MHB Is the Fourier Transform of an L1 Function Always Continuous and Bounded?

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mathmari
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Hey! :o

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:

1. $\hat{f}$ is continuous
2. $\lim_{t\rightarrow \pm \infty}\hat{f}(t)=0$
3. $||\hat{f}||_{\infty}\leq ||\hat{f}||_{1}$

Could you give me some hints how I could do that?? (Wondering)
 
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1. Let $ \in \Bbb R$ and let $t_n$ be a sequence of real numbers converging to $t$. Show that $\hat{f}(t_n) \to \hat{f}(t)$ by the dominated convergence theorem applied to the sequence $f_n(x) = f(x)e^{ixt_n}$.

2. First prove the result when $f$ is a continuous function with compact support. For the general case, use the fact that space of continuous functions on $\Bbb R$ with compact support is dense in $L^1(\Bbb R)$.

3. Show that $|\hat{f}(t)| \le \|f\|_1$ for every $t$.
 
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