$f(\alpha x)-f(\beta x) \to 0$ faster than $\frac{1}{\log x}$ for all functions $f$

I was solving the following problem (# 19-46 in Spivak's Calculus, 4th ed.):

Suppose that $\dfrac{f(x)}{x}$ is integrable on every interval $[a,b]$ for $0<a<b$, and that $\lim_{x\to 0}f(x)=A$ and $\lim_{x\to \infty}f(x)=B$. Prove that for all $\alpha, \beta>0$ we have $$\int_0^\infty \frac{f(\alpha x)-f(\beta x)}{x}dx = (A-B)\log\frac\beta\alpha.$$

Having not seen this exercise before, it struck me strange that the improper integral $\int_0^c\frac{f(\alpha x)-f(\beta x)}{x}dx$ must exist. The comparison theorem then implies that $f(\alpha x)-f(\beta x)$ must go to zero faster than $\frac{1}{\log x}$. On the other hand, I claim we can make $f(\alpha x)-f(\beta x)$ go to zero slower than $\frac{1}{\log^{1+\epsilon}x}$, for instance if we choose $f:x\mapsto \dfrac{1}{\log^\delta(x)}$, for $\delta$ small.

Since we can make the sum $\sum\limits_i (f(\alpha x_i)-f(\beta x_i))$ telescope if we choose $x_i = \left( \dfrac{\alpha}{\beta} \right)^i$ (supposing WLOG $\beta> \alpha$), the previous observation is closely related to the fact that $\sum\limits_i \dfrac{1}{\log(\omega^i)}$ does not converge for $0<\omega<1$, but $\sum\limits_i \dfrac{1}{\log^{1+\epsilon}(\omega^i)}$ does converge.

It's strange that $\dfrac{1}{\log x}$ is like the lower "speed limit" for how fast $f(\alpha x)-f(\beta x)$ can go to zero. Is there an intuitive reason why this is so, other than the ones I just pointed out? What are results for other functions $f(\phi(x))-f(\psi(x))$, where $\lim_{x\to 0}\phi(x)=\lim_{x\to 0}\psi(x)=0$?

• Shouldn't it be $\lim_{x\rightarrow 0}f(x)=A$ in the formulation? – user2097 Sep 2 '14 at 23:06
• @user2097 Fixed, thanks – Eric Auld Sep 3 '14 at 0:51
• I did not say anything about actual solution of the exercise, since this did not seem to be a part of your question. – user147263 Sep 3 '14 at 14:59
• @Thursday Your example with the $c_n$ seems to provide a counterexample, would you agree? The solution in the solutions manual uses substitution and therefore assumes continuity of $f$. Perhaps you could include how you determined that if $f$ is monotone, then the improper integral must exist on $[0,1]$. – Eric Auld Sep 3 '14 at 17:10
• My examples can be made continuous in the usual way: replace rectangles with trapezoids or triangles. Therefore, everything I said applies to the continuous case as well. You may want to consider carefully the distinction between improper Riemann integral and Lebesgue integral. E.g., $\int_0^{\infty} \sin(x^2) \,dx$ converges as an improper Riemann integral, but this function is not Lebesgue integrable on $[0,\infty)$. – user147263 Sep 3 '14 at 17:44

The comparison theorem then implies that $f(\alpha x)−f(\beta x)$ must go to zero faster than $1/\log x$

Not really. Since $f$ was not assumed monotone, the integrand need not have constant sign, and therefore, the comparison theorem need not apply. Another thing to keep in mind is that the integral here is an improper Riemann integral. The function $$\frac{f(\alpha x)-f(\beta x)}{x}$$ need not be Lebesgue integrable on $[0,1]$.

Indeed, take $\alpha=2$, $\beta=1$. Pick a sequence $c_n$ converging to $0$. Define $f(x) = c_n$ when $2^{1-2n}\le x< 2^{2-2n}$ and $f(x)=0$ otherwise. Then $|f(2x)-f(x)| =|c_n|$ when $2^{-2n}\le x<2^{2-2n}$. Note that $|f(2x)-f(x)|$ can tend to zero as slowly as we want. Consequently, $$\int_0^1 \frac{|f(2x)-f(x)|}{x}\,dx \ge \sum_n \frac{2^{2-2n}-2^{-2n}}{2^{2-2n}} |c_n| = \frac34 \sum_n |c_n|$$ can easily diverge.

Even when $f$ is monotone (hence the integrand has constant sign and is Lebesgue integrable), it is incorrect to conclude

that $f(\alpha x)−f(\beta x)$ must go to zero faster than $1/\log x$

Rather, you can conclude that $|f(\alpha x)−f(\beta x)|$ cannot be always bounded from below by a multiple of $1/|\log x|$. This still allows $|f(\alpha x)−f(\beta x)|$ to be not-very-small at times. I'll modify the above example: pick a sequence of indices $n_k$. Let $c_n = 2^{-k}$ when $n_k\le n<n_{k+1}$. Define $f(x) = c_n$ when $2^{-n}\le x< 2^{1-n}$. Then $f$ is monotone. On the interval $2^{-n}\le x< 2^{1-n}$ we have $$f(2x)-f(x) =\begin{cases} 2^{-k-1},\quad &n=n_k \\ 0 & \text{otherwise}\end{cases}$$
Note that $x$ can be much smaller than $2^{-k-1}$ since we can make $n_k$ grow as fast as we want.

Posting another answer to address the actual exercise, rather than the examples inspired by it.

By writing $f = f\chi_{[0,1]}+f\chi_{(1,\infty)}$, we can deal with one limit at a time. So, let's say $f$ is zero outside of some bounded interval. Then for every $x>0$ the function $$F(x) = \int_x^\infty \frac{f(t)}{t}\,dt \tag1$$ is well defined. It is also absolutely continuous and satisfies $F'(x) = f(x)/x$ for almost every $x$. (Depending on the results assumed as known, you may or may not need this.)

The change of variable formula works fine for integrable functions: we need some smoothness of the map by which we change the variable, but no smoothness or continuity is needed from the integrand. Hence, $$\int_x^\infty \frac{f(\alpha t)}{t}\,dt = \int_{\alpha x}^\infty \frac{f(s)}{s}\,ds = F(\alpha x) \tag2$$ One way to justify (2) is to observe that $F(\alpha x)$ is absolutely continuous and satisfies $\frac{d}{dx}F(\alpha x) = -f(\alpha x)/x$. This qualifies it as an antiderivative of $-f(\alpha x)/x$, and the Fundamental theorem of calculus gives (2).

We are to prove that $$\lim_{x\to 0}\int_x^\infty \frac{f(\alpha t)-f(\beta t)}{t}dt = A\log\frac\beta\alpha$$ Using (2), rewrite the left side as $$\lim_{x\to 0} \int_{\alpha x}^{\beta x} \frac{f(t)}{t}dt$$ Finally, use some kind of Mean value theorem for integrals, or simply remark that an inequality of the form $A-\epsilon \le f(t)\le A+\epsilon$ holds on the interval of integration when $x$ is sufficiently small, which implies $$(A-\epsilon)\log \frac{\beta}{\alpha} \le \int_{\alpha x}^{\beta x} \frac{f(t)}{t}dt\le (A+\epsilon)\log \frac{\beta}{\alpha}$$ And $\epsilon$ was arbitrary, so...

• Thank you! Do you happen to have a reference for a proof of the substitution rule in integration without resorting to $f$ continuous? Every single proof I'm finding online assumes $f$ is continuous (probably designed to handle simpler cases). – Eric Auld Sep 3 '14 at 18:30
• @EricAuld There is also a compromise between how much one assumes about $f$ vs the maps that changes variable. In Theorem 3.2 here the integrand is assumed bounded (but need not be continuous). You can get rid of that assumption when the change is $x\mapsto \alpha x$, because then the absolute continuity of composition is clear. – user147263 Sep 3 '14 at 18:34