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A student raised the question today: can one interchange the integrals in

$$ \int_0^1 \left(\int_0^x f(y) \, dy \right)^2 \, dx $$

for sufficiently nice functions $f$. I answered no, with the reasoning that integrals are fundamentally linear operations.

However now that I think about it, it is certainly possible that some kind of non-linear analogy to Fubini-Tonelli exists and I'm simply not aware of it. Indeed, it seems quite plausible that some interchanged expression involving square roots is equal to the expression above, and perhaps in general something involving the inverse of the non-linear function of the inner integral.

Is anyone aware of a non-linear Fubini-Tonelli theorem?

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In the case of integral powers it is easiest if you write $$ \int_0^1\left(\int_0^x f(y)dy\right)^2 dx = \int_0^1\int_0^x\int_0^x f(y)f(z)dz~dy~dx$$ for which you can apply the usual Fubini-Tonelli theorem as long as you keep track of which wedge you are integrating over. –  Willie Wong Jul 20 '11 at 11:35
    
BTW, this type of "tensor power expansion" is used in, for example, some elementary proofs of Sobolev embedding theorems. –  Willie Wong Jul 20 '11 at 11:37
    
@Glen: $f(y)$ or $f(x,y)$? If $f(y)$, can one interchange the integrals in the linear version? –  Shai Covo Jul 20 '11 at 13:12
    
@Shai In this question let's leave it at $f(y)$, since that's where all the discussion has headed. I had $f(x,y)$ in mind, which one could probably guess from my comments, but my mistake for not writing it correctly in the first place :). –  Glen Wheeler Jul 20 '11 at 13:35
    
Thanks for this response. However, with $f(y)$ it does not hold in general $\int_0^1 {(\int_0^x {f(y)\,dy} )\,dx} = \int_0^1 {(\int_y^1 {f(x)\,dx} )\,dy}$ (take for example $f(y)=y$). That is, one cannot change the order of integration already in the linear version. –  Shai Covo Jul 20 '11 at 13:44
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2 Answers

up vote 5 down vote accepted

Following up on Willie Wong's comment: $$ \begin{align} & \phantom{= {}} \int_0^1 \left( \int_0^x f(y)\,dy \right)^2\,dx = \int_0^1 \int_0^x \int_0^x f(y)f(z)\,dy\,dz \, dx = \int_0^1\int_0^1 \int_{\max\{y,z\}}^1 f(y)f(z)\, dx \, dy\, dz \\ & = \int_0^1\int_0^1 \left( f(y)f(z) \int_{\max\{y,z\}}^1 1\, dx \right) dy \, dz = \int_0^1\int_0^1 f(y)f(z) \left(1-{\max\{\,y,z\,\}}\right) \, dy \, dz. \end{align} $$

The integration with respect to $x$ now appears as $1-\max\{\,y,z\,\}$ on the inside; the square of the integral with respect to $y$ now appears as $\int_0^1\int_0^1 \cdots\,dy\,dz$ on the outside.

As for "dimensional" correctness, $dy$ and $dz$ are both in the same units as $dy$ in the original integral; $f(y)$ and $f(z)$ are both in the same units as $f(y)$ in the original integral; and $x$ has to be in the same units as $y$ in the original integral; so $x$ is in the same units as $1-\max\{\,y,z\,\}$. So the units are of the form $a^2 b^3$ on both sides of the "$=$".

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Let me state Minkowski's integral inequality:

Let $1\leq p <\infty$. Let $F$ be a measurable function on the product space $(X,\mu)\times (T,\nu)$, where $\mu,\nu$ are $\sigma$-finite. We have

$\left[\int_{T}\left(\int_{X}\left|F(x,t)\right|d\mu(x)\right)^pd\nu(t)\right]^{\frac{1}{p}}\leq\int_{X}\left[\int_{T}\left|F(x,t)\right|^pd\nu(t)\right]^{\frac{1}{p}}d\mu(x)$.

If I may say so, the following proof is quite elegant (in fact, the above result is Exercise 1.1.6., page 12 of Classical Fourier Analysis by Loukas Grafakos, and the proof below is my solution to this exercise):

Proof. Let $G:T\to [0,\infty]$ be defined by $G\left(t\right)=\int_{X} \left|F\left(x,t\right)\right|d\mu\left(x\right)$. Then Minkowski's integral inequality is nothing but a reformulation of the assertion that $\left\|G\right\|_{L^p\left(T\right)}\leq \int_{X} [\int_{T} \left|F\left(x,t\right)\right|^p d\nu\left(t\right)]^{\frac{1}{p}} d\mu\left(x\right)$. To prove this assertion, we first note the well-known fact that $\left\|G\right\|_{L^p\left(T\right)}=\sup_{g\in L^{q}\left(T\right)} \left\|Gg\right\|_{L^1\left(T\right)}$; i.e., the norm of $G$ as an operator on $L^q\left(T\right)$ equals the $L^p$-norm of $G$. Now compute that for $g\in L^q\left(T\right)$, H\"older's inequality gives

$\left\|Gg\right\|_{L^1(T)} = \int_T \left|\int_X \left|F\left(x,t\right)\right|g\left(t\right)d\mu\left(x\right)\right|d\nu\left(t\right)$ $= \int_X \int_T \left|F\left(x,t\right)\right|\left|g\left(t\right)\right| d\nu\left(t\right)d\mu\left(x\right)$ $\leq \int_X \left\|F\right\|_{L^p\left(T\right)} \left\|g\right\|_{L^q\left(T\right)} d\mu\left(x\right)$ $= \int_X \left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}} d\mu\left(x\right)$,

where the second equality follows from Fubini's theorem and where we have used the notation $\left\|F\right\|_{L^p\left(T\right)}=\left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}}$. This proves Minkowski's integral inequality. Q.E.D.

The idea of the proof is to remove the exponents (of course, in the case $p=1$, the result is Tonelli's theorem). This can be done using duality. I hope this helps!

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This is certainly related. One can use several standard techniques to estimate the integral (from above and below) and then the cases of equality gives expressions equivalent to that above in special cases. But Fubini-Tonelli is fundamentally an equality. Still, thanks for the answer. +1 –  Glen Wheeler Jul 20 '11 at 11:04
    
Well, you certainly cannot have an equality from naively exchanging $L^1$ and $L^2$. Minkowski's inequality is sharp; IIRC you can check Hardy-Littlewood-Polya, where the condition for equality is given: equality only if $F(x,t)$ in the notation above factors as $X(x)T(t)$. –  Willie Wong Jul 20 '11 at 11:42
    
The following point of view may be helpful: the usual Minkowski inequality is the triangle inequality for the $L^p$ norm. As for every norm, the triangle inequality can be generalized (by simple induction) to bound the norm of a finite sum of vectors/functions by the sum of their norms. One can then generalize this to infinite sums by (carefully) taking limits). Minkowski's integral inequality generalizes it to "arbitrary uncountable sums", that is integrals w.r.t. some measure (counting measure in the above cases). –  Mark Jul 20 '11 at 13:04
    
@Glen: Dear Glen, you are right, of course. I have noted the Minkowski integral inequality as a result relevant to your question. I thought that it might be helpful. However, I certainly did not claim it is an equality (and it is not, in general). –  Amitesh Datta Jul 21 '11 at 1:06
    
I'd like to post the following link to a related question on this site: math.stackexchange.com/q/33659/8157 –  Giuseppe Negro Aug 11 '11 at 15:15
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