Divergent function of ratio must be logarithm Given. Consider two functions $F(t)$ and $r(t,x)$ such that $\lim_{t\to\infty} F(t) = \infty$ and $\lim_{t\to\infty} r(t,x)$ is finite for any $x$.  ($x$ and $t$ are always positive in what follows.)  Suppose they sum to a function only of the ratio $x/t$:
\begin{align}
F(t) + r(t,x) = f (x/t) \qquad (1)
\end{align}
Claim. $f(x/t)$ must diverge logarithmically at large $t$.  That is,
\begin{align}
\lim_{t\to\infty} \frac{f(x/t)}{\ln t} = const\ (\text{independent of } x) \qquad (2)
\end{align}
Flawed proof.
The paper cited below does the following.  Define $r(x)= \lim_{t\to\infty} r(t,x)$.  Then we can write:
\begin{align}
f(x/t) = F(t) + r(x) + \phi(t,x) \qquad (3)
\end{align}
where $\phi(t,x) = r(t,x) - r(x)$ goes to zero as $t\to\infty$.
Now it seems reasonable (this is the incorrect step!) that $F(t)+r(x)$ and $\phi(t,x)$ should separately be functions only of the ratio $x/t$, since they have different behavior at large $t$.  In particular,
\begin{align}
F(t) + r(x) = W(x/t) \qquad (4)
\end{align}
It is then straightforward to show that $W$ is a logarithm, which completes the proof.  We differentiate both sides of $(4)$ with $x$:
\begin{align}
r'(x) = \frac{1}{t}W'(x/t)    \qquad (5)
\end{align}
Then set $x=1, t=\frac{1}{y}$:
\begin{align}
r'(1) = y\ W'(y)    \qquad (6)
\end{align}
so $W(y) = r'(1) \ln y + $ const.
Counterexample to (4). Consider $F(t) = \ln t + \frac{1}{t}, r(t,x) = -\ln x - \frac{1}{t}.$  Then $r(x) = -\ln x$ and $F(t) + r(x) = \ln \frac{t}{x} + \frac{1}{t}$, so $(4)$ fails.
Can this proof be saved?  If not, how else can it be done?
Reference.  "A hint of renormalization" (American Journal of Physics, hep-th/0212049).  See equation (30) and Appendix C.  My $t$ is his $\Lambda$.
 A: Under very weak conditions, we can show that $f$ diverges logarithmically. Measurability of $f$ (or $F$) is enough to obtain the result (and, if measurability is not required, then there are counterexamples). We can show that
$$
\lim_{t\to\infty}\frac{F(t)}{\log t}=\lim_{t\to\infty}\frac{f(x/t)}{\log t}=\textrm{const}.
$$
The first equality is clear from the fact that $r(t,x)=f(x/t)-F(t)$ is bounded in the limit $t\to\infty$. It is possible for the constant to be zero, in which case I would say that $f$ diverges sub-logarithmically rather than logarithmically. This is the case, for example, if $F(t)=f(t^{-1})=\log\log\max(e,t)$.
Defining $g\colon\mathbb{R}\to\mathbb{R}$ by $g(t)=f(e^{-t})$, we need to show that $g(t)/t$ tends to a limit as $t\to\infty$.
For any $t^\prime > 0$, we have
$$
g(t+t^\prime)-g(t)=f(e^{-t^\prime-t})-F(e^t)-f(e^{-t})+F(e^t)=r(e^t,e^{t^\prime})-r(e^t,1).
$$
So, this converges to a finite limit as $t\to\infty$. Setting $\lambda=\lim_{t\to\infty}(g(t+1)-g(t))$, then for any $\epsilon>0$ we have $-\epsilon\le g(t+1)-g(t)-\lambda\le\epsilon$ for all large enough $t$ -- say, for all $t \ge t_0$. Setting $t_n=t_0+n$ then, for each $t\ge t_0$ we can find positive integer $n$ with $t\le t_n\le t+1$
\begin{align}
\lvert g(t)-\lambda t\rvert&\le\sum_{k=1}^n\lvert g(t_k)-g(t_{k-1})-\lambda\rvert+\lambda\lvert t-n\rvert+\lvert g(t_0)\vert+\lvert g(t)-g(t_n)\rvert\\
&\le n\epsilon+\lambda\lvert t-t_n+t_0\rvert+\lvert g(t_0)\vert+\lvert g(t)-g(t_n)\rvert\\
&\le (t_n-t_0)\epsilon+\lambda\lvert t_0-1\rvert+\lvert g(t_0)\vert+\sup_{s\in[0,1]}\lvert g(t+s)-g(t)\rvert
\end{align}
As long as we can show that $\sup_{s\in[0,1]}\lvert g(t+s)-g(t)\rvert$ is uniformly bounded by some $K > 0$ for all large enough $t$, the right hand side of this inequality equals $t\epsilon$ plus a bounded term. So, $\limsup_{t\to\infty}\lvert g(t)/t-\lambda\rvert$ is bounded by $\epsilon$. Taking $\epsilon$ arbitrarily small shows that $g(t)/t\to0$.
It remains to be shown that $\sup_{s\in[0,1]}\lvert g(t+s)-g(t)\rvert$ is uniformly bounded over all large enough $t$. Here, I will use measurability. First, as $g(t+s)-g(t)$ converges to a finite limit for each fixed $s$, $\limsup_{t\to\infty}\lvert g(t+s)-g(t)\rvert$ is bounded. By monotone convergence, the Lebesgue measure of $\{s\in[0,1]\colon\limsup_{t\to\infty}\lvert g(t+s)-g(t)\rvert < K\}$ tends to $1$ as $K$ goes to infinity. In particular, it is positive for some $K$. Then, using monotone convergence again, there exists a $t^*$ such that $\{s\in[0,1]\colon\sup_{t\ge t^*}\lvert g(t+s)-g(t)\rvert < K\}$ has positive measure. That is, there is an $A\subseteq[0,1]$ of positive measure such that $\lvert g(t+s)-g(t)\rvert < K$ for all $t\ge t^*$ and $s\in A$.
Now, I use the fact that, for a set $A$ of positive measure, the sum $A+A$ contains an open interval. The sum of intervals of lengths $r,s$ is an interval of length $r+s$. So, $A_n\equiv\{s_1+\cdots+s_n\colon s_1,\ldots,s_n\in A\}$ contains an interval of length greater than $1$ for large enough $n$. Say, $[a,a+1]\subseteq A_n$. Then, for all $s\in[a,a+1]$, we have $s=s_1+\cdots+s_n$ for $s_1,\ldots,s_n\in A$, so
$$
g(t+s)-g(s)=\sum_{k=1}^n\left(g(t+s_1+\cdots+s_k)-g(t+s_1+\cdots+s_{k-1})\right)
$$
which is bounded by $nK$. Therefore, for $t\ge t^*+a$
\begin{align}
\sup_{s\in[0,1]}\left\lvert g(t+s)-g(t)\right\rvert
&\le\sup_{s\in[a,a+1]}\left\lvert g(t-a+s)-g(t-a)\right\rvert+\left\lvert g(t-a+a)-g(t-a)\right\rvert,\\
&\le nK+nK
\end{align}
which is uniformly bounded as required.
