§ 6. Convexity inequalities

1. The convexity theorem

Theorem 1

— Let $X$ be a locally compact space, $\mu$ a positive measure on $X$ , $F$ a real Banach space, $D$ a closed convex set in $F$ , and $f$ a function on $X$ such that $f(X) \subset D$ . For every non-negligible integrable numerical function $g \geqslant 0$ such that $fg$ is integrable, the point $\frac{\int fg\, d\mu}{\int g\, d\mu}$ belongs to $D$ .

For, let $F'$ be the dual of $F$ and let $\langle z, a' \rangle \leqslant \alpha$ ( $a' \in F', \alpha \in \mathbf{R}$ ) be a relation defining a closed half-space containing $D$ . Since $fg$ is integrable, so is the numerical function $\langle fg, a' \rangle = \langle f, a' \rangle g$ , and

\int \langle fg, a' \rangle\, d\mu = \left\langle \int fg\, d\mu, a' \right\rangle

(§ 4, No. 2, Cor. 1 of Th. 1); but, by hypothesis, $\langle f(x), a' \rangle \leqslant \alpha$ for all $x \in X$ , therefore $\langle f(x)g(x), a' \rangle \leqslant \alpha g(x)$ ; on integrating, we have

\left\langle \int fg\, d\mu, a' \right\rangle \leqslant \alpha \int g\, d\mu .

This proves that the point $\frac{\int fg\, d\mu}{\int g\, d\mu}$ belongs to every closed half-space containing $D$ ; but, by the Hahn–Banach theorem, $D$ is the intersection of the closed half-spaces containing it (TVS, II, §5, No. 3, Cor. 1 of Prop. 4), whence the theorem.

Corollary

— If the positive measure $\mu$ has total mass equal to 1 and if $f$ is integrable, then $\int f\, d\mu$ belongs to the closed convex envelope of $f(X)$ in $F$ .

It suffices to take for $g$ the constant function 1.

2. Inequality of the mean

We are going to sharpen Th. 1 for numerical measurable functions (finite or not).

Definition 1

— Let X be a locally compact space, $\mu$ a measure on X. Given a numerical function f (finite or not), defined locally almost everywhere in X, we call maximum in measure, or $\mu$ -maximum (resp. minimum in measure, or $\mu$ -minimum) of the function f, and denote by $M_\infty(f)$ (resp. $m_\infty(f)$ ), the infimum (resp. supremum) of the set of numbers $\alpha$ such that $f(x) \leq \alpha$ (resp. $f(x) \geq \alpha$ ) locally almost everywhere (for $\mu$ ).

It follows at once from the definition that $m_\infty(f) = -M_\infty(-f)$ , thus from every property of the maximum in measure one deduces a corresponding property of the minimum in measure.

For every $\alpha > M_\infty(f)$ , the set of $x \in X$ such that $f(x) > \alpha$ is locally negligible; now, the set of $x \in X$ such that $f(x) > M_\infty(f)$ is the union of the sets where $f(x) > r_n$ , with $r_n$ running over the set of rational numbers $> M_\infty(f)$ ; therefore $f(x) \leq M_\infty(f)$ locally almost everywhere (§ 5, No. 2). Similarly $f(x) \geq m_\infty(f)$ locally almost everywhere; it follows that $m_\infty(f) \leq M_\infty(f)$ if the measure $\mu$ is nonzero; moreover, the relation $m_\infty(f) = M_\infty(f)$ is equivalent to saying that $f$ is equal to a constant locally almost everywhere. It is clear that if the measure $\mu$ is nonzero, then

\inf_{x \in X} f(x) \leq m_\infty(f) \leq M_\infty(f) \leq \sup_{x \in X} f(x).

If two functions $f, g$ are equal locally almost everywhere, then $m_\infty(f) = m_\infty(g)$ and $M_\infty(f) = M_\infty(g)$ .

Finally, if $f$ and $g$ are two functions such that $f + g$ is defined locally almost everywhere, then

(1)

M_\infty(f + g) \leq M_\infty(f) + M_\infty(g)

provided the second member is defined, as follows at once from Def. 1; similarly, if

f

and

g

are both

\geq 0

and are such that

fg

is defined locally almost everywhere, then

(2)

M_\infty(fg) \leq M_\infty(f) M_\infty(g)

provided the second member is defined.

If $M_\infty(f) < +\infty$ , then $f(x) < +\infty$ locally almost everywhere, but not necessarily almost everywhere. A numerical function $f$ is said to be bounded in measure (for the measure $\mu$ ) if it is defined and finite almost everywhere and if, moreover, the numbers $m_\infty(f)$ and $M_\infty(f)$ are both finite (the latter condition amounts to saying that $M_\infty(|f|) < +\infty$ ).

