differentiation under the integral sign

Question

Let $f \in L^1(R)$. Consider the function: $$ F(x) = \int_R e^{ixt}f(t) dt $$

1. If $|t|^kf(x) \in L^\infty(R)$ for all $k \ge 1$, show that $F$ is infinitely differentiable.

2. Suppose in addition that $f$ is continuous, show that $\lim_{|x|\rightarrow \infty} F(x) = 0$.

For the first part, it's easy to show that the first derivative of $F$ exist. Basically, we only need to show the following equality: $$ \lim_{|h|\rightarrow 0} \int_{R} \frac{e^{i(x+h)t} - e^{ixt}}{h} f(t)dt = \int_{R}\lim_{|h|\rightarrow 0} \frac{e^{i(x+h)t} - e^{ixt}}{h} f(t)dt $$ Since $f \in L^1$, the above can be achieved by the Dominated Convergence Theorem. Similarly, to show $F$ is twice differentiable, we need to establish the following: $$ \lim_{|h|\rightarrow 0} \int_{R} \frac{e^{i(x+h)t} - e^{ixt}}{h} itf(t)dt = \int_{R}\lim_{|h|\rightarrow 0} \frac{e^{i(x+h)t} - e^{ixt}}{h} itf(t)dt $$

My question arises here: if we adapt the idea for showing $F$ is differentiable, then we need $itf(t)$ to be in $L^1$. But the assumption in the question is that $itf(t) \in L^\infty$. Thanks.

Answer 1

Hint: $\vert t\vert^k f(t)$ is bounded for all $k$. So, for example, to show that $tf(t)\in L^1$, consider estimating $\vert t\vert f(t)$ using the fact that $\vert t\vert^3 f(t)$ is bounded. You might also need to break up the integration into $\vert t\vert<1$ and $\vert t\vert>1$.