# $(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$

Let $$(X, \mathfrak{B}, \mu)$$ be a measurable space, possibly not $$\sigma$$-finite, and $$f_1, \cdots, f_n \colon X\to (-\infty, +\infty)$$ be integrable functions on $$X$$. Does $$(\int f_1d\mu)^2+\cdots+(\int f_nd\mu)^2\leq(\int \sqrt{f_1^2+\cdots+f_n^2}d\mu)^2$$ holds? (Since $$\sqrt{f_1^2+\cdots+f_n^2}\leq |f_1|+\cdots+|f_n|$$, note that integrand in RHS is integrable.)

My first attempt was to apply the Fubini's theorem and Cauchy-Schwarz to the LHS: \begin{align}(LHS)&=(\int f_1(x)d\mu(x))(\int f_1(y)d\mu(y))+\cdots+(\int f_n(x)d\mu(x))(\int f_n(y)d\mu(y))\\&=\int f_1(x)f_1(y)+\cdots+f_n(x)f_n(y) d(\mu\otimes\mu)(x,y)\\ &\leq\int \sqrt{f_1^2(x)+\cdots+f_n^2(x)}\sqrt{f_1^2(y)+\cdots+f_n^2(y)}d(\mu\otimes\mu)(x,y)\\&=(RHS)\end{align}

However this approach is valid only if $$X$$ is $$\sigma$$-finite.

Note that the inequation is equivalent to the following: If $$f\colon X\to \mathbb{R}^n$$ is integrable, $$|\int f d\mu|\leq \int|f| d\mu$$

• Jensen's inequality would apply en.wikipedia.org/wiki/Jensen's_inequality – orangeskid May 29 at 6:46
• @orangeskid: The issue nessy brought up is what happens with the underlying measure $\mu$ is not $\sigma$--finte. In such case, Jensen's not applicable. – Oliver Diaz May 29 at 6:58
• @nessy: I have worked out a much simpler solution with uses a few facts from linear algebra, specifically p norms in $\mathbb{R}^n$. – Oliver Diaz May 29 at 10:39
• @Oliver Diaz: I finally understood why we need $\sigma$ finite. It is because we replace the measure with an equivalent probability measure. However, we can restrict ourselves with $\sigma$-finite measures when we deal with the integral of a single $L^1$ function, since we can ignore the $0$ set ( and the rest must be $\sigma$-finite). – orangeskid May 29 at 20:08
• @orangeskid: The $\sigma$-finite case is solved already by nessy. Her concern was the non-$\sigma$-finte case, where her using of Fubini's theorem does not apply and certainly Jensen's either. Her insight however gives more ore less the path to follow, i.e. $\|\int f\|\leq\int\|f\|$ where $\|\;\|$ is Eucliean norm in $\mathbb{R}^n$. It turns out that the aforemetioned inequality holds in very general settings. See solution below – Oliver Diaz May 29 at 20:27

Here are a couple of strategies that work in general and make no use of any type of local integrability properties of the underlying measure ($$\sigma$$-finiteness or not).

Consider the space $$L$$ of functions $$f:X\rightarrow\mathbb{R}^n$$ which are integrable in each component and define $$\|f\|^*=\int\|f\|_2\,d\mu$$, where $$\|\;\|_2$$ is the Euclidean norm on $$\mathbb{R}^n$$. This is defines a norm on $$L$$ since $$\|f\|^*\leq\sum^n_{k=1}\int|f|_j\,d\mu<\infty$$. Also, $$\int|\|f\|_2-\|g\|_2|\,d\mu\leq\int\|f-g\|_2\,d\mu=\|f-g\|^*$$

Consider $$\mathcal{E}$$ the collection of (integrable) simple functions on $$(X,\mathscr{B},\mu)$$ and define $$\mathbb{R}^n\otimes\mathcal{E}=\{\sum^m_{k=1}u_k\phi_k: u_k\in\mathbb{R}^n, \phi_k\in\mathcal{E}, m\in\mathbb{N}\}$$