Proposition 1 (Inequality of the mean)

— *Let f be a measurable numerical function that is bounded in measure. For every integrable numerical function $g \geqslant 0$ , the function $fg$ (defined almost everywhere) is integrable and

m_\infty(f) \int g d|\mu| \leqslant \int fg d|\mu| \leqslant M_\infty(f) \int g d|\mu|.

Moreover, two of the three members of the inequality (3) cannot be equal unless, in the set of $x \in X$ such that $g(x) \neq 0$ , f is equal to $M_\infty(f)$ almost everywhere or equal to $m_\infty(f)$ almost everywhere.

Indeed, $fg$ is measurable (§ 5, No. 3, Cor. 5 of Th. 1); moreover, the inequality $m_\infty(f)g(x) \leqslant f(x)g(x) \leqslant M_\infty(f)g(x)$ holds, not only locally almost everywhere, but even almost everywhere, because the set of points $x \in X$ where $g(x) \neq 0$ is a countable union of integrable sets (§ 5, No. 6, Lemma 1). It follows that $fg$ is integrable (§ 5, No. 6, Th. 5) and the inequality (3) holds. On the other hand, the function $M_\infty(f)g - fg$ is almost everywhere defined and equal to $(M_\infty(f) - f)g$ ; it is therefore $\geqslant 0$ almost everywhere in $X$ ; since the relation $M_\infty(f) \int g d|\mu| = \int fg d|\mu|$ is equivalent to $\int (M_\infty(f) - f)g d|\mu| = 0$ , it can hold only if the function $(M_\infty(f) - f)g$ is negligible, which completes the proof.

Setting aside the trivial case that $\int g d|\mu| = 0$ , the inequality (3) may be deduced from Th. 1 of No. 1 applied to the interval $D = [m_\infty(f), M_\infty(f)]$ . One can bring to Th. 1 of No. 1 the complements analogous to those of Prop. 1, that specify the case in which the point $(\int fg d\mu)/(\int g d\mu)$ belongs to the boundary of D (Exer. 2).

3. The spaces $L^\infty_F$

Definition 2

For every mapping $f$ of $X$ into a Banach space $F$ , one sets $N_\infty(f) = M_\infty(|f|)$ ; $f$ is said to be bounded in measure (for the measure $\mu$ ) if $N_\infty(f)$ is finite. The set of mappings of $X$ into $F$ that are measurable and bounded in measure is denoted $\mathscr{L}_F^\infty(X, \mu)$ (or $\mathscr{L}_F^\infty(\mu)$ , or simply $\mathscr{L}_F^\infty$ ).

A function $f$ in $\mathscr{L}_F^\infty$ may thus be characterized by the fact that there exists a bounded measurable function equal locally almost everywhere to $f$ .

It follows immediately from (1) that

N_\infty(f + g) \leqslant N_\infty(f) + N_\infty(g);

on the other hand, $N_\infty(\alpha f) = |\alpha| N_\infty(f)$ for every scalar $\alpha$ . The set $\mathscr{L}_F^\infty$ is therefore a linear subspace of the space of all mappings of $X$ into $F$ , and $N_\infty(f)$ is a semi-norm on this vector space. Let $(f_n)$ be a sequence of functions in $\mathscr{L}_F^\infty$ that converges to $f \in \mathscr{L}_F^\infty$ for the topology defined by the semi-norm $N_\infty(f)$ ; for every integer $m$ , there exist a locally negligible set $H_m$ and an integer $n_0$ such that $|f(x) - f_n(x)| \leq 1/m$ for every integer $n \geq n_0$ and every $x \notin H_m$ (every countable union of locally negligible sets being locally negligible); the union $H$ of the $H_m$ is locally negligible, and one sees that $f_n(x)$ tends uniformly to $f(x)$ on the complement of the locally negligible set $H$ ; the converse is immediate.

It is clear that every function equal locally almost everywhere to a function in $\mathscr{L}_F^\infty$ belongs to $\mathscr{L}_F^\infty$ . In particular, the locally negligible functions defined on $X$ with values in $F$ form a linear subspace $\mathscr{N}_F^\infty$ of $\mathscr{L}_F^\infty$ , characterized by the relation $N_\infty(f) = 0$ (the closure of 0 for the topology defined by $N_\infty(f)$ ). The Hausdorff space associated with $\mathscr{L}_F^\infty$ , that is, the quotient space $\mathscr{L}_F^\infty / \mathscr{N}_F^\infty$ , is denoted $L_F^\infty(X, \mu)$ (or $L_F^\infty(\mu)$ or $L_F^\infty$ ); its topology is defined by the norm deduced from $N_\infty$ by passage to the quotient; the norm of a class $\dot{f} \in L_F^\infty$ is denoted $N_\infty(\dot{f})$ , or also $\| \dot{f} \|_\infty$ . When $F = \mathbf{R}$ (resp. $\mathbf{C}$ ), we write $\mathscr{L}^\infty$ and $L^\infty$ in place of $\mathscr{L}_\mathbf{R}^\infty$ and $L_\mathbf{R}^\infty$ (resp. $\mathscr{L}_\mathbf{C}^\infty$ and $L_\mathbf{C}^\infty$ ) if this can cause no confusion.

