Let $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be a measurable function. I would like to prove the following inequality:
$$\left(\int_{\mathbb{R}}\left\lvert \int_0^t f(s, x)\,ds\right\rvert^q\, dx \right)^{\frac{1}{q}} \le \int_0^t \left(\int_{\mathbb{R}}\lvert f(s, x)\rvert^q\,dx\right)^{\frac{1}{q}}\, ds, $$
under the minimal assumption that all integrals make sense and the rightmost term above is finite. The idea was to rewrite the inequality this way
$$\left\lVert \int_0^t f(s,\cdot)\, ds\right\rVert_q \le \int_0^t \lVert f(s, \cdot) \rVert_q\, ds,$$
which looks so obviously true... But I'm afraid of some pitfall here. In fact, we cannot guarantee continuous dependence of $f$ on the first variable, so $\int_0^t f(s, \cdot)\, ds$ is not the usual Riemann integral in a Banach space.
What do you think: is this approach leading somewhere or I'd better try another one? (Which one, just in case? :-) )
EDIT: Answer I've found a very satisfactory answer in Hardy-Littlewood-Polya's Inequalities (@Willie Wong: thank you!). I'm glad to expose it here (with slightly different language, in case you ask).
Theorem Let $\Omega_t, \Omega_x$ be $\sigma$-finite measure spaces and $f \colon \Omega_t \times \Omega_x \to [0, \infty]$ be a measurable function. If $1 < p < \infty$ then $$\left\lVert \int_{\Omega_t} f(s, \cdot)\, ds\right\rVert_p \le \int_{\Omega_t}\lVert f(s, \cdot) \rVert_p \,ds,$$ where $\lVert \cdot \rVert_p$ refers to $L^p(\Omega_x)$.
Lemma Let $\Omega$ be a $\sigma$-finite measure space and $J \colon \Omega \to [0, \infty]$ a measurable function. If $1 < p < \infty$ and $F \ge 0$ then the following statements are equivalent:
- $\lVert J \rVert_p \le F$;
- $\forall g \in L^{p'}(\Omega), g \ge 0, \int_{\Omega} g^{p'}dx \le 1$ we have $\int_{\Omega}Jg\, dx \le F$.
Proof of Theorem Let $J(y)=\int_{\Omega_t}f(s, y)\, ds$. $J$ is a measurable positive function on $\Omega_x$. Take $g \in L^{p'}(\Omega_x), g \ge 0, \int_{\Omega_x}g(y)dy \le 1$. Then by Fubini's theorem and Hölder's inequality we have
$$\int_{\Omega_x}J(y)g(y)\, dy = \int_{\Omega_t}ds \int_{\Omega_x}f(s, y)g(y)dy\le \int_{\Omega_t}\left(\int_{\Omega_x}f(s, y)^p dy\right)^{\frac{1}{p}}\, ds, $$
that is, $\int_{\Omega_x}J(y)g(y)dy\le \int_{\Omega_t} \lVert f(s, \cdot) \rVert_p\, ds$ and so $\lVert J \rVert_p \le \int_{\Omega_t} \lVert f(s, \cdot)\rVert_p\, ds$ by the lemma. ////
The general principle here is very interesting: if you want to prove an inequality like $\int J^p\, dx \le \text{something}$, you can get past that annoying exponent $p$ by proving $\int Jg\, dx \le \text{something}$ for all suitable $g$.
References Hardy-Littlewood-Polya, Inequalities: my Theorem is their Theorem 202, my Lemma is their Theorem 191.
EDIT 2020
Let us see how to prove the Lemma. It looks MUCH tougher than it actually is. We don't actually need any functional analysis to prove it, no dual spaces or anything like that. The proof that 1. $\Rightarrow$ 2. is literally just an immediate application of the inequality of Hölder. The proof that 2. $\Rightarrow$ 1. is based on the obvious computation
$$
\lVert J^{p-1}\rVert_{p'}=\lVert J\rVert_p^{p-1}, $$
which motivates us to write
$$
\lVert J\rVert_p^p=\lVert J\rVert_p^{p-1}\int J(x)\frac{J^{p-1}(x)}{\lVert J^{p-1}\rVert_{p'}}\, dx, $$
and now we can apply the assumption 2 to bound the integral, because $\frac{J^{p-1}(x)}{\lVert J^{p-1}\rVert_{p'}}$ has $p'$ norm equal to 1. We conclude
$$
\lVert J\rVert_p^p\le \lVert J\rVert_p^{p-1} F, $$
from which 1. immediately follows. $\Box$
Remark. As stressed by the book Analysis of Lieb and Loss, in the case of $L^p$ spaces the abstract functional analysis is not necessary, and it actually sometimes obscures what is going on. This is a good example of that phenomenon.
This edit comes nine years after the original question, one of my first ones. I am still here. Looks like I got hooked for real. :-)
