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

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

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

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

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.