Proposition 2

The space* $\mathscr{L}_F^\infty$ is complete; the space $L_F^\infty$ is a Banach space.

For, let $(f_n)$ be a Cauchy sequence in $\mathscr{L}_F^\infty$ ; for every integer $n$ , there exists an integer $k_n$ such that $N_\infty(f_r - f_s) \leq 1/n$ for $r \geq k_n$ and $s \geq k_n$ ; thus, there exists a locally negligible set $A_{rs}$ such that $|f_r(x) - f_s(x)| \leq 1/n$ for all $x \notin A_{rs}$ . If $A_n$ is the union of the sets $A_{rs}$ (for $r \geq k_n$ and $s \geq k_n$ ), then $A_n$ is locally negligible and, for every $x \notin A_n$ , $|f_r(x) - f_s(x)| \leq 1/n$ for all indices $r \geq k_n, s \geq k_n$ . Let $A$ be the locally negligible set formed by the union of the $A_n$ , and set $g_n(x) = f_n(x)$ for $x \notin A$ , $g_n(x) = 0$ for $x \in A$ ; then $g_n$ belongs to $\mathscr{L}_F^\infty$ and, by the definition of $A$ , the sequence $(g_n)$ converges uniformly on $X$ to a function $g$ . It follows that the function $g$ is measurable (§ 5, No. 4, Th. 2); moreover, $g$ is bounded on the set of $x \in X$ where $|g_{k_1}(x)| \leq N_\infty(g_{k_1})$ and, since the complement of this set is locally negligible, $g$ belongs to $\mathscr{L}_F^\infty$ . It is clear that in $\mathscr{L}_F^\infty$ , the sequence $(g_n)$ has limit $g$ , and the same is therefore true of the sequence $(f_n)$ , since $N_\infty(f_n - g_n) = 0$ for all $n$ . The second part of the proposition may be deduced immediately from this.

Remark

— 1) Every bounded continuous function $f$ on $X$ with values in $F$ belongs to $\mathscr{L}_F^\infty$ , and

N_\infty(f) \leq \| f \| = \sup_{x \in X} |f(x)| .

In order that $N_\infty(f) = \|f\|$ for every bounded continuous function $f$ , it is necessary and sufficient that the support of the measure $\mu$ be equal to $X$ . For, if there exists a continuous function $f$ with negligible compact support and not identically zero, then $N_\infty(f) = 0$ and $\|f\| > 0$ . Conversely, if the support of $\mu$ is equal to $X$ then, for every bounded continuous function $f$ and every number $\alpha < \|f\|$ , the set of $x \in X$ such that $|f(x)| > \alpha$ is open and nonempty, hence has outer measure $> 0$ , which shows that $N_\infty(f) = \|f\|$ .

When the support of $\mu$ is equal to $X$ , we may therefore identify the normed space $C^b(X; F)$ , of bounded continuous functions on $X$ with values in $F$ , with a subspace of the space $L_F^\infty$ . Since $L_F^\infty$ is not in general Hausdorff, the subspace $C^b(X; F)$ is not in general closed in $L_F^\infty$ , but its canonical image in $L_F^\infty$ is a closed subspace of $L_F^\infty$ (which can moreover be identified with $C^b(X; F)$ in the case contemplated). In general, $C^b(X; F)$ is distinct from $L_F^\infty$ , that is, for an arbitrary bounded measurable function $f$ , there does not in general exist a continuous function $g$ equal to $f$ locally almost everywhere (§ 5, Exer. 12). This implies that the space $\mathscr{H}(X; F)$ of mappings of $X$ into $F$ , continuous with compact support, is in general not dense in $L_F^\infty$ , whereas it is dense in each of the spaces $L_F^p$ for $1 \leq p < +\infty$ (§ 3, No. 4. Def. 2).

2) It is immediate that the topology defined by the semi-norm $N_\infty$ is finer than the topology induced on $L_F^\infty$ by the topology of convergence in measure (§ 5, No. 11).

