(Minkowski’s inequality).

The inequality (1) is obvious when one of the terms of the second member is equal to $+\infty$. In the contrary case, $f$ and $g$ are finite almost everywhere ($§ 2$, No. 3, Prop. 7). If $f_1$ and $g_1$ are finite positive functions equivalent to $f$ and $g$, respectively, then $f_1^p, g_1^p$ and $(f_1 + g_1)^p$ are equivalent to $f^p, g^p$ and $(f + g)^p$, respectively, and since equivalent positive functions have the same upper integral ($§ 2$, No. 3, Prop. 6), we are reduced to proving the inequality (1) in the case that $f$ and $g$ are everywhere finite; but in this case the inequality is a special case of the general Minkowski inequality proved in Ch. I, No. 2, Prop. 3.

We shall also have occasion to make use of the following elementary inequality: if $p \geqslant 1$ then, for any numbers $a \geqslant 0,\ b \geqslant 0$,

The inequality is obvious if $a = b = 0$ or if one of the numbers $a, b$ is $+\infty$; if $a, b$ are finite and $a + b > 0$, it may be written

which follows from the fact that

(3)

(4)

For, the relation (3) follows at once from Def. 1 and the fact that $|\mu|^*$ is positively homogeneous; on the other hand, since $|f + g| \leq |f| + |g|$, the inequality (4) follows from Minkowski’s inequality (1) and the fact that $|\mu|^*$ is increasing.

We extend Def. 1 to the case of numerical functions, finite or not, defined on $X$, by again setting

for such a function $f$. One sees immediately that the relations (3) and (4) also hold for these functions when $f + g$ is defined on $X$ and $\alpha \neq 0$. Moreover:

(5)

Set $f = \sum_{n=1}^\infty f_n$; $f$ is the upper envelope of the increasing sequence of functions $g_n = \sum_{k=1}^n f_k$; the definition of $N_p(f)$, and Th. 3 of §1, No. 3, show that $N_p(f) = \sup_n N_p(g_n)$. But $N_p(g_n) \leq \sum_{k=1}^n N_p(f_k)$ by Prop. 1, whence the inequality (5).

The proposition follows at once from Th. 1 of §2, No. 3.

If f and g are two equivalent mappings of X into F, then $N_p(f) = N_p(g)$ for all $p \geq 1$ (§ 2, No. 3, Prop. 6); thus, $N_p(f)$ depends only on the class $\tilde{f}$ of f, and one sets, by definition, $N_p(\tilde{f}) = N_p(f)$. Since the classes of mappings of X into F form a vector space (§ 2, No. 4), the relations (3) and (4) may also be written

One defines similarly $N_p(\tilde{f})$ for every class of equivalent numerical functions (finite or not).

One can then define $N_p(f)$ for a function with values in F (resp. in $\overline{\mathbf{R}}$) defined almost everywhere in X, by setting $N_p(f) = N_p(\tilde{f})$; it is then clear that the relations (3) and (4) again hold (assuming $\alpha \neq 0$ and $f + g$ defined almost everywhere, in the case of numerical functions, finite or not).

If $0 < p < 1$, one again sets $N_p(f) = (\int^* |f|^p d|\mu|)^{1/p}$, but the inequalities (4) and (5) are no longer valid (cf. Ch. I, Exer. 6 and Ch. IV, §6, Exer. 13).

It comes to the same thing to say that, on the set of mappings $f \in \mathscr{F}_\mathbf{F}^p$ whose support is contained in a fixed compact set, the topology of uniform convergence is finer than the topology of convergence in mean of order $p$.

For, let $h$ be a continuous mapping of X into $[0,1]$, with compact support, equal to 1 on K (Ch. III, §1, No. 2, Lemma 1). For every $\varepsilon > 0$, there exists an $M \in \mathscr{B}$ such that, for every mapping $f \in M$, $|f(x) - f_0(x)| \leq \varepsilon h(x)$ for all $x \in X$. From this, it follows that $N_p(f - f_0) \leq \varepsilon N_p(h)$, whence the proposition.

