talk about elements of L^{p} assuming that there is no distinction between two functions which are in the relation ρ. Using this widely accepted convention we’ll continue to talk about L^{p} as a space of functions although in reality the elements of L^{p} are classes of equivalence of functions.

It’s very easy to check the following: Theorem The space L^{p} is a real linear space. Proof: I. It’s clear that for f ∈ L^{p}, α ∈ R we have αf ∈ L^{p}. Let’s check now that if f, g ∈ L^{p} then f + g ∈ L^{p}. Indeed, if f, g ∈ L^{p} then f + g is measurable and

|f + g|^{p} ≤ (|f| + |g|)^{p} ≤ 2^{p} · max{|f|^{p}, |g|^{p}}

and since |f|^{p} and |g|^{p} are integrable this implies that f + g ∈ L^{p}. √ Definition For 1 < p < ∞ we define the conjugate of p to be the number q so that

\frac{1}{p} + \frac{1}{q} = 1

A pair of very useful and widely used inequalities are the following: Inequalities I. Hölder’s inequality For f ∈ L^{p}, (1 < p < ∞) and g ∈ L^{q} we have that

∫_{X} |fg|dμ ≤ (∫_{X} |f|^{p})^{\frac{1}{p}} (∫_{X} |g|^{q} dμ)^{\frac{1}{q}}

II.Minkowski’s inequality If 1 ≤ p < ∞ and f, g ∈ L^{p} then:

(∫_{X} |f + g|^{p}dμ)^{\frac{1}{p}} ≤ (∫_{X} |f|^{p} dμ)^{\frac{1}{p}} + (∫_{X} |g|^{p} dμ)^{\frac{1}{p}}

The proof are left as exercises for the reader. The following hints might be useful: for the Hölder inequality one could use the so called Young’s inequality which says that for two positive numbers a, b we have ab ≤ \frac{a^{p}}{p} + \frac{b^{q}}{q} where p and q are conjugate. Also for the proof of Minkowski’s inequality one could use Hölder’s inequality, see for instance [3].

It is now very easy to check that

||f||_{p} = (∫_{X} |f|^{p})^{\frac{1}{p}}

is a norm and thus L^{p} is a normed linear space with respect to the above norm. The more complicated part is to check that this space is complete with respect to this norm, i.e. it is a Banach space. But this is the result of the following:

Theorem The space L^{p} , (1 ≤ p < ∞) is a Banach space. Proof: Let f_{n} be a Cauchy sequence with respect to the above norm, which is to say that for any ϵ > 0

there exists n_{0}(ϵ) so that for m, n ≥ n_{0}(ϵ)

||f_{n} − f_{m}||_{p} < ϵ

The sequence (f_{n}) contains a subsequence f_{n_{k}}, n_{1} < n_{2} < … so that

||f_{n_{k+1}−f_{n_{k}}} ||_{p} ≤ \frac{1}{2^{k}} , (k = 1, 2, 3, …)

Let

g_{j} = Σ_{k=1}^{j}|f_{n_{k+1}−f_{n_{k}}} |, g = Σ_{k=1}^{∞}|f_{n_{k+1}−f_{n_{k}}} |

Using the last two relations and Minkowski’s inequality we get that ||g_{j}||_{p} ≤ 1 for j = 1, 2, . . .. Applying the Fatou’s lemma to the sequence (g_{j}^{p}), we obtain that ||g||_{p} ≤ 1, which implies that g is

finite a.e. on X. As a result the series

f_{n_{1}}(x) + Σ_{k=1}^{∞}(f_{n_{k+1}}(x) − f_{n_{k}}(x))

is absolutely convergent a.e. and let’s denote by f(x) its sum in those points where it is convergent, and let f(x) = 0 in the points x where it is not convergent.

We’ll prove now that f ∈ L^{p} and f_{n} → f in L^{p}. To this extent let’s observe that the partial sum S_{j}(j = 1, 2, …) of the telescopic series from above is precisely f_{n_{j}}(j = 1, 2, …), hence

f(x) = lim_{k→∞} f_{n_{k}}(x)(a.e.)

Let m > n_{0} for which ||f_{n} − f_{m}||_{p} < ϵ and let’s denote by h_{k} = |f_{n_{k}} − f_{m}|^{p}, (k ∈ N). Applying now to the sequence (h_{k})_{k∈N} the Fatou’s lemma, and taking into account the fact that f(x) is the limit of f_{n_{k}}(x) we get:

∫_{X} |f − f_{m}|^{p} dμ ≤ lim inf_{k→∞} ∫_{X} |f_{n_{k}} − f_{m}|^{p} dμ ≤ ϵ^{p}

The last relation insure that f − f_{m} ∈ L^{p} which implies that f = (f − f_{m}) + f_{m} ∈ L^{p} and, at the same time ||f − f_{m}||_{p} → 0 for m → ∞ which finishes the proof. √

An immediate consequence is: Corollary The space L^{2}(X) with the scalar product:

< f, g >= ∫_{X} f(x) · g(x) dμ

is a Hilbert space.

In order to see what’s the relation between simple functions and functions in L^{p} let’s denote by S^{p} the space of all simple functions which are of p^{th} power integrable. One can immediately see that S^{p} is a linear subspace of L^{p}. The we have:

Theorem The space S^{p} is dens in L^{p}, (1 ≤ p < ∞). Proof: Let f ∈ L^{p}. Since f = f^{+} − f^{−} and since positive functions can be approximated by an

increasing sequence of simple functions, we have that there exists an increasing sequence g_{n} of simple, nonnegative functions, converging pointwise to f^{+} and analogously, a sequence h_{n} converging pointwise to f^{−}.