4. Hölder’s inequality

In this No., $p$ and $q$ will denote two real numbers such that $1 \leq p \leq +\infty$ , $1 \leq q \leq +\infty$ , bound by the relation $1/p + 1/q = 1$ ; thus $q = p/(p-1)$ if $1 < p < +\infty$ , $q = +\infty$ if $p = 1$ , and $q = 1$ if $p = +\infty$ ; $p$ and $q$ will be called conjugate exponents. Note that the relation $1 \leq p \leq 2$ is equivalent to $2 \leq q \leq +\infty$ ; $p = q$ only when $p$ and $q$ are equal to 2.

Theorem 2 (Hölder’s inequality)

— Let $f$ and $g$ be two numerical functions that are finite almost everywhere and are such that $f$ is equal almost everywhere to a function in $L^p$ and $g$ to a function in $L^q$ . Then, the function $fg$ (defined almost everywhere) is integrable, and

N_1(fg) \leq N_p(f) N_q(g).

Let $f_1$ (resp. $g_1$ ) be a function in $L^p$ (resp. $L^q$ ) to which $f$ (resp. $g$ ) is equal almost everywhere; $fg$ is equal almost everywhere to the function $f_1 g_1$ , which is everywhere defined and finite, and which is measurable, being the product of two measurable functions (§ 5, No. 3, Cor. 5 of Th. 1). If $1 < p < +\infty$ , Hölder’s inequality for the upper integral (Ch. I, No. 3, Prop. 4) yields the inequality (4), and the relation $N_1(fg) < +\infty$ then shows that $fg$ is integrable (§ 5, No. 6, Th. 5). If $p = 1$ , $q = +\infty$ , the inequality (4) and the fact that $fg$ is integrable are immediate consequences of the inequality of the mean (No. 2, Prop. 1); thus the theorem is proved in all cases.

Corollary 1

— Let $F, G, H$ be three Banach spaces, and let $(u, v) \mapsto \Phi(u, v)$ be a continuous bilinear mapping of $F \times G$ into $H$ such that $|\Phi(u, v)| \leq |u| \cdot |v|$ . If $f \in \mathscr{L}_F^p$ and $g \in \mathscr{L}_G^q$ , then the function $\Phi(f, g)$ is integrable and

\left| \int \Phi(f, g)\, d\mu \right| \leq \int |\Phi(f, g)|\, d|\mu| \leq N_p(f) N_q(g).

For, $\Phi(f, g)$ is measurable (§ 5, No. 3, Cor. 5 of Th. 1); since $|\Phi(f, g)| \leq |f| \cdot |g|$ , the corollary follows from Th. 2 and the integrability criterion of § 5, No. 6, Th. 5.

Two special cases of Cor. 1 are important in applications:

Corollary 2

— Let $F$ be a real (resp. complex) Banach space, $F'$ its strong dual (TVS, III, § 3, No. 1), and let $(z, z') \mapsto \langle z, z' \rangle$ be the canonical bilinear form on $F \times F'$ . If $f \in \mathscr{L}_F^p$ and $g \in \mathscr{L}_{F'}^q$ , then the real (resp. complex) function $\langle f, g \rangle$ is integrable and

\left| \int \langle f, g \rangle\, d\mu \right| \leq \int |\langle f, g \rangle|\, d|\mu| \leq N_p(f) N_q(g).

For, $|\langle z, z' \rangle| \leq |z| \cdot |z'|$ .

When $F$ is a real or complex Hilbert space, one knows that it can be canonically identified with its dual $F'$ (TVS, V, § 1, No. 7). Since the space $L_F^2$ is complete, we have the following result:

Corollary 3

— Let $\mu$ be a positive measure on $X$ , $F$ a real (resp. complex) Hilbert space. On the space $L_F^2$ , the symmetric (resp. Hermitian) form

(\widetilde{f}, \widetilde{g}) \mapsto \int \langle f, g \rangle\, d\mu

defines a Hilbert space structure, for which the norm is equal to $\| \widetilde{f} \|_2$ .

Corollary 4

— Let $F$ be a Banach space, $f$ a function in $\mathscr{L}_F^p$ , and $g$ a numerical function belonging to $\mathscr{L}^q$ ; then, the function $fg$ is integrable and

\left| \int fg\, d\mu \right| \leq \int |fg|\, d|\mu| \leq N_p(f) N_q(g).

Corollary 5

— Let $f_1, f_2, \ldots, f_n$ be n positive integrable functions, and let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be n numbers $> 0$ such that $\sum_{i=1}^n \alpha_i = 1$ ; under these conditions, the function $f_1^{\alpha_1} f_2^{\alpha_2} \cdots f_n^{\alpha_n}$ is integrable and

\int f_1^{\alpha_1} f_2^{\alpha_2} \cdots f_n^{\alpha_n} d|\mu| \leqslant \left( \int f_1 d|\mu| \right)^{\alpha_1} \left( \int f_2 d|\mu| \right)^{\alpha_2} \cdots \left( \int f_n d|\mu| \right)^{\alpha_n}.

