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I am looking for a set of functions $f\left(\sum_i x_i\right)$ that satisfies either or both of the following properties ($x_i\in\mathbb{R}^+$):

  • $f\left(\sum_i x_i\right)=\prod_i g\left(x_i\right)$
  • $f\left(\sum_i x_i\right)=\sum_i h\left(x_i\right)$

As an example, one may consider exponentials and linear transformations, respectively:

  • $\exp\left(x+y\right)=\exp\left(x\right)\exp\left(y\right)$
  • $a\times\left(x+y\right)=a\times x + a\times y$

Note that in the examples $f,g$ and $f,h$ are the same functions, respectively. This does not need to be the case in general. A counter-example is the quadratic function:

  • $\left(x+y\right)^2=x^2+y^2+\underbrace{2xy}_\mathrm{problematic\ part}$

It appears that this is related to homomorphisms (wikipedia).

Cheers, Till

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So why are the exponential and linear functions not suitable for your purposes? What exactly are you asking for? –  Nate Eldredge Jan 28 '12 at 14:41
    
The exponential and linear functions are suitable but I would like to determine all functions that have such properties. –  Till Hoffmann Jan 28 '12 at 14:47
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First of all, in necessarily $f(x)=g(x)$ in the first case and $f(x)=h(x)$ in the second case, so you are really looking functions $f$. If $f$ must be continuous, then $f(x)=e^{ax}$ in the first case and $f(x)=ax$ in the second case, for some constant $a$. –  Thomas Andrews Jan 28 '12 at 14:55
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It also depends on what your domain and range are supposed to be. If it was the complex numbers, the function $f(a+bi)=a-bi$ is an additive answer to your function (and is continuous.) –  Thomas Andrews Jan 28 '12 at 14:57
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But my point is you can prove that $f(x)=g(x)$ and $f(x)=h(x)$, respectively, from the conditions. Shown in my answer below. –  Thomas Andrews Jan 28 '12 at 15:28

1 Answer 1

up vote 2 down vote accepted

Assuming your domain and range are the real numbers.

If you only want continuous functions, the only solutions are $f(x)=g(x)=e^{ax}$ and $f(x)=0$ for the multiplicative case and $f(x)=h(x)=ax$ for the additive case.

For the additive case, first note that $f(x)=f(x+0)=h(x)+h(0)$, and that $f(x)=f(x+0+0)=h(x)+h(0)+h(0)$. So $h(0)=0$ and $f(x)=h(x)$.

Second, note that if $n\in\mathbb N$, $f(nx)=nf(x)$. This is true just by writing $nx = x + x + ... + x$.

Also note that $f(-x)=-f(x)$, since $f(x)+f(-x) = f(x+(-x))=f(0)=0$.

So, for any $n\in \mathbb Z$, $f(nx)=nf(x)$.

Now, let $q=\frac{m}{n}$ be any rational number, $m,n\in\mathbb Z$, $n\neq 0$.

Then $mf(x) = f(mx) = f(nqx)= n f(qx)$. Dividing both sides by $n$, we see that $f(qx)=qf(x)$ for any rational number $q$.

In particular, if we set $a=f(1)$, then for any rational $q$, $f(q)=f(q\times 1) = q f(1)= aq$.

If $f$ is continuous, then $f(x)=ax$ for all $x$.

If $f$ is not necessarily continuous, then all we know is that $f$ is something called a linear function from $\mathbb R$ to $\mathbb R$, where the real numbers are seen as a vector space over the field of rational numbers. There are tons of such functions, but the discontinuous ones are discontinuous everywhere.

For the multiplicative example, first note that $f$ has to be everywhere non-negative, because $f(x)=f(\frac{x}2)f(\frac{x}2)$.
If $f(z)=0$ anywhere, then $f(x)=f(z+(x-z))=f(z)f(x-z)=0$ for any $x$, so $f$ must be zero everywhere. Otherwise, $f(x)$ is always positive, and therefore we can define $f_0(x)=\log f(x)$. This is an "additive" solution, so, if it is continuous, we know $f_0(x)=ax$ for some $a$, and hence $f(x)=e^{ax}$.

If we don't know it is continuous, then all we can say is that $f_0(x)$ is a linear function as defined above.

Addendum Given that you are restricting yourself to $\mathbb R^+$, you can extend any such function on $\mathbb R^+$ to a function on all of $\mathbb R$ in the additive case easily enough by defining $f_1(x)=f(x)$ for $x>0$, $f_1(x)=-f(-x)$ for $x<0$ and $f_1(0)=0$. Then $f_1(x)$ is additive on the entire reals, so we can use my argument above.

The multiplicative case is slightly harder, first because we need to change our proof that if $f(z)=0$ for any $z$ then $f(x)=0$ for all $x$. Our proof, if written for $\mathbb R^+$, only works for $x>z$, but what we can show easily is that $f(\frac{z}n)=0$ for any natural $n$, and, for any $x>0$, there is an $n$ such that $\frac{z}n<x$. So again if $f$ is zero anywhere, it is zero everywhere.

Then we can extend any multiplicative $x$ to the entire reals either by setting it to $0$ or setting it to $\frac{1}{f(x)}$. for $x$ negative.

share|improve this answer
    
That is great, thank you! Do you happen to have any literature references so I can read up on the above? –  Till Hoffmann Jan 28 '12 at 15:37
    
Depends, what is your level of mathematical knowledge? The notion that $\mathbb R$ is a vector space over $\mathbb Q$ is part of abstract algebra, and it even requires some set theory to try to talk about this pair. –  Thomas Andrews Jan 28 '12 at 15:45
    
My mathematical background is from theoretical physics (i.e. a bit of set/group theory but not quite what may be necessary for a formal treatment). –  Till Hoffmann Jan 28 '12 at 16:18

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