Fourier Transform

Moderate Decrease

Let $f$ be a function on $\mathbb{R}$, $f$ is said to be moderate decrease if $f$ is continuous and there exist a constant $A>0$, so that $|f(x)|\le \frac{A}{1+x^2}$ for all $x\in \mathbb{R}.$ We denote the set of moderate decrease function on $\mathbb{R}$ as $\mathcal{M}(\mathbb{R})$.

Lemma

For $f\in \mathcal{M}(\mathbb{R})$, the limit exists so we can define ${\int }_{-\infty }^{\infty }f(x)\text{d}x={\lim }_{N\to \infty }{\int }_{-N}^{N}f(x)\text{d}x.$

Proof The limit exist because $I_N={\int }_{-N}^{N}f$ is a Cauchy sequence, let $M>N$, $|I_N-I_M|=|{\int }_{N\le |x|\le M}f(x)dx|\le {\int }_{N\le |x|\le M}\frac{A}{x^2}\le \frac{2A}{N}\to 0$

Remark

In fact we can replace the exponent of degree $2$ in definition of moderate decrease by $1+ϵ$, with $ϵ>0$, that is $|f(x)|\le \frac{A}{1+|x{|}^{1+ϵ}}.$ We choose $ϵ$ as 1 simply by convention.

Proposition

The integral of function of moderate decrease satisfies following property (i) Linearity: if $f,g\in \mathcal{M}(\mathbb{R})$and$a,b\in \mathbb{C}$, then ${\int }_{-\infty }^{\infty }(af+bg)\text{d}x=a{\int }_{-\infty }^{\infty }f\text{d}x+b{\int }_{-\infty }^{\infty }g\text{d}x$ (ii) Translation invariant: for every $h\in \mathbb{R}$, we have ${\int }_{-\infty }^{\infty }f(x-h)\text{d}x={\int }_{-\infty }^{\infty }f(x)\text{d}x$ (iii) Scaling under dilations: for $\delta >0$, $\delta {\int }_{-\infty }^{\infty }f(\delta x)\text{d}x={\int }_{-\infty }^{\infty }f(x)\text{d}x$ (iv) Continuity: if $f\in \mathcal{M}(\mathbb{R})$, then ${\int }_{-\infty }^{\infty }|f(x-h)-f(x)|\text{d}x\to 0\text{as}h\to 0.$

Proof (i) is immediate, (ii) is equivalent to prove ${\lim }_{N\to \infty }{\int }_{-N}^{N}f(x-h)dx-{\int }_{-N}^{N}f(x)dx=0,$ since ${\int }_{-N}^{N}f(x-h)dx={\int }_{-N-h}^{N-h}f(x)dx$, $|{\int }_{-N-h}^{N-h}f(x)dx-{\int }_{-N}^{N}f(x)dx|=|{\int }_{-N-h}^{-N}f(x)dx-{\int }_{N-h}^{N}f(x)dx|$ $\le |{\int }_{-N-h}^{-N}f(x)dx|+|{\int }_{N-h}^{N}f(x)dx|\le \frac{Ah}{1+N^2}dx+\frac{Ah}{1+(N-h{)}^2}dx\le \frac{A^{\prime }}{1+N^2},$ for large enough $N$ and some A’ (iii) can be proved in similar way of (ii). For (iv), through triangle inequality, we get ${\int }_{-\infty }^{\infty }|f(x-h)-f(x)|dx\le {\int }_{-N}^N|f(x-h)-f(x)|dx+{\int }_{|x|\ge N}|f(x-h)|dx+{\int }_{|x|\ge N}|f(x)|dx,$ $f(x-h),f(x)\in L^1(\mathbb{R})$, then for fixed $ϵ>0$, we can find large enough $N$ that ${\int }_{|x|\ge N}|f(x-h)|dx<\frac{ϵ}{4},{\int }_{|x|\ge N}|f(x)|dx<\frac{ϵ}{4},$ for fixed $N$, since $f$ continuous, it’s uniformly continuous in the interval $[-N-1,N+1]$, so ${\sup }_{|x|\le N}|f(x-h)-f(x)|\to 0$ as $h$ tends to $0$, so we can take $h$ that supremum is less than $\frac{ϵ}{4N}$, so ${\int }_{-N}^N|f(x-h)-f(x)|dx<\frac{ϵ}{2},$ combining all these, we get ${\int }_{-\infty }^{\infty }|f(x-h)-f(x)|dx<ϵ.$

Fourier Transform

If $f\in \mathcal{M}(\mathbb{R})$, we define its Fourier Transform for $\xi \in \mathbb{R}$, by $\hat{f}(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }\text{d}x.$

Remark

However, nothing in the definition above guarantees that $\hat{f}$ is of moderate decrease, or has a specific decay. In particular, it is not clear in this context how to make sense of the integral ${\int }_{-\infty }^{\infty }\hat{f}{e}^{2\pi ix\xi }\text{d}x$ and the resulting Fourier inversion formula. To remedy this, we introduce a more refined space of functions---- Schwarz Space.