In my current research, I'm confronted with the justification of some facts, and I don't know how to proceed in proving them, so I need to know if there exist some theorems (precisely three theorems) which allow me to do so.

The problem I am investigating is the following: I have an explicit real valued function $f$, DEFINED and CONTINUOUS on each point of $D=]0,1[^4 $. As it is customary to do, let $f(u,v,w,t)$ be the value that this function takes at point of $(u,v,w,t)\in D$: $f$ cannot be defined at the boundary of $D$ and I can't extend its domain of definition $D$ in order to define it on the whole $\overline{D}=[0,1]^4 $, the closure of $D$. I know that $$ f(u,v,w,t)=\sum_{n=0}^{+\infty} f_n(u,v,w,t)\quad\forall (u,v,w,t)\in D $$ where $\{f_n\}_{n\in\Bbb N}$ is a sequence of functions defined and continuous over $D$ which can be extended as continuous function on $\overline{D}$. This makes me think that, for all integers $n$, $$ \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t\quad \text{ exists.} $$ And now the questions.

  1. What theorem (be it a necessary and sufficient or only a sufficient condition) would allow me to prove the following formula? $$ \begin{split} \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 & f(u,v,w,t)\mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t \\ =&\sum_{n=0}^{+\infty} \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 f_n(u,v,w,t)\mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t \end{split} $$

  2. What theorem (be it a necessary and sufficient or only a sufficient condition) would allow me to perform any change of the order of integration respect to any of the variables involved, in order to have for example that $$ \begin{split} \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 &f(u,v,w,t)\mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t\\ = & \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 f(u,v,w,t) \mathrm{d}w \mathrm{d}t \mathrm{d}u \mathrm{d}v \\ = & \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 f(u,v,w,t) \mathrm{d}v \mathrm{d}u \mathrm{d}t \mathrm{d}w \quad ? \end{split} $$

  3. Finally, suppose that one further hypothesis is made over $f$: $f$ depend on a parameter $ a \geq 0$, call it $f_a$ and suppose that $\forall (u,v,w,t) \in ]0,1[^4$ the mapping $a \mapsto f_a(u,v,w,t)$ is $C^{\infty}$ over $ \mathbb{R}_+$: what theorem (again be it a necessary and sufficient or only a sufficient condition) allow me to say that $$ g:a \mapsto \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 f_a(u,v,w,t) \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t \in C^{2}(\Bbb R^+) $$ and $$ g''(a)=\displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 \displaystyle \frac{\mathrm{d}^2}{\mathrm{d}a^2} f_a(u,v,w,t) \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t $$ i.e. would allow me to differentiate twice under the integral symbol?

I know what theorem allowing me to have 1) 2) and 3) in case that $f$ is defined over an interval of $\Bbb{R}$ and so in case a simple integral

Could anybody help me please? Does there exist a freely accessible reference over the internet where I can find such theorems?


closed as off-topic by abx, Ben McKay, user44191, Sean Lawton, Jan-Christoph Schlage-Puchta Jun 12 at 10:56

This question appears to be off-topic. The users who voted to close gave these specific reasons:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – abx, Sean Lawton, Jan-Christoph Schlage-Puchta
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The condition $$ \sum_{n=1}^\infty \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t < +\infty $$ will let you do 1,2. Alternatively, $f_n(u,v,w,t)\ge 0$ will also let you do 1,2 wit the proviso that you have to allow value $+\infty$ for both sides.

  • $\begingroup$ thanks mister Gerarld for your help. for having 1) and 2), do i need , further more the condtion you wrote , $ \displaystyle \sum_{n \geq 0} \displaystyle \int_{0}^1 \int_{0}^1 \int_{0}^1 \int_{0}^1 | f_n(u,v,w,t)| \mathrm{d}u \mathrm{d}v \mathrm{d}w \mathrm{d}t $ converge? thanks $\endgroup$ – mamiladi Jun 11 at 1:32
  • $\begingroup$ @mamiladi ... you are right, I forgot the $\sum$ in there. I added it. $\endgroup$ – Gerald Edgar Jun 11 at 11:57

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