This space will play the role of elementary functions in the construction of the real valued integral. It is easy to check that $$\mathbb{R}^n\otimes\mathcal{E}$$ is dense in $$(L,\|\;\|^*)$$; furthermore, any function in $$\mathbb{R}^n\otimes\mathcal{E}$$ can be expressed as $$\Phi=\sum^{M}_{j=1}v_j\mathbb{1}_{A_j}$$ where $$v_j\in \mathbb{R}^n$$, $$A_j\in\mathscr{B}$$, $$\mu(A_j)<\infty$$, and $$M\in\mathbb{N}$$. Consider now the elementary integral $$\int\Big(\sum^m_{k=1}u_k\phi_k\Big):=\sum^m_{j=1}u_k\int\phi_k\,d\mu$$

Since $$\Phi=\sum_{u\in\mathbb{R}^n}u\mathbb{1}_{\{\Phi=u\}}$$ (notice that the sum over $$\mathbb{R}^n$$ is actually finite), $$\int\Phi =\sum^m_{j=1}u_j\mu(A_j)=\sum_{u\in\mathbb{R}^n}u\int\mathbb{1}_{\{\Phi=u\}}\,d\mu\tag{1}\label{one}$$

which means that the elementary integral extended to $$\mathbb{R}^n\otimes\mathcal{E}$$ does not depend on any particular representation of $$\Phi$$. Now $$\Big\|\int\Phi\Big\|_2\leq\sum_{u\in\mathbb{R}^n}\|u\|_2\int\mathbb{1}_{\{\Phi=u\}}\,d\mu=\int\Big(\sum_{u\in\mathbb{R}^n}\|u\|_2\mathbb{1}_{\{\Phi=u\}}\Big)\,d\mu=\int\|\Phi\|_2\,d\mu=\|\Phi\|^*\tag{2}\label{two}$$ $$\eqref{two}$$ is the inequality you are looking for but only for functions in $$\mathbb{R}^n\otimes\mathcal{E}$$. For all functions in $$L$$ one can use some density arguments.

1. Notice that $$\|\;\|_2$$ can be replaced by $$\|\;\|_p$$ ($$p\geq1$$).

2. Your problem is an example of an integral defined on vector--valued functions.

3. The arguments used, with some technical additions (Daniell integration, and measurability issues) can be used to construct Bochner's integral where $$\mathbb{R}^n$$ is replaced by a Banach space.

Another, much simpler solution may be obtained by applying linear functionals to the vector $$\int f=\sum^n_{j=1}e_j\int f_j\,d\mu$$ where $$e_1,\ldots,e_n$$ is the standard basis of $$\mathbb{R}^n$$. As above, w $$\|\,\|_p$$ is $$p$$-norm in $$\mathbb{R}^n$$. We use the fact that $$(\mathbb{R}^n,\|;\|_p)$$ and $$(\mathbb{R}^n,\|\,\|_q)$$ are dual to each other when $$\tfrac1p+\tfrac1q=1$$.

If $$\Lambda:\mathbb{R}^n\rightarrow\mathbb{}$$ is linear, then $$\Lambda x =x\cdot u$$ for some unique $$u\in\mathbb{R}$$. Thus

\begin{aligned} \Lambda \Big(\int f\Big) &= u\cdot\Big(\int f\Big)=\sum^n_{j=1}u_j\int f_j\,d\mu =\int u\cdot f\,d\mu \end{aligned} and so, by Hölder's inequality (in $$\mathbb{R}^n$$) \begin{aligned} \left|\Lambda \Big(\int f\Big)\right|&\leq\int|u\cdot f|\,d\mu\\ &\leq\int\|u\|_q\|f\|_p\,d\mu=\|u\|_q\int\|f\|_p\,d\mu \end{aligned} The result than follows by taking $$\sup$$ over all linear functionals $$\Lambda$$ with functional norm $$\|\Lambda\|:=\sup_{\|x\|_p=1}|\Lambda x|\leq1$$, or equivalently, by taking $$\sup$$ over all vectors $$u\in\mathbb{R}^n$$ with $$\|u\|_q=1$$. Thus