Let f_{n} = g_{n} − h_{n} which is clearly a simple function. We’ll show that f_{n} converges in the L^{p} norm to f. Let’s start by observing that

|f_{n}| = |g_{n} − h_{n}| ≤ |g_{n}| + |h_{n}| ≤ f^{+} + f^{−} = |f|, ∀n ∈ N,

which implies that

|f_{n}|^{p} ≤ |f|^{p}

On the other hand ∫_{X} |f|^{p} dμ < ∞ by hypothesis, hence f_{n} ∈ L^{p}. Let ϕ_{n} = |f − f_{n}|^{p} and let’s observe that this sequence satisfies all the conditions of the dominated

convergence theorem. Indeed:

(a)ϕ_{n} ∈ L^{1}, ∀n ∈ N, since f, f_{n} ∈ L^{p} implies that ϕ_{n} = |f − f_{n}|^{p} ∈ L^{1}. (b), ϕ_{n} = |f − f_{n}|^{p} ≤ 2^{p}|f|^{p} ( since |f_{n}|^{p} ≤ |f|^{p}), so there exists the integrable function g = 2^{p}|f|^{p} which

dominates the terms of the sequence ϕ_{n}.

(c)ϕ_{n} → 0 because |f − f_{n}| → 0 when n → ∞. Thus, according to the dominated convergence theorem, one gets that lim_{n→∞} ∫_{X} ϕ_{n} dμ = 0, which is

lim_{n→∞} ∫_{X} |f − f_{n}|^{p} dμ = 0

or

lim_{n→∞} ||f − f_{n}||_{p} = 0

which says precisely that S^{p} is dense in L^{p} in the sense of the L^{p} norm √.

4.3 The Space L^{∞} Let (X, A, μ) be a measure space and f : X → R̄ be an A− measurable function. We’re defining the essential upper bound , or the essential supremum of f to be:

ess sup f = inf{α ∈ R̄; μ({x; f(x) > α}) = 0

which is the smallest α ∈ R̄ for which f ≤ α(μ − a.e.). Let’s denote by

||f||_{∞} = ess sup |f|

and by L^{∞}(X, A, μ) or L^{∞}(X) or L^{∞} the space of functions f : X → R̄ for which ||f||_{∞} < ∞ and by L^{∞}(X, A, μ) or L^{∞}(X) or simply L^{∞} the quotient space of L^{∞} with respect to the equivalence relation ρ previously defined.

Keeping the convention made in the case of L^{p} spaces (of identifying the classes of equivalence with any representative) one can easily verify that L^{∞} is a normed real space with respect to the norm ||.||_{∞} (the triangle inequality, i.e. Minkowski’s inequality for p = ∞ is immediately verified since:

|f + g| ≤ |f| + |g| ≤ ||f||_{∞} + ||g||_{∞}(a.e.)

which implies that

||f + g||_{∞} ≤ ||f||_{∞} + ||g||_{∞}

Now, let’s see an interpretation of what the convergence in the ||.||_{∞} norm means: Theorem The sequence (f_{n}) ⊂ L^{∞} converges to f ∈ L^{∞} in the sense of the ||.||_{∞} norm if and only if

there exists a set A ∈ A with μ(A) = 0 so that f_{n} → f uniformly on X − A.

Proof: ” ⇒ ” If f_{n} → f in L^{∞} then for any k ∈ N, there exists an n_{0} so that

||f_{n} − f||_{∞} < \frac{1}{k}

for n ≥ n_{0}, which means that there exists a set A_{k} ∈ A so that μ(A_{k}) = 0 and

|f_{n}(x) − f(x)| < \frac{1}{k}

for n ≥ n_{0} and x ∈ X − A_{k}(k = 1, 2, …)

Let’s denote by A = ∪_{k=1}^{∞}A_{k}; one can see that μ(A) = 0 and f_{n} → f uniformly on X − A.

” ⇐ ” Let’s assume that ∃A ∈ A with μ(A) = 0 so that f_{n} → f uniformly on X − A so, for any ϵ > 0 there exists n_{0} ∈ N so that for n ≥ n_{0}

|f_{n}(x) − f(x)| < ϵ, ∀x ∈ X − A

which implies that ||f_{n} − f||_{∞} < ϵ, for n ≥ n_{0} i.e.

f_{n} → f

in L^{∞}, which finishes the proof. √

Finally, we discover that L^{∞} is a Banach space as well as the other L^{p} spaces, i.e. it is complete with respect to the norm ||.||_{∞}.

Theorem The space L^{∞} is a Banach space. Proof: Let (f_{n}) ⊂ L^{∞} be a Cauchy sequence. Then there exists A ∈ A with μ(A) = 0 so that f_{n}−f_{m} → 0

uniformly on X − A, for n, m → ∞.

Thus the sequence f_{n} converges uniformly on X − A to a function f. If we extend the function f on A as well, taking f(x) = 0 for ∀x ∈ A we get that f ∈ L^{∞} and f_{n} → f in the L^{∞} norm √.

Remarks: I.Without any difficulty the reader can show that Hölder’s inequality remains true for p = ∞, i.e. for f ∈ L^{1} and g ∈ L^{∞} we have

∫_{X} |fg| dμ ≤ ||g||_{∞}||f||_{1}

II. We have that the space of simple functions which are bounded is dense in L^{∞} as well.

Refrences:

[1] Halmos R. ”Measure Theory”

[2] Hewitt E. and Stromberg K. ”Real and abstract analysis”

[3] Rudin, W. ”Real and complex analysis”