How can I find a function that is in $L^1(\mathbb{R})$ with its derivative also but its limit tends to zero?

Question

I am trying to find for a function that would full-fill these conditions below: $$f \in L^1(\mathbb{R})$$ $$f' \in L^1(\mathbb{R})$$

but its $\lim_{t \to \infty}=0$.

I've tried to find a function which dominator is larger than nominator but usually these are improper integrals and do not converge!

Can somebody suggest any function? Thanks in advance!

Answer 1

The hardest part of this problem has been determining what the problem is. Here is my guess at the problem and hints for an answer. Melina will let us know if this is relatively close.

Problem. Supose that $f:[0,\infty)\to\mathbb{R}$ is a continuous function for which both $f$ and $f'$ belong to $L_1([0,\infty)$.

Part (a) Show that it is not necessarily the case that $f(x)\to 0$ as $x\to \infty$.

Part (b) Show that, if we also assume that $f$ is absolutely continuous, then it is true that $f(x)\to 0$ as $x\to \infty$.

Hint for (b): Let $\epsilon>0$. Then there is a $T$ so that $$|f(s)-f(t)| = \left| \int_s^t f'(t)\,dt \right| < \epsilon$$ for all $T \leq s <t$. From that and the fact that $f$ belongs to $L_1([0,\infty)$ one can deduce that $f(x)\to 0$ as $x\to \infty$.

Hint for (a): There is a standard and well-known example of a nonegative, continuous function $f:[0,\infty)\to\mathbb{R}$ for which $f(n)=1$ for each $n=1,2,3, ...$ and yet the integral $\int_0^\infty f(t)\,dt$ exists (either as a Lebesgue integral or as an improper Riemann integral --which amounts to the same thing here). Modify that example so that the function $f$ is singular, i.e., so that $f'=0$ almost everywhere.

Note 1.: As @zhw points out there is the possibility (probability?) that the original version of the problem assumed that $f$ is everywhere differentiable, in which case (since we know $f'$ is integrable) we can use part (b) of the problem.

Note 2. Melina has confirmed that her assignment assumed even more, namely that $f$ is to be continuously differentiable. This reduces the problem to an elementary analysis problem and all integrals can be interpreted as Riemann or improper Riemann integrals. The indication that $f$ and $f'$ belong to $L_1(\mathbb R)$ does not normally suggest that $f'$ would have to exist everywhere. We also do not need to mention absolute continuity (although it wouldn't be incorrect to do so). If $f'$ is everywhere continuous then the hint for (b) just uses the baby fundamental theorem of the calculus stated for continuous functions.