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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!

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  • $\begingroup$ Any comments will be welcomed $\endgroup$ – Pedro Gomes Apr 18 '18 at 18:02
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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*}

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  • $\begingroup$ The constant depends on the authors. $\endgroup$ – user284331 Apr 18 '18 at 18:06
  • $\begingroup$ Thanks! What do you mean by the constant? Any thoughts on my second question? $\endgroup$ – Pedro Gomes Apr 18 '18 at 18:07
  • $\begingroup$ The $1/\sqrt{2\pi}$. Yes, that formula is to derive the coefficients of Fourier series in a compact way. $\endgroup$ – user284331 Apr 18 '18 at 18:09
  • $\begingroup$ When I tried to derive the Fourier coefficient I used a projection over $e^{it\xi}$ for the space of periodic functions in the interval $[-\pi,\pi]$. Unfortunately I am not understanding on how the coefficients differ from the transform. Is the transform the coefficient? Thanks! $\endgroup$ – Pedro Gomes Apr 18 '18 at 18:12
  • $\begingroup$ Yes, you are right, I should say it precisely, for the Fourier series coefficients, the interval is $[-\pi,\pi]$, not $(-\infty,\infty)$, and this version is to use for the theory of singular integrals. $\endgroup$ – user284331 Apr 18 '18 at 18:14

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