Functions such that $\sup_{t\in\mathbb{R}}\int_{\mathbb{R}}e^{a(x)t}|f(x)|dx<\infty$ and ...

Question

Can we find a bounded function $a:\mathbb{R}\to\mathbb{R}$ and a function $f\in L^1(\mathbb{R})$ with $f\neq 0$ such that $$\sup_{t\in\mathbb{R}}\int_{\mathbb{R}}e^{a(x)t}|f(x)|dx<\infty$$ and $$\inf_{t\in\mathbb{R}}\int_{\mathbb{R}}e^{a(x)t}|f(x)|dx=0.$$ I tried to consider $a(x)=a$ a constant, but it didn't work, because we will get $$\sup_{t\in\mathbb{R}}e^{at}\int_{\mathbb{R}}|f(x)|dx<\infty$$ which is finite only if $a=0$, but in this case $$\inf_{t\in\mathbb{R}}e^{at}\int_{\mathbb{R}}|f(x)|dx=\int_{\mathbb{R}}|f(x)|dx\neq0.$$

Answer 1

The first condition is restrictive. Define the set $S_l:=\{x,a(x)\geqslant 1/l\}$. Then we have for $t\gt 0$, $$\int_{S_l} e^{a(x)t}|f(x)|\mathrm dx\geqslant e^{t/l}\int_{S_l}|f(x)|\mathrm dx,$$ hence $$\int_{S_l}|f(x)|\mathrm dx\leqslant e^{-t/l}\sup_u\int_{\mathbf R}e ^{a(x)u} |f(x)|\mathrm dx.$$ Letting $t$ going to infinity, we derive that $f(x)=0$ for almost every $x \in S_l$, hence up to sets of zero measure, the inclusion $$\{a(x)\gt 0\}\subset\{f(x)=0\}.$$ Similarly, we get the same inclusion with $-a$ instead of $a$, hence $$\{a(x)\neq 0\}\subset\{f(x)=0\}.$$ But this is not compatible with the second condition.