2
$\begingroup$

When working on an applied math topic, I have come across the following general problem.

Let $f(x_1, x_2, ..., x_n)$ be a real function of $n$ real variables $x_1, x_2, ..., x_n$ which is defined on a unit $n$-cube $I_n = [0, 1]^n$. Let $A_n$ be the integral of this function over $I_n$ (thus $A_n$ is the average value of this function on $I_n$). Now consider a function $g(x) := f(x, x, ..., x)$. It is defined on the unit interval $I_1$. Let $B$ be the integral of the function $g(x)$ over $I_1$.When it is true that $A_n < B$?

This general problem was motivated by the following specific question which is of great interest to me. Consider a function $f(x_1,x_2)= 1/(1+a_1x_1+a_2x_2)$ defined on a unit square $I_2=[0,1]^2$. It depends on two real parameters, $a_1$ and $a_2$ (ensuring that the denominator in the expression for the function $f$ is non-zero). Let $A_2$ be the integral of this function over $I_2$, and let $B$ be the integral of $g(x):= f(x,x)$ over the unit interval $I_1$. When the inequality $A_2 < B$ holds?

Having performed some tedious calculations I was able to investigate this problem in 2 special cases: $a_1=a_2$ (the inequality $A_2 < B$ holds) and $a_1=-a_2$ (the opposite inequality holds). I wonder if the analysis could be performed using some general ideas (using perhaps such properties as convexity upward/downward or monotonicity). Do you know if there is some general theory/results studying this type of problems?

Thank you very much for reading my question.

$\endgroup$
3
$\begingroup$

The approach suggested by Fedor is interesting but I see an issue with it. Namely:

The integral (over the square $ I_2 =\{(x, y): x, y \in [0, 1]\} $) of the left-hand side of the inequality listed by Fedor equals $2A_2$ (in my notation I follow Fedor, and instead of using $x_1$ and $x_2$ I use $x$ and $y$). But is the integral of the right-hand side over this square equal $2B$?

The integral $B$ is defined as the integral of $1/[1+ (a_1+a_2)x]$ from $x = 0$ to $x=1$. My calculation of this integral gives $B=(a_1+a_2)^{-1}\ln(1+a_1+a_2)$.

On the other hand, when we integrate the right-hand side of this inequality over $I_2$, we integrate $1/[1+(a_1+a_2)(x+y)/2]$. And this definite integral, according to my calculation, is not equal to $2B$.

So I think the original question is still open.

$\endgroup$
  • $\begingroup$ You are correct and I was not. Hopefully now I also am, see new edition of my answer.) $\endgroup$ – Fedor Petrov Mar 16 '17 at 4:46
1
$\begingroup$

(New version, previous was completely wrong.)

Let us compare $f(x,y)+f(y,x)=g(a_1x+a_2y)+g(a_2x+a_1y)$ and $f(x,x)+f(y,y)=g(a_1x+a_2x)+g(a_1y+a_2y)$, where $g(t)=1/(1+t)$. Since $g$ is convex on $(0,+\infty)$ (I assume that $1+\min(a_1,a_2,a_1+a_2)>0$, else integrals diverge), comparing $g(A)+g(B)$ and $g(C)+g(D)$ with $A+B=C+D$ is equivalent to comparing $\max(A,B)$ and $\max(C,D)$. That is, if $a_1$ and $a_2$ have the same sign, we get $f(x,x)+f(y,y)\geqslant f(x,y)+f(y,x)$, otherwise $f(x,x)+f(y,y)\leqslant f(x,y)+f(y,x)$. Integrating over the square we compare your integrals.

$\endgroup$
  • $\begingroup$ @ Fedor. Thank you very much, Fedor, for the input. However, as Michael B. has pointed out, there seems to be an issue with your answer. $\endgroup$ – Peter5 Mar 15 '17 at 22:32
  • $\begingroup$ Oh, such a shameful mistake. $\endgroup$ – Fedor Petrov Mar 16 '17 at 4:32
  • $\begingroup$ @ Fedor. Don’t worry about a flaw in the previous version of your answer. Let me list a couple quotes: “There are no mistakes or failures -- only lessons” (DW), “Being human means you will make mistakes. And you will make mistakes, because failure is God's way of moving you in another direction” (OW). The new version is correct. Again, thank you for working on the question! $\endgroup$ – Peter5 Mar 16 '17 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.