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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.

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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 $\vert t\vert>1$.

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  • $\begingroup$ I see what you mean. But we only know then |t|f(t) is bounded on |t|>1 which is not enough to show that it is in L^1(R). $\endgroup$ Jul 11, 2013 at 5:49
  • $\begingroup$ 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. $\endgroup$
    – icurays1
    Jul 11, 2013 at 5:57
  • $\begingroup$ 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. $\endgroup$ Jul 11, 2013 at 6:11
  • $\begingroup$ Yep, you got it. $\endgroup$
    – icurays1
    Jul 11, 2013 at 6:15

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