Bounds on derivative of integrable, monotonically decreasing, differentiable functions on $\mathbb R_+$

Question

The following three conditions have shown up as hypotheses in some recent work, and despite not having been able to find an example, we assume the third is not implied by the former two. We're hoping some fresh eyes will get help us get unstuck.

Let $\mathbb R_+$ denote the nonnegative reals, and let $f\colon \mathbb R_+ \to \mathbb R_+$.

Our conditions are

1. $f$ is integrable;
2. $f$ is differentiable and monotonically decreasing; and
3. there exists $x_0 \ge 0$ such that for all $x > x_0$ we have $f(x) > - x \,f'(x)$.

We are interested in examples of functions $f$ satisfying 1 and 2 but not 3, hoping that they will give us some insight into how to get rid of condition 3.

Answer 1

The function $$f(x)=\exp(-\exp(x))$$ seems to do the job. Integrability, differentiability are clear, the derivative is $\exp(x)\exp(-\exp(x))$, and the function $$ g(x)=f(x)+xf'(x)=\exp(-\exp(x))(1-x\exp(x))$$ has exactly one null at $W(1)\approx0.567$ (Lambert function), where it goes from positive to negative:

Answer 2

Am I missing something extremely obvious?

Let $f(x) = x^{-\alpha}$ for any $\alpha > 1$, when $x > 1$. And put anything reasonable for $x \leq 1$. Near infinity we have

$$ f(x) = x^{-\alpha} < \alpha x^{-\alpha} = (-x) (-\alpha x^{-\alpha - 1}) = -x f'(x) $$

$f$ is clearly differentiable, monotonic, and integrable.

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In fact, your third condition can be rearranged to read that for all sufficiently large $x$, the function $g(x) = xf(x)$ satisfies $g'(x) > 0$. This appears rather hard to satisfy for $f$ integrable.

Answer 3

For an explicit example, you might try $$ f(x) = \sum_{n=1}^\infty \dfrac{1}{n^2} (1 - \tanh(n^2(x-n)))$$ noting that $f'(n) < -1$.