Since the Hausdorff space associated with $\mathscr{F}_\mathbf{F}^p$ is a normed space, it suffices to prove that every Cauchy sequence $(f_n)$ in $\mathscr{F}_\mathbf{F}^p$ has a limit for the topology of convergence in mean of order $p$ (GT, IX, §2, No. 6, Prop. 9). By hypothesis, for every $\varepsilon > 0$ there exists an integer $m_0$ such that the relations $m \geq m_0,\ n \geq m_0$ imply $N_p(f_n - f_m) \leq \varepsilon$. One may therefore define, by induction on $k$, a strictly increasing sequence $(n_k)$ of integers $\geq 0$ such that $N_p(f_{n_{k+1}} - f_{n_k}) \leq 2^{-k}$. If we show that the series with general term $g_k = f_{n_{k+1}} - f_{n_k}$ ($k \geq 1$) is convergent in mean of order $p$, then it will have a sum $g \in \mathscr{F}_\mathbf{F}^p$, and $f = g + f_{n_1}$ will be the limit of the sequence $(f_{n_k})$ in $\mathscr{F}_\mathbf{F}^p$; $f$ will then be a cluster point of the sequence $(f_n)$; since this sequence is a Cauchy sequence, it will have $f$ as limit, and Prop. 5 will have been proved (GT, II, §3, No. 2, Cor. 2 of Prop. 5).

Prop. 5 is thus a consequence of the following proposition:

Consider the positive function (finite or not) $g(x) = \sum_{n=1}^\infty |f_n(x)|$. By the countable convexity theorem (No. 2, Th. 1),

thus $g$ is finite almost everywhere (§ 2, No. 3, Prop. 7), which means that the series with general term $f_n(x)$ is absolutely convergent almost everywhere. Since $\mathbf{F}$ is complete, this series is convergent almost everywhere and, for every $x \in X$, $|f(x)| \leqslant \sum_{n=1}^\infty |f_n(x)| = g(x)$, whence

which proves that $f$ belongs to $\mathscr{F}_\mathbf{F}^p$. On the other hand, for every integer $n$,

almost everywhere, whence $N_p \left( f - \sum_{k=1}^n f_k \right) \leqslant \sum_{k=n+1}^\infty N_p(f_k)$. By hypothesis, the series with general term $N_p(f_n)$ is convergent; therefore, for every $\varepsilon > 0$, there exists an integer $n$ such that $\sum_{k=n+1}^{\infty} N_p(f_k) \leq \varepsilon$, and the inequality (8) proves that $f$ is the sum of the series with general term $f_n$, for the topology of convergence in mean of order $p$.

Propositions 5 and 6 are thus completely proved.

In other words, the functions in $\mathscr{L}_\mathbf{F}^p$ are the limits of sequences of continuous functions with compact support, for the topology of convergence in mean of order $p$.

For, $f$ is finite almost everywhere and $N_p(f - g) \leq N_p(h - g) \leq \varepsilon$; Prop. 7 therefore shows that $f$ is $p$-th power integrable.

Since, by definition, $\mathscr{L}_\mathbf{F}^p$ is a closed subspace of $\mathscr{F}_\mathbf{F}^p$, and since the latter space is complete (No. 3, Prop. 5), we have the following result (GT, II, §3, No. 4, Prop. 8):

In the space $L_\mathbf{F}^p$, the norm $N_p(\tilde{\mathbf{f}})$ of a class is again denoted $\| \tilde{\mathbf{f}} \|_p$.

Th. 2 can be sharpened as follows:

As in the proof of Prop. 5 of No. 3, it suffices to define the sequence $(n_k)$ by induction, in such a way that $N_p(\mathbf{f}_{n_{k+1}} - \mathbf{f}_{n_k}) \leq 2^{-k}$; parts 2° and 3° then follow from Prop. 6 of No. 3 and the fact that $\mathscr{L}_\mathbf{F}^p$ is closed in $\mathscr{F}_\mathbf{F}^p$. On the other hand, if $h(x)$ is the sum of the series with general term $|\mathbf{f}_{n_{k+1}}(x) - \mathbf{f}_{n_k}(x)|$, Th. 1 of No. 2 shows that $N_p(h) < +\infty$; therefore, by the definition of $|\mu|^*$, there exists a lower semi-continuous function $g \geq h + |\mathbf{f}_{n_1}|$ such that $N_p(g) < +\infty$, which completes the proof.

