Existence of a class of function

Question

Consider a function $\gamma: [0, \infty) \to (0, \infty)$ satisfying these following conditions:

i) $\gamma$ decreases on $[0, \infty)$ and $\displaystyle \lim_{t\to \infty} \gamma(t) = 0$;

ii) $\gamma\not \in L^1$;

iii) $t\dot \gamma \in L^1$.

By part integration, we get $$\int_0^T t\dot \gamma(t) dt = T\gamma(T) - \int_0^T \gamma(t)dt.$$

Since $t\dot \gamma \in L^1$ and $\gamma \not \in L^1$, we conclude $\displaystyle \lim_{t\to \infty} t\gamma(t) = \infty$.

I could not find any $\gamma$, so I wondered if such a function exists.

Thank you very much for any hint or solution.

Answer 1

Such functions do not exist.

If $t \mapsto t\dot{\gamma}(t)$ belongs to $L^1$, then $\dot{\gamma}$ is locally integrable. Since $\lvert\dot{\gamma}(t)\rvert \leqslant \lvert t\dot{\gamma}(t)\rvert$ for $t \geqslant 1$, $\dot{\gamma}$ is integrable over $[1,+\infty)$, and since $\dot{\gamma}(t) \leqslant 0$ and $\gamma(0)$ is finite, $\dot{\gamma}$ is also integrable over $[0,1]$, so it follows that $\dot{\gamma} \in L^1([0,+\infty))$. Now we have

$$\gamma(x) = -\int_x^{+\infty} \dot{\gamma}(t)\,dt = \int_x^{+\infty} \lvert \dot{\gamma}(t)\rvert\,dt,$$

and hence

\begin{align} \int_0^{+\infty} \gamma(x)\,dx &= \int_0^{+\infty} \int_x^{+\infty} \lvert\dot{\gamma}(t)\rvert\,dt\,dx \\ &= \int_0^{+\infty} \int_0^t\,dx\, \lvert\dot{\gamma}(t)\rvert\,dt \\ &= \int_0^{+\infty} t\lvert\dot{\gamma}(t)\rvert\,dt \\ &< +\infty \end{align}

by Tonelli's theorem.

Thus $i)$ and $iii)$ imply $\gamma \in L^1([0,+\infty))$.

Answer 2

As a complement to Daniel Fischer's answer, note that if you restrict your domain to $(0,\infty)$ (or increase your codomain to $(0,\infty]$) then such functions do exist, one example being $$\gamma(t)=\begin{cases} t^{-1/2}&\text{if }0<t<1,\\ t^{-2}&\text{if }t\ge1. \end{cases}$$ You can easily construct a smooth version if you prefer, since all that matters is what happens at $0$ and $+\infty$.