Understanding the Fourier transform

Question

I am studying the properties of the Fourier transform. Fourier transform: $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{it\xi}dt\:\:,\:\: \xi\in\mathbb{R}$$

Notes: If $f\in L_1(\mathbb{R})\\||f||=\int_\limits{-\infty}^{\infty}|f(t)|dt<\infty$

$\hat{f}(\xi)\in C_B(\mathbb{R})\\||g||_{C_B(\mathbb{R})}=\sup_{t\in\mathbb{R}}|g(t)|$

Then $\sup_{\xi\in\mathbb{R}}|\hat{f}(\xi)|\leqslant\frac{1}{\sqrt{2\pi}}||f||_1$

Observation: I tried to understand this notes on Fourier transform in the following way:

I want to prove $\sup_{\xi\in\mathbb{R}}|\hat{f}(\xi)|\leqslant\frac{1}{\sqrt{2\pi}}||f||_1$ and I started using: $\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{it\xi}dt\leqslant\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}e^{it\xi}dt=\frac{1}{\sqrt{2\pi}}||f||_1\int_\limits{-\infty}^{\infty}e^{it\xi}dt$

However I guess after computation that $\int_\limits{-\infty}^{\infty}e^{it\xi}dt=0$.

Questions:

1) What do you think of what I have done so far? How can I prove the property?

2) Fourier transform here presented is $\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{it\xi}dt$. However in books I find $2\pi$ instead of $\sqrt{2\pi}$. Does this formula intends to compute the coefficients? How should I derive it with functional analysis tools?

Thanks in advance!

Answer 1

That is not valid, for $e^{it\xi}\notin L^{1}$.

Rather, the trick is to use absolute value into the integrals:

\begin{align*} |\widehat{f}(\xi)|&=\left|\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{it\xi}dt\right|\\ &\leq\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|f(t)e^{it\xi}|dt\\ &=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|f(t)|dt\\ &=\dfrac{1}{\sqrt{2\pi}}\|f\|_{L^{1}(-\infty,\infty)}. \end{align*}