For, the product $f_1^{\alpha_1} f_2^{\alpha_2} \cdots f_n^{\alpha_n}$ is measurable, being the product of measurable functions (§ 5, No. 3, Th. 1 and its Cor. 5); since the inequality (8) is true for upper integrals (Ch. I, No. 2, Cor. of Prop. 2), the function $f_1^{\alpha_1} f_2^{\alpha_2} \cdots f_n^{\alpha_n}$ is integrable (§ 5, No. 6, Th. 5), whence the corollary.

Cor. 2 of Th. 2 is sharpened by the following proposition:

Proposition 3

— Let $\mu$ be a positive measure on X, F a real or complex Banach space, F' its strong dual, and $(z, z') \mapsto \langle z, z' \rangle$ the canonical bilinear form on $F \times F'$ .

1° For every function $f \in \mathscr{L}_F^p$ ( $1 \leq p \leq +\infty$ ),

N_p(f) = \sup \left| \int \langle f, g \rangle \, d\mu \right|

as g runs over the set of functions in $\mathscr{L}_{F'}^q$ , such that $N_q(g) \leq 1$ .

2° For every function $g \in \mathscr{L}_{F'}^q$ ( $1 \leq q \leq +\infty$ ),

N_q(g) = \sup \left| \int \langle f, g \rangle \, d\mu \right|

as f runs over the set of functions in $\mathscr{L}_F^p$ such that $N_p(f) \leq 1$ .

Let us first prove the relation (9); we distinguish between two cases.

(i) $1 \leq p < +\infty$ . The relation (9) being trivial when $N_p(f) = 0$ (because f and $\langle f, g \rangle$ are then negligible), we can always suppose, on multiplying f by a scalar, that $N_p(f) = 1$ . Suppose first that f is an integrable step function, $f = \sum_{k=1}^n a_k \varphi_{A_k}$ , where the $A_k$ are pairwise disjoint (§ 4, No. 9, Lemma). Thus $\sum_{k=1}^n |a_k|^p \mu(A_k) = 1$ by hypothesis. For every $\varepsilon > 0$ , there exists (for every index k) a vector $a'_k \in F'$ such that $|a'_k|^q = |a_k|^p$ if $p > 1$ (resp. $|a'_k| = 1$ if $p = 1$ ) and $\langle a_k, a'_k \rangle \geq (1 - \varepsilon)|a_k| \cdot |a'_k|$ (TVS, IV,

§1, No. 3, Prop. 8). Setting $g = \sum_{k=1}^n a_k' \varphi_{A_k}$ , we have $\sum_{k=1}^n |a_k'|^q \mu(A_k) = 1$ if $p > 1$ (resp. $\sup_{1 \leq k \leq n} |a_k'| = 1$ if $p = 1$ ), thus $N_q(g) = 1$ ; on the other hand,

\int \langle f, g \rangle d\mu = \sum_{k=1}^n \langle a_k, a_k' \rangle \mu(A_k) \geq (1 - \varepsilon) \sum_{k=1}^n |a_k| \cdot |a_k'| \mu(A_k)

and, since $|a_k'| = |a_k|^{p/q} = |a_k|^{p-1}$ if $p > 1$ (resp. $|a_k'| = 1$ if $p = 1$ ), we have

\int \langle f, g \rangle d\mu \geq (1 - \varepsilon) \sum_{k=1}^n |a_k|^p \mu(A_k) = (1 - \varepsilon) N_p(f) = 1 - \varepsilon,

which proves the relation (9) in this case.

Let us pass to the case that $f$ is any element of $\mathscr{L}_F^p$ such that $N_p(f) = 1$ . For every $\varepsilon > 0$ , there exists a step function $f_1 \in \mathscr{L}_F^p$ such that $N_p(f - f_1) \leq \varepsilon$ (§ 4, No. 10, Cor. 1 of Prop. 19). By what we have just seen, there exists a function $g \in \mathscr{L}_F^q$ , such that $N_q(g) = 1$ and

\int \langle f_1, g \rangle d\mu \geq N_p(f_1) - \varepsilon \geq 1 - 2\varepsilon.

Now,

\int \langle f, g \rangle d\mu = \int \langle f_1, g \rangle d\mu + \int \langle f - f_1, g \rangle d\mu

and, by (6),

\left| \int \langle f - f_1, g \rangle d\mu \right| \leq N_p(f - f_1) N_q(g),

whence

\left| \int \langle f, g \rangle d\mu \right| \geq 1 - 3\varepsilon,

which proves (9).