For, there exists a subsequence $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k}(x))$ converges almost everywhere to $g(x)$, where $g$ is a function in $\mathscr{L}_F^p$ such that $(f_n)$ converges in mean of order $p$ to $g$. The hypotheses therefore imply that $f(x) = g(x)$ almost everywhere, whence the corollary.

1° the sequence $(g_n)$ converges in mean of order $p$ to $f$; 2° for almost every $x \in X$, the sequence $(g_n(x))$ converges to $f(x)$.

For, since the space $\mathscr{L}_F^p$ is metrizable, there exists a Cauchy sequence $(f_n)$ in $\mathscr{L}_F^p$ consisting of functions in $\mathscr{E}$ and convergent in mean of order $p$ to $f$ (GT, IX, §2, No. 6, Prop. 8); it suffices to apply Th. 3 to this sequence.

Cor. 2 is applicable in particular to the case where $\mathscr{E}$ is taken to be the space $\mathscr{K}_F$ of continuous functions with compact support.

Let $f \in \mathscr{L}_F^p$; for every $\varepsilon > 0$, there exists a function $g \in \mathscr{K}_F$ such that $N_p(f - g) \leq \varepsilon$; since $|u \circ f - u \circ g| \leq \|u\| \cdot |f - g|$, we have

and since $u \circ g$ is continuous with compact support, the theorem is proved.

This follows from the fact that the mapping $t \mapsto a t$ of $\mathbf{R}$ into F is continuous.

This follows at once from Cors. 1 and 2 of Th. 4.

The set $\mathscr{K}_F$ of continuous mappings of X into F with compact support is by definition dense in $\mathscr{L}_F^p$. On the other hand, every function $g \in \mathscr{K}_F$ may be uniformly approximated by functions of the form $\sum_k a_k f_k$, where the $f_k$ are continuous functions with support contained in a fixed compact neighborhood of the support of $g$ (Ch. III, §1, No. 2, Lemma 2); it follows (No. 3, Prop. 4) that $g$ is in the closure in $\mathscr{L}_F^p$ of the set of $\sum_k a_k f_k$, whence the proposition.

For every $\varepsilon > 0$ there exists a continuous function $g$ with compact support, such that $N_p(f - g) \leq \varepsilon$; since $||f| - |g|| \leq |f - g|$, we have $N_p(|f| - |g|) \leq \varepsilon$, which proves that $|f| \in \mathscr{L}^p$. On the other hand, if $f_1, f_2$ are two functions in $\mathscr{L}_F^p$ then $N_p(|f_1| - |f_2|) \leq N_p(f_1 - f_2)$, which shows that $f \mapsto |f|$ is a uniformly continuous mapping.

The condition is sufficient since $f = f^+ - f^-$; it is necessary, because if $f \in \mathscr{L}^p$ then $|f| \in \mathscr{L}^p$ (Prop. 11).

The first part of the proposition follows at once from Prop. 11 of No. 5; since the set of $\tilde{f} \geq 0$ is also the set of $\tilde{f}$ such that $| \tilde{f} | = f$, it is closed, because $\tilde{f} \mapsto | \tilde{f} |$ is a continuous mapping and $L^p$ is Hausdorff.

We thus see that the topology on $L^p$ defined by the norm $\| \tilde{f} \|_p$ is compatible with the ordered vector space structure of $L^p$ (TVS, II, § 2, No. 7).

The supremum of $H$ in $L^p$ is then the limit (in the Banach space $L^p$) of the section filter of $H$.

