# Infinite dimensional integral inequality

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:

1. $\lVert J \rVert_p \le F$;
2. $\forall g \in L^{p'}(\Omega), g \ge 0, \int_{\Omega} g^{p'}dx \le 1$ 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.

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Looks like the perfect occasion to learn about the Bochner integral. – t.b. Apr 18 '11 at 16:10
For $t < \infty$, if the RHS is well defined, using Holder's inequality you have that $\int_0^t\int |f(s,x)| dx ds \leq$ C_t RHS. So your function is integrable in the product measure. By Fubini then $\int_0^tf(s,x)ds$ is (Lebesgue) measurable in $x$, and so the $L^q$ norm makes sense. BTW, the integral Minkowski's inequalities are theorems 200 - 202 in Hardy-Littlewood-Polya. – Willie Wong Apr 18 '11 at 16:40
@Theo: Thank you! This makes for a quick and clean avenue of attack. @Willie Wong: I'm reading the direct proof of Hardy-Littlewood-Polya's Inequalities. I'm liking very much this idea of a book focused solely on inequalities. However, I find it a bit difficult to read because of unusual notation and typography. Can you give me some alternative reference a bit less...err... dated? I hope this does not sound too much of a blasphemy. – Giuseppe Negro Apr 18 '11 at 17:58
@dissonance Do yourself a favor and get Michael Steele's The Cauchy-Schwarz Master Class. You won't regret it! – Byron Schmuland Apr 18 '11 at 18:22
There's also a proof in F. Jones' Lebesgue Integration on Euclidean Space, Section 11.E. – lvb Apr 18 '11 at 18:37

So, first thing to try is to set

$$f(x,t) = \sum_{j = 1}^N g_j(x) 1_{F_j}(y)$$

for pairwise disjoint $F_j$. For this the inequality is easy to verify.

Now we would like to take limits, but the question is: Are the measurable functions pointwise limits of function of the form of $f$? It turns out this is the case as has been shown by Nate Eldredge on a question of mine (I tried to prove the same). Sequence of measurable functions.

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