(ii) $p = +\infty$ . We may again restrict ourselves to the case that $N_\infty(f) > 0$ . Let $\alpha$ be any number such that $0 < \alpha < N_\infty(f)$ ; by hypothesis, the set of $x \in X$ such that $|f(x)| > \alpha$ is measurable and is not locally negligible, therefore it contains a compact set $K$ of measure $> 0$ . Since $f$ is measurable, there exists a compact set $K_1 \subset K$ of measure $> 0$ , such that the restriction of $f$ to $K_1$ is continuous. It follows that, for every $\varepsilon > 0$ , there exists a partition of $K_1$ into a finite number of integrable sets, in each of which the oscillation of $f$ is $\leq \varepsilon$ ; at least one of these sets $A$ has measure > 0. Let $a$ be one of the values of $f$ in $A$ ; then $|a| > \alpha$ and $|f(x) - a| \leq \varepsilon$ for all $x \in A$ . There exists a vector $a' \in F'$ such that $|a'| = 1$ and $|\langle a, a' \rangle| \geq |a| - \varepsilon$ ; the function $g = \varphi_A \cdot a'/\mu(A)$ is integrable and $N_1(g) = 1$ ; on the other hand,

\int \langle f, g \rangle d\mu = \frac{1}{\mu(A)} \int \langle f, a' \rangle \varphi_A d\mu.

Now, one can write

\int \langle f, a' \rangle \varphi_A d\mu = \langle a, a' \rangle \mu(A) + \int \langle f - a, a' \rangle \varphi_A d\mu,

and since

|\langle f - a, a' \rangle \varphi_A| \leq \varepsilon \varphi_A,

we see that

\left| \int \langle f, g \rangle d\mu \right| \geq |\langle a, a' \rangle| - \varepsilon \geq |a| - 2\varepsilon > \alpha - 2\varepsilon;

since $\varepsilon$ is arbitrary and $\alpha$ is any number < $N_\infty(f)$ , the relation (9) is also verified in this case.

One argues in exactly the same manner to prove the relation (10), on considering separately the case $1 \leq q < +\infty$ and the case $q = +\infty$ , and using the fact that for every $z' \in F'$ , $|z'| = \sup_{|z| \leq 1} |\langle z, z' \rangle|$ by the definition of the norm in $F'$ .

Remark

— 1) Let $\mathscr{E}$ be a dense linear subspace of $\mathscr{L}_{F'}^q$ ; then the formula (9) holds when $g$ runs over the intersection of $\mathscr{E}$ with the set $B$ of functions in $\mathscr{L}_{F'}^q$ such that $N_q(g) \leq 1$ . For, it suffices to observe that the interior $\overset{\circ}{B}$ of $B$ is dense in $B$ and that $\overset{\circ}{B} \cap \mathscr{E}$ is dense in $\overset{\circ}{B}$ , since $\overset{\circ}{B}$ is open. This remark applies in particular to the set $\mathscr{E} = \mathscr{K}(X; F')$ of continuous functions with compact support (with values in $F'$ ) when $1 \leq q < +\infty$ , that is, $1 < p \leq +\infty$ . But in this case, the formula (9) is true as $g$ runs over $B \cap \mathscr{K}(X; F')$ , even for $p = 1$ . For, we may, as above, restrict ourselves to the case that $f$ is a step function. We saw then that if $N_1(f) = 1$ , then for every $\varepsilon > 0$ there exists a step function $g \in \mathscr{L}_{F'}^\infty$ such that $|g(x)| \leq 1$ for all $x \in X$ and $|\int \langle f, g \rangle d\mu| \geq 1 - \varepsilon$ . There exists a finite number of pairwise disjoint compact sets $K_i$ such that $g$ has a constant value $a'_i$ on each $K_i$ and such that, if $K$ is the union of the $K_i$ , then $\int |f| \varphi_{\mathscr{C}_K} d\mu \leq \varepsilon$ . Let $U_i$ be a neighborhood of $K_i$ such that the sets $U_i$ are pairwise disjoint, and let $h_i$ be a continuous mapping of $X$ into $[0, 1]$ with support contained in $U_i$ and equal to 1 on $K_i$ . Setting $h = \sum a'_i h_i$ , we have $h(x) = g(x)$ on $K$ and $|h(x)| \leq 1$ on $X$ , therefore

\int |\langle f, h \rangle| \varphi_{\mathscr{C}_K} d\mu \leq \varepsilon

and consequently $|\int \langle f, h \rangle d\mu| \geq 1 - 3\varepsilon$ , which proves our assertion. Analogous remarks can be made for the formula (10).

