# Is $f(x)=k \log x$ the only solution to the equation $\sum_{i=1}^{n}f(x_{i})=f(\prod_{i=1}^{n}x_{i})$?

For $$x_{i}\in \mathbb{R}_{+}$$

Is $$f(x)=C\log x$$ $$(C\in \mathbb{R})$$ the only solution to the equation:

$$\sum_{i=1}^{n}f(x_{i})=f(\prod_{i=1}^{n}x_{i})$$

I have tried to solve this problem in the following way:

$$p:\forall x_{1},x_{2}\in \mathbb{R}_{+}$$ $$f(x_{1})+f(x_{2})=f(x_{1}x_{2})$$

$$q:\forall x_{i}\in \mathbb{R}_{+}$$ $$\sum_{i=1}^{n}f(x_{i})=f(\prod_{i=1}^{n}x_{i})$$

I think it is obvious that $$p\Leftrightarrow q$$

(If I am wrong please correct me.)

So the solution to

$$x_{i}\in \mathbb{R}_{+}$$

$$\sum_{i=1}^{n}f(x_{i})=f(\prod_{i=1}^{n}x_{i})$$

can be expressed as

$$x_{1},x_{2}\in \mathbb{R}_{+}$$

$$f(x_{1})+f(x_{2})=f(x_{1}x_{2})$$

So the previous question can be changed into a question in group theory:

Is $$f:\mathbb{R}_{+}\rightarrow \mathbb{R}$$ $$f(x)=Clnx$$ $$(C\in \mathbb{R})$$ the only Isomorphic mapping from $$<\mathbb{R}_{+},*>$$ to $$<\mathbb{R},+>$$ ?

But I cannot prove it in this way of thinking this question.

• The $f$ should inside the sum, please correct – Nicky Hekster Feb 24 at 15:39
• @NickyHekster Thanks,corrected. – Veritas Julius Feb 24 at 15:43
• $<\mathbb R_+,+>$ is not a group. Maybe you mean $<\mathbb R,+>$? – Dog_69 Feb 24 at 15:52
• @Dog_69 Thank you I have already correct it. – Veritas Julius Feb 24 at 16:00

I doubt this is true in general, but at least it is correct under the extra assumption that $$f$$ is continuous; this is why.

Normalizing, we can assume that $$f(e)=1$$. Then $$f(e^n)=f(e\dotsb e)=f(e)+\dotsc+ f(e)=nf(e)=n$$ for any positive integer $$n$$. As a result, $$n=f(e^n)=f(e^{n/q}\dotsb e^{n/q})=qf(e^{n/q})$$, implying $$f(e^{n/q})=n/q$$. Thus, $$f(e^c)=c$$ for any rational $$c>0$$, and by continuity, $$f(e^x)=x$$ for any $$x>0$$. Substituting $$y=e^x$$ we get $$f(y)=\ln y$$, where $$y>1$$. The case $$y<1$$ (corresponding to $$x<0$$) is dealt with similarly.

If $$f$$ is a solution then $$g=e^f$$ satisfies: $$\prod_{i=1}^ng(x_i)=\prod_{i=1}^ne^{f(x_i)}=e^{\sum_{i=1}^nf(x_i)}=e^{f(\prod_{i=1}^nx_i)}=g(\prod_{i=1}^nx_i)$$ In particular $$g(xy)=g(x)g(y)$$ You can see here the only continuous solution is $$g(x)=x^c$$ hence $$f(x)=\log g(x)=\log x^c=c\log x$$

I'm assuming you mean $$\mathbb{R}_+^*$$.

In this case, since $$(\mathbb{R}_+^*, \times)$$ is isomorphic to $$(\mathbb{R},+)$$ (with $$\ln$$), and actually diffeomorphic, the question becomes : are there other morphisms $$(\mathbb R, +)\to (\mathbb R, +)$$ than multiplication by a certain constant.

Of course it's a usual exercise that any such continuous morphism is $$x\mapsto ax$$ (take $$a=f(1)$$, prove the equality on the integers by induction, then on the rationals using $$q\times \frac{p}{q} = p$$, then on the reals by density and continuity). Actually with a bit more work you can prove that it's even the only such measurable morphism.

But it turns out if you leave all topological considerations on the side, there are plenty of other morphisms. This is because $$(\mathbb R, +)$$ is an infinite dimensional $$\mathbb Q$$-vector space (it actually has dimension $$\mathfrak{c}$$, the cardinal of the reals). Thus a morphism $$\mathbb{ R\to R}$$ is the same as a $$\mathbb Q$$-linear map, and so we can compute $$\hom(\mathbb{R,R}) = \hom(\mathbb{Q^{(\mathfrak{c})}, Q^{(\mathfrak{c})}}) = \displaystyle\prod_\mathfrak{c}\hom(\mathbb{Q,Q^{(\mathfrak{c})}}) = \displaystyle\prod_\mathfrak{c}\mathbb{Q^{(\mathfrak{c})}} = (\mathbb{Q^{(\mathfrak{c})}})^\mathfrak{c} = \mathbb{R}^\mathfrak{c}$$ (where all the $$=$$ should be isomorphisms); so that's a bunch more than $$\mathbb{R}$$ !

In terms of cardinality this is easily seen to be of cardinality $$2^\mathfrak{c}$$ : most group morphisms $$\mathbb{R\to R}$$ aren't of the form $$x\mapsto ax$$, actually those of this form are "extremeley rare"; but all the measurable/continuous ones are.