# 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.

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. :-)

• Looks like the perfect occasion to learn about the Bochner integral.
– t.b.
Commented Apr 18, 2011 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. Commented Apr 18, 2011 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. Commented Apr 18, 2011 at 17:58
• @dissonance Do yourself a favor and get Michael Steele's The Cauchy-Schwarz Master Class. You won't regret it!
– user940
Commented Apr 18, 2011 at 18:22
• There's also a proof in F. Jones' Lebesgue Integration on Euclidean Space, Section 11.E.
– lvb
Commented Apr 18, 2011 at 18:37

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