2) Let $\mu$ be a positive measure on $X$ , $f$ a measurable function $\geq 0$ (finite or not) whose support is contained in a countable union of compact sets $K_n$ . Then, for every $p$ such that $1 \leq p \leq +\infty$ ,

(11) \quad N_p(f) = \sup \int^* |fg| d\mu,

as $g$ runs over the set of functions in $\mathscr{H}(X; \mathbf{R})$ such that $N_q(g) \leq 1$ . For, the formula (11) is a special case of (9) when $N_p(f) < +\infty$ , since $f$ is then equivalent to a function in $\mathscr{L}^p$ (§ 5, No. 6, Th. 5). If $N_p(f) = +\infty$ , for every integer $n > 0$ set $f_n = \inf(n, f \varphi_{K_n})$ . Then

N_p(f_n) = \sup \int^* |f_n g| d\mu \leq \sup \int^* |f g| d\mu,

whence, on passing to the limit (assuming, as we may, that the sequence $(K_n)$ is increasing), we have $\sup \int^* |f g| d\mu = +\infty$ (§ 1, No. 3, Th. 3).

Corollary

— *Let $\mu$ be a positive measure on $X$ , $F$ a Banach space, $F'$ its strong dual, and $g$ any function in $\mathscr{L}^q_{F'}$ . Then, the linear form on $L^p_F$ , deduced from the linear form $f \mapsto \int \langle f, g \rangle d\mu$ on $\mathscr{L}^p_F$ by passage to the quotient, is continuous and has norm $N_q(g)$ .

5. Application: relations between the spaces $L^p_F$ ( $1 \leq p \leq +\infty$ )

Proposition 4

— *Let $f$ be a measurable function with values in a Banach space $F$ ; the set $I$ of numbers $p$ , such that $1 \leq p \leq +\infty$ and $N_p(f)$ is finite, is either empty or is an interval of $\overline{\mathbf{R}}$ . If $I$ is nonempty, the restriction to $I$ of the mapping $p \mapsto N_p(f)$ is continuous; moreover, if $f$ is not negligible, $\log N_p(f)$ is a convex function of $1/p$ on $\bar{I}$ .

We already know (Ch. I, No. 3, Prop. 5) that the set $J$ of finite numbers $p \geq 1$ such that $N_p(f) < +\infty$ is either empty or is an interval, and that $\log N_p(f)$ is a convex function of $1/p$ on $J$ (when $f$ is not negligible); this of course implies the continuity of $p \mapsto N_p(f)$ on $J$ .

If $J$ is empty then either $I = \varnothing$ or $I = \{+\infty\}$ , and the proposition is obvious in this case; assume henceforth that $J$ is nonempty. The proposition is also obvious if $f$ is negligible; assume henceforth that $f$ is not negligible. If $s \in J$ then, for every finite number $p > s$ , $|f|^p = |f|^s |f|^{p-s}$ , and the inequality of the mean shows that

(12) \quad N_p(f) \leq (N_s(f))^{s/p} (N_\infty(f))^{(p-s)/p}.

Letting $p$ tend to $+\infty$ , it follows that

\limsup_{p \to +\infty} N_p(f) \leq N_\infty(f).

This proves that if $+\infty \in I$ then $J$ contains arbitrarily large numbers; thus $I$ is indeed an interval of $\overline{\mathbf{R}}$ , and $\overline{I} = \overline{J}$ . The proposition will be proved if we show that $p \mapsto N_p(f)$ is continuous on $\overline{J}$ , and it suffices to establish continuity at the end-points of $J$ . We can suppose, moreover, that $J$ does not reduce to a point. Let $r$ and $s$ be the left and right end-points of $J$ ( $r < s \leq +\infty$ ). Let $A$ be the (measurable) set of $x \in X$ such that $|f(x)| \geq 1$ ; then

\int |f|^p d|\mu| = \int |f|^p \varphi_A d|\mu| + \int |f|^p \varphi_{C_A} d|\mu|.

As $p \in J$ tends to $r$ , $|f|^p \varphi_A$ tends to $|f|^r \varphi_A$ while decreasing, and $|f|^p \varphi_{C_A}$ tends to $|f|^r \varphi_{C_A}$ while increasing. Thus $\int |f|^p \varphi_{C_A} d|\mu|$ tends to $\int^* |f|^r \varphi_{C_A} d|\mu|$ (§ 1, No. 3, Th. 3). On the other hand, $|f|^p \varphi_A$ is integrable for $p \in J$ , and $\int |f|^p \varphi_A d|\mu|$ tends to $\int |f|^r \varphi_A d|\mu|$ (§ 4, No. 3, Prop. 4). Therefore $\int |f|^p d|\mu|$ tends to $\int^* |f|^r d|\mu|$ , which proves the continuity of $p \mapsto N_p(f)$ at $r$ .

