In $\mathbb{R}$ with Lebesgue measure, we take $f\in L^1$ and we set $\hat{f}(t)=\int f(x) e^{ixt} dx$, for each $x$ $\ \ \ (i^2=-1)$
Show that:
- $\hat{f}$ is continuous
- $\lim_{t\rightarrow \pm \infty}\hat{f}(t)=0$
$||\hat{f}||_{\infty}\leq ||\hat{f}||_{1}$
Since $f\in L^1$ we have that $f$ is measurable and $\int |f| d\mu <+\infty$ which means that $f$ is integrable. From that it follows that $f$ is continuous. The exponential is also continuous, so the integral is continuous. That means that $\hat{f}$ is continuous. Is this correct??
Do we change the order of the limit and the integral??
$||\hat{f}||_{\infty}$ is the supremum, right?? $||\hat{f}||_1=\int|\hat{f}| d\mu$ but how can show the inequality??