13
$\begingroup$

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 $\Lambda$.

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .