Integral convergence implies pointwise

Question

This is a simplified version of a question I am looking at, but embarrassingly I can't do this one anyway.

Let's assume $f(x)$ is a decreasing positive function, $f(0)$ is infinite and, moreover, $f(x)$ is comparable to $1/x$ near $x=0$. We look only at positive $x$'s.

Question. Assume we know that the limit $$ \lim_{a\to 0^+}\int\limits_{a}^{2a} f(t)dt $$ exists as a positive finite number. Does it imply that $$ \lim_{n\to\infty} f(2^{-n})\cdot 2^{-n} $$ exists?

Note: it definitely does not imply that the limit of $xf(x)$ exists.

Answer 1

If you are ok with $f$ being not continuous and not strictly decreasing, here's a simple enough counter-example. (With a bit more work you can make it continuous and strict, but it distracts from the point.)

Let $f(x)$ be the step-function defined by

$$ f(x) = \begin{cases} 2^{k} & x \in ( 2^{-k}, 2^{1-k}) \\ 2^{k} & x = 2^{-k}, \quad k \text{ is odd} \\ 2^{k+1} & x = 2^{-k}, \quad k\text{ is even} \end{cases} $$

You have that the integral $$ \int_a^{2a} f(t) ~dt = 1 $$ for any $a$.

You have that $2^{-k} f(2^{-k})$ alternating between 1 and 2 and does not converge.

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Incidentally, the "comparability to $1/x$" in your question is superfluous. Since you assumed decreasing, you have that

$$ a f(2a) \leq \int_a^{2a} f(x) ~dx \leq a f(a) $$

which implies that

$$ \liminf xf(x) \geq \lim \int_a^{2a} f(x) ~dx $$

$$ \limsup xf(x) \leq 2 \lim\int_a^{2a} f(x)~ dx$$

so comparability is automatic, and also the existence of a sequence $x_n \to 0$ such that $x_n f(x_n)$ converges. (Of course $(2^{-n})$ need not be such a sequence.)