# differentiation under the integral sign

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.

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 $\vert t\vert>1$.
• According to your question, we have $\vert t\vert^k f(t)$ bounded for all $k\geq 1$. When $\vert t\vert<1$, $\vert t\vert^k<1$ and hence $\vert t\vert ^kf(t)< f(t)$. Since $f\in L^1$, you're OK near 0. Then,use my hint to control $\vert t\vert f$ away from 0. Jul 11, 2013 at 5:57
• I got it. Say $|t|^3f(t) < B$ for some B away from 0. Then, $|t|f(t) < B/t^2$ away from 0 which is a $L^1$ function. Thanks a lot. Jul 11, 2013 at 6:11