$$\left\|\int f\right\|_p \leq \int\|f\|_p\,d\mu$$

• I like the last proof; it seems to work for a general (semi)norm $\|\cdot \|$. By Hahn-Banach, given any $v (= \int_X f)$ there exists a functional of norm $1$, call it $L$, so that $\|v\| = L(v)$. Then apply your method and get $\|v\| \le \|L \| \cdot \int_X \|f\|$. This should be the "canonical" proof. Nice, thank you. – orangeskid May 30 at 2:12
• The first proof is more or less how one constructs Bochner’s integral through the Daniell procedure. the later is close to how one construct Bochner integral through the Dunford integral. – Oliver Diaz May 30 at 3:39
• I took a look at the definition of Pettis integral. The second solution should also work for this integral. – orangeskid May 30 at 5:06
• Yes, the Pettis integral. I confuse him with Dunford, as in the Dunford-Pettis theorem. Any way, that type of integral is useful even when functions take values on more abstract linear spaces (at least locally convex). The advantage of Bochner+Daniell is that one can derive a natural notion of measurability (Caratheodory's cut idea does not work here). – Oliver Diaz May 30 at 14:48

First, assume that $$(X,\mu)$$ is a $$\sigma$$ finite space. Then there exists a probability measure $$\nu$$ on $$X$$ that is equivalent to $$\mu$$, that is $$\mu = \rho \cdot \nu$$ where $$\rho>0$$ is a measurable function, $$\rho>0$$. We have for every $$f\in L^1(X, \mu)$$ $$\int_X f d\mu = \int_X f \, d\, \rho \nu = \int_X \rho f\, d \nu$$

Now, let $$\phi$$ be a convex function on $$\mathbb{R}^n$$ that is also positively homogeneous ( a sublinear function). Then we have $$\int_X \phi( f) d \mu= \int_X \rho \phi(f) d\nu = \int_X \phi(\rho f) d\nu \ge \phi(\int_X \rho f d\nu ) = \phi( \int_X f d\mu)$$

The inequality above is Jensen's inequality, for the convex functions $$\phi$$ and the function $$L^1$$ $$\rho f$$ on the probability space $$(X,\nu)$$.

We can reduce to the case $$X$$ $$\sigma$$-finite as follows: Consider $$X' = \{x\in X | f(x) \ne 0\}$$. Since $$f$$ is $$L^1$$, all the subsets $$\{x |\ |f(x)|\ge 1/n\}$$ are have finite measure. Hence $$X'$$ is $$\sigma$$-finite. We can reduce all our integrals to integrals over $$X'$$.

Now, how to find the probability measure $$\nu$$ equivalent to $$\mu$$. Let $$X= \sqcup_n X_n$$ where $$\mu(X_n) <\infty$$. Now, find $$\eta>0$$ such that $$\int_X \eta\, d\mu = 1$$, for instance $$\eta=\sum_{n\ge 1}\frac{1}{2^n} \cdot \frac{\chi(X_n)}{\mu(X_n)}$$ Put $$\nu = \eta \cdot \mu$$.

$$\bf{Added:}$$ I think the natural solution is the second one of @Oliver Diaz, let's restate it in general terms.

Consider $$\|\cdot \|$$ a seminorm on $$\mathbb{R}^n$$ (or, more generaly, a sublinear function). We want to show the inequality $$\| \int_X f d\mu \| \le \int_X \|f\| d \mu$$

Denote by $$v \colon = \int_X f d\mu$$. By Hahn-Banach theorem, there exists a linear functional $$L\colon \mathbb{R}^n \to \mathbb{R}$$ such that $$L(v) = \|v\|$$, and $$L(w)\le \|w\|$$ for all $$\|w\|\in \mathbb{R}^n$$. We get $$\|\int_X f d\mu \| = L(\int_X f d\mu)=\int_X L(f) d\mu \le \int_X \|f\| d\mu$$

• The problem requires a "vector" version of Jensen's inequality (for the $\sigma$--finite case). The function here is an $\mathbb{R}^n$--valued integrable function. A proof of a version for Jensen's on this setting though not difficult, it does require more than the standard real line version. – Oliver Diaz May 30 at 14:40