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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, $$|\int f d\mu|\leq \int|f| d\mu$$

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  • $\begingroup$ Jensen's inequality would apply en.wikipedia.org/wiki/Jensen's_inequality $\endgroup$ – orangeskid May 29 at 6:46
  • $\begingroup$ @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. $\endgroup$ – Oliver Diaz May 29 at 6:58
  • $\begingroup$ @nessy: I have worked out a much simpler solution with uses a few facts from linear algebra, specifically p norms in $\mathbb{R}^n$. $\endgroup$ – Oliver Diaz May 29 at 10:39
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    $\begingroup$ @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). $\endgroup$ – orangeskid May 29 at 20:08
  • $\begingroup$ @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 $\endgroup$ – Oliver Diaz May 29 at 20:27
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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.


Comments:

  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

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


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  • $\begingroup$ 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. $\endgroup$ – orangeskid May 30 at 2:12
  • $\begingroup$ 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. $\endgroup$ – Oliver Diaz May 30 at 3:39
  • $\begingroup$ I took a look at the definition of Pettis integral. The second solution should also work for this integral. $\endgroup$ – orangeskid May 30 at 5:06
  • $\begingroup$ 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). $\endgroup$ – Oliver Diaz May 30 at 14:48
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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 $$\|\int_X f d\mu \| = L(\int_X f d\mu)=\int_X L(f) d\mu \le \int_X \|f\| d\mu$$

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  • $\begingroup$ 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. $\endgroup$ – Oliver Diaz May 30 at 14:40

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