The condition is clearly necessary, since $\tilde{f} \mapsto \| \tilde{f} \|_p$ is an increasing function on the set of elements $\geq 0$ of $L^p$. To see that it is sufficient, we first observe that it implies that the image of $H$ under the mapping $\tilde{f} \mapsto \| \tilde{f} \|_p$ has a limit in $\mathbf{R}$, by the monotone limit theorem; the image of the section filter $\mathfrak{F}$ of $H$ under this mapping is therefore a base of a Cauchy filter on $\mathbf{R}$. The proof will be complete if we show that $\mathfrak{F}$ itself is a base of a Cauchy filter on $L^p$; for, $\mathfrak{F}$ will then converge in $L^p$, since $L^p$ is complete (No. 4, Th. 2), and the proposition will follow from TVS, II, § 2, No. 7, Prop. 18.

To see that $\mathfrak{F}$ is a base of a Cauchy filter, we shall make use of the following lemma:

When $f$ and $g$ are continuous with compact support, the relation (9) may be written

and is then a consequence of the elementary inequality $(g - f)^p \leq g^p - f^p$ (No. 1, formula (2)). To pass from this to the general case, it suffices to observe that the two members of (9) are continuous functions on $\mathscr{L}^p \times \mathscr{L}^p$, and that every function $f \geq 0$ in $\mathscr{L}^p$ is the limit (for convergence in mean of order $p$) of a sequence of continuous functions $\geq 0$ with compact support, by the continuity of the mapping $g \mapsto |g|$ on $\mathscr{L}^p$ (Prop. 11).

The lemma having been established, for every $\varepsilon > 0$ there exists by hypothesis an $\tilde{f} \in \mathrm{H}$ such that, for every $\tilde{g} \geq \tilde{f}$ belonging to $\mathrm{H}$, we have $(\|\tilde{g}\|_p)^p - (\|\tilde{f}\|_p)^p \leq \varepsilon$; from this it follows that $(\|\tilde{g} - \tilde{f}\|_p)^p \leq \varepsilon$; thus, if $\tilde{g}_1$ and $\tilde{g}_2$ are two elements in $\mathrm{H}$ that are $\geq \tilde{f}$, then $\|\tilde{g}_1 - \tilde{g}_2\|_p \leq 2\varepsilon^{1/p}$, which proves that $\mathfrak{F}$ is a Cauchy filter base on $L^p$ and completes the proof of Prop. 14.

This follows from the continuity of the norm $\|\tilde{f}\|_p$ in $L^p$, and the monotone limit theorem.

Every directed set $\mathrm{H}$ in $L^p$ (for the relation $\leq$), consisting of classes $\geq 0$ and bounded above in $L^p$, has a supremum: for, if $\tilde{h}$ is an upper bound for $\mathrm{H}$ in $L^p$, then $\|\tilde{f}\|_p \leq \|\tilde{h}\|_p$ for all $\tilde{f} \in \mathrm{H}$, and Prop. 14 is applicable. This proves the corollary (Ch. II, §1, No. 3, Prop. 1).

The conclusions of Prop. 14 no longer hold when they are formulated for the functions in $\mathscr{L}^p$ instead of their classes. To be precise, if $M$ is a subset of $\mathscr{L}^p$, consisting of functions $\geq 0$, directed for the relation $\leq$, and such that $\sup_{f \in M} N_p(f) < +\infty$, the class of the upper envelope $g$ of $M$ is not necessarily identical to the supremum in $L^p$ of the classes of the functions f \in M ; in particular, g is not necessarily p-th power integrable, and even if $g \in \mathscr{L}^p$, $N_p(g)$ can be distinct from $\sup_{f \in M} N_p(f)$ (cf. §1, No. 3, Remark 1 following Th. 3).

Nevertheless, we have the following theorem:

The condition being obviously necessary, we need only prove that it is sufficient. Now, if the condition is satisfied then Prop. 14 shows that the sequence $(\tilde{f}_n)$ is a Cauchy sequence in $L^p$, therefore the sequence $(f_n)$ is a Cauchy sequence in $\mathscr{L}^p$; since $f_n(x)$ tends to $f(x)$ for all $x \in X$, $f$ is p-th power integrable and is the limit of the sequence $(f_n)$ for the topology of convergence in mean of order $p$ (No. 4, Cor. 1 of Th. 3). Therefore $N_p(f_n)$ tends to $N_p(f)$ since $N_p$ is a continuous function on $\mathscr{L}^p$.