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

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The discussion focuses on the properties of the Fourier Transform of an L1 function in the context of Lebesgue measure. It establishes that the Fourier Transform, defined as $\hat{f}(t)=\int f(x) e^{ixt} dx$, is continuous and approaches zero as t approaches infinity. Additionally, it asserts that the supremum norm of the Fourier Transform is bounded by the L1 norm of the function. The hints provided suggest using the dominated convergence theorem and the density of continuous functions with compact support in L1 to prove these properties. Overall, the discussion emphasizes the continuity, boundedness, and limiting behavior of the Fourier Transform for L1 functions.
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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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