The same reasoning may be applied at the point $s$ if $s < +\infty$ . Finally, suppose that $s = +\infty$ . In view of (13), it suffices to prove that

\liminf_{p \to +\infty} N_p(f) \geq N_\infty(f).

Now, let $a$ be a number such that $0 < a < N_\infty(f)$ . Since, by hypothesis, there exist finite values of $p$ such that $N_p(f) < +\infty$ , the set $A$ of $x \in X$ such that $|f(x)| \geq a$ , which is measurable and non-negligible, is integrable by virtue of the inequality $\varphi_A \leq (|f|/a)^p$ ; moreover, we infer from this inequality that $N_p(f) \geq a \cdot (|\mu|(A))^{1/p}$ ; letting $p$ tend to $+\infty$ , it follows that $\liminf_{p \to +\infty} N_p(f) \geq a$ , which completes the proof.

Corollary

— If $r, s, p$ *are three numbers such that

1 \leq r < p < s \leq +\infty,

then the intersection $\mathscr{L}_F^r \cap \mathscr{L}_F^s$ is contained in $\mathscr{L}_F^p$ .

Note that in general the topologies induced on the intersection $\mathscr{L}_F^r \cap \mathscr{L}_F^s$ by the topologies of the $\mathscr{L}_F^p$ ( $r < p < s$ ) are distinct. If no further hypothesis is made on $\mu$ , the topologies induced on $\mathscr{L}_F^r \cap \mathscr{L}_F^s$ by those of $\mathscr{L}_F^r$ and $\mathscr{L}_F^s$ are in general not comparable (in other words, the ratio $N_r(f)/N_s(f)$ can take arbitrarily large values and arbitrarily small values in $\mathscr{L}_F^r \cap \mathscr{L}_F^s$ ; cf. Exer. 8).

Prop. 4 may be sharpened when $\mu$ is a bounded measure:

Proposition 5

— Let $\mu$ be a bounded measure, and let $f$ be a $\mu$ -measurable function with values in a Banach space $F$ . The set I of numbers $p$ such that $1 \leq p \leq +\infty$ and $N_p(f)$ is finite, is either empty or is an interval with left end-point $p = 1$ and containing this point; moreover, $(|\mu|(X))^{-1/p} N_p(f)$ is an increasing function of $p$ on I.

This is an immediate consequence of Prop. 4 above and of the Cor. of Prop. 4 of Ch. I, No. 3.

Corollary

— If the measure $\mu$ is bounded, the relation $r < s$ implies $\mathscr{L}_F^s \subset \mathscr{L}_F^r$ ; moreover, the topology of convergence in mean of order $s$ is finer than the topology of convergence in mean of order $r$ (on $\mathscr{L}_F^s$ ).

One can show that in general the topology of convergence in mean of order $s$ is strictly finer than the topology of convergence in mean of order $r$ (Exer. 8).

Proposition 6

— Let $X$ be a discrete space, $\mu$ the measure on $X$ defined by placing a mass $+1$ at each point of $X$ . If $f$ is a mapping of $X$ into the Banach space $F$ , the set I of numbers $p$ such that $1 \leq p \leq +\infty$ and $N_p(f)$ is finite, is either empty or is an interval with right end-point $+\infty$ and containing this point; moreover, $N_p(f)$ is a decreasing function of $p$ on I.

For, $\mu^*(|f|) = \sum_{x \in X} |f(x)|$ for every function $f$ (§ 1, No. 1, Example), and $N_\infty(f) = \|f\| = \sup_{x \in X} |f(x)|$ ; if there exists a number $\alpha > 0$ such that $|f(x)| \geq \alpha$ for infinitely many values of $x \in X$ , then $N_p(f) = +\infty$ for every finite $p$ ; in the contrary case, there exists an $x_0 \in X$ such that $|f(x_0)| = \|f\|$ , whence

N_\infty(f) = |f(x_0)| \leq N_p(f)

for every finite $p$ . Since the function $\log N_p(f)$ is convex with respect to $1/p$ and takes its smallest value at the point $+\infty$ , it is necessarily a decreasing function of $p$ on I (FRV, I, § 4, No. 3, Prop. 5), which completes the proof.

Corollary

— If $X$ is discrete and the measure $\mu$ is defined by a mass $+1$ at each point of $X$ , then the relation $r < s$ implies $\mathscr{L}_F^r \subset \mathscr{L}_F^s$ ; moreover, the topology of convergence in mean of order $r$ is finer than the topology of convergence in mean of order $s$ (on $\mathscr{L}_F^r$ ).

Exercises

See the exercises for § 6.