I am trying to go through the proof of Lemma 2.6.2 of Mehta, Random matrices, 3rd edition, Elsevier Academic Press. The lemma states that:
If three continuous and differentiable functions $f_{k}(x)$, $k=1,2,3$, satisfy the equation:
\begin{equation}\tag{1} f_{1}(xy)=f_{2}(x)+f_{3}(y), \end{equation}
then they are necessarily of the form $a\ln(x)+b_{k}$, with $b_{1}=b_{2}+b_{3}$.
Now let me write the first few lines of the proof given in the book:
Differentiating equation (1) with respect to $x$, we have:
\begin{equation}\tag{2} f_{1}'(xy)=\frac{1}{y}f_{2}'(x) \end{equation}
which, on integration with respect to $y$, gives:
\begin{equation}\tag{3} \frac{1}{x}f_{1}(xy)=f_{2}'(x)\ln(y)+\frac{1}{x}g(x) \end{equation}
where $g(x)$ is still arbitrary.
Now the proof is a few more lines but I am stuck at equation (3).
My try:
I am guessing that by prime the author means partial derivatives with respect to $x$. Then if I take total derivative (mathematicians might feel all this is very problematic, sorry!) of equation (1) with respect to $x$, I get:
\begin{align} \frac{d}{dx}f_{1}(xy)=&\frac{d}{dx}f_{2}(x)\\ \Rightarrow y\frac{\partial}{\partial x}f_{1}(xy)=&f_{2}'(x)\\ \Rightarrow f_{1}'(xy)=&\frac{1}{y}f_{2}'(x) \end{align} which is equation (2).
Now if I integrate this equation with respect to y, I get:
\begin{equation}\tag{4} \int\frac{\partial}{\partial x}f_{1}(xy)dy=f_{2}'(x)\ln(y)+cf_{2}'(x) \end{equation} for some integration constant $c$. But I am not able to proceed further, specifically, I am not able to get equation (3) of Mehta no matter what I do. I tried to use the relation:
\begin{equation} \frac{d}{d(xy)}f_{1}(xy)=\frac{\partial}{\partial x}f_{1}\frac{dx}{d(xy)}+\frac{\partial}{\partial y}f_{1}\frac{dy}{d(xy)} \end{equation}
using the chain rule for partial derivatives and rewriting the integrand on LHS in equation (4) but that has not been of much help. Any help would be highly appreciated, thanks.
Firstly, $f_k$ are single variable functions for all $k = 1,2,3$ so primes indicate regular derivatives. Here, $f_1'(xy)$ should be interpreted as inputting the single variable $xy$ into the derivative of $f_1$.
Now, note that in equation (4), your constant of integration $c$ can depend on $x$ as you treated $x$ as a constant when integrating. You may want to write it as a function $c = c(x)$. If $c(x)$ is differentiable, so is $h(x) = xc(x)$ by the product rule. Since $c(x)$ is arbitrary, we can write $c(x) = \frac{1}{x}xc(x) = \frac{1}{x}h(x)$, where $h(x)$ is still arbitrary.
For LHS, try substituting $u = xy$ (so that $dy = \frac{1}{x}du$).
Can you take it from here? Let me know if you need a little more.