Talk:Linear Operator on General Logarithm

I'm not sure how to look at linear operators, that's why I'm confused LF. Is the domain a subset of the reals, or is the domain a set of real functions? That's why I didn't say $\phi: \R \to \R$, because I wanted $\phi$ to have the connotation of a function on functions. As to $y$, that was a mistake, you're correct it should be $y>0$ not $\phi(y) > 0$ In any event, it needs to be clear that we're dealing with a function of $y$ here and $\ln a$ is a constant multiple. --GFauxPas 17:27, 16 January 2012 (EST)

You are correct in considering linear operators in $\R^\R \to \R^\R$ (where $\R^\R$ means the functions from $\R\to\R$), because pointwise addition and scalar ($\R$-) multiplication yield the space $\R^\R$ to be a module over $\R$. As $\R$ is a field, $\R^\R$ is even a vector space over $\R$. To avoid further confusion, it would be best to consider this space. Make sure to specify that you mean by $y>0$ that $\forall x\in \R: y(x) > 0(x)=0$. Hopefully, this stuff makes sense to you. --Lord_Farin 03:34, 17 January 2012 (EST)