Proving that a Function is the Fourier Transform of an L1 Function

In this post, we will prove the following fact:

Fact: The function $g:\mathbb{R}\to\mathbb{C}$ defined as $g(\xi)=(1+|\xi|^2)^{-\epsilon}$ for some $\epsilon>0$ is the Fourier Transform of an $L^1(\mathbb R)$ function.

Motivation: The sole reason I wish to prove this fact is because of the elegance of the proof. It uses a slick “dyadic separation” argument that is fairly common in Fourier analysis and this proof is, perhaps, the best example of using such an argument. So, enjoy!

Proof:

Let $\chi$ a smooth function that is $1$ in $[-1,1],$ $0$ outside $(-2,2)$ and $0\le \chi(x)\le 1$ for all $x\in\mathbb{R}.$ Construction of such a function has been shown in $\S$1.46 (Grandpa Rudin).

We define two functions $\psi_0$ and $\psi$ as $\psi_0:x\mapsto\chi(x)$ and $\psi:x\mapsto \chi(x)-\chi(2x).$ Obviously, both $\psi_0$ and $\psi$ are compactly supported smooth functions (since $\chi$ is such a function). Let $K$ be compact set which contains both their support (the union of their supports suffices).

It is easy to see that $\psi_0(x)+\sum_{k=1}^\infty\psi(x/2^k)=1$ for all $x\in\mathbb{R}.$ This is because, $$\psi_0(x)+\sum_{k=1}^N\psi(x/2^k)=[\chi(x)]+[\chi(x/2)-\chi(x)]+\dots+[\chi(x/2^N)-\chi(x/2^{N-1})]=\chi(x/2^N)$$

Hence, by continuity of $\chi,$$$\psi_0(x)+\sum_{k=1}^\infty\psi(x/2^k)=\lim_{N\to\infty}\chi(x/2^N)=\chi\left(\lim_{N\to\infty}x/2^N\right)=\chi(0)=1$$

So we can decompose the given function $g:x\mapsto (1+|x|^2)^{-\epsilon}$ as $g=\sum_{k=0}^\infty g_k$ where $g_0:x\mapsto\psi_0(x)g(x)$ and $g_k:x\mapsto\psi(x/2^k)g(x)$ for $k\in\mathbb{N}.$ Let $h_k:x\mapsto g_k(2^kx)$ for all $k\in\mathbb{N}\cup\{0\}.$

Fix some $k\in\mathbb{N}.$ Then, for any $x\in\mathbb{R}$ such that $|x|\ge 1/2,$ we have

$$|h_k(x)|=\left|\psi(x)\frac{1}{(1+2^{2k}x^2)^{\epsilon}}\right|\le \left|\psi(x)2^{-2k\epsilon}x^{-2\epsilon}\right|$$

Now, $|\psi(x)|\le \sup_{x\in K}|\psi(x)|$ since $\text{supp}(\psi)\subset K$ and since $\psi$ is smooth, $\sup_{x\in K}|\psi(x)|<\infty.$ On the other hand, $|x|^{-2\epsilon}\le 2^{2\epsilon}.$ Hence, $$|h_k(x)|\le 2^{-2k\epsilon}\left[2^{2\epsilon}\sup_{x\in K}|\psi(x)|\right]\text{ if }|x|\ge 1/2$$

For any $x\in\mathbb{R}$ such that $|x|< 1/2,$ note that both $x,2x\in (-1,1)$ which means that $\psi(x)=\chi(x)-\chi(2x)=0.$ Hence, $h_k(x)=0$ in this case. So, setting $C_0=2^{2\epsilon}\sup_{x\in K}|\psi(x)|,$ we see that $$|h_k(x)|\le C_02^{-2k\epsilon}\text{ for all }x\in\mathbb{R}$$

Again, using a similar argument, for any $x\in\mathbb{R}$ such that $|x|\ge 1/2$ we have using the product rule, $$|h_k'(x)|\le \left|\psi'(x)\frac{1}{(1+2^{2k}x^2)^{\epsilon}}\right|+\left|\psi(x)\cdot\epsilon\cdot(1+2^{2k}x^2)^{-\epsilon-1}\cdot2^{2k}\cdot2x\right|$$

Just like before, we have $\left|\psi'(x)\frac{1}{(1+2^{2k}x^2)^{\epsilon}}\right|\le 2^{-2k\epsilon}\left[2^{2\epsilon}\sup_{x\in K}|\psi’(x)|\right]$ and $$\left|\psi(x)\cdot\epsilon\cdot(1+2^{2k}x^2)^{-\epsilon-1}\cdot2^{2k}\cdot2x\right|\le \sup_{x\in K}|\psi(x)|\cdot 2^{-2k\epsilon-2k}\cdot 2^{2k+1}\cdot \epsilon\cdot |x|^{-2\epsilon-1}$$

$$\le \sup_{x\in K}|\psi(x)|\cdot 2^{2k+1}2^{-2k\epsilon-2k}\cdot \epsilon\cdot 2^{2\epsilon+1}=2^{-2k\epsilon}\left[2^{2\epsilon+2}\epsilon\sup_{x\in K}|\psi(x)|\right]$$

Letting, $C_1=2^{2\epsilon}\sup_{x\in K}\left|\psi’(x)\right|+2^{2\epsilon+2}\epsilon\sup_{x\in K}|\psi(x)|$ and noting that $h_k'(x)=0$ since both $\psi(x)=\psi’(x)=0$ if $|x|<1/2$ (note that the derivative is $0$ here since the function is $0$ everywhere in the open set $\{x\in\mathbb{R}:|x|<1/2\}$) we have

$$|h’_k(x)|\le C_12^{-2k\epsilon}\text{ for all }x\in\mathbb{R}$$

Using another iteration of the product rule, we get $|h_k''(x)|\le C_22^{-2k\epsilon}$ for some $C_2\ge 0.$ We have a similar calculations for all three cases if $k=0.$ Hence, we conclude $$\sup_{x\in\mathbb{R}}|h_k^{(n)}(x)|\le C_n2^{-2k\epsilon}\text{ where }C_n<\infty\text{ for all }n\in\{0,1,2\}$$

Since $\text{supp}(h^{(n)}_k)\subset K$ for all $n\in\{0,1,2\}$ and $k\in\mathbb{N}\cup\{0\},$ they must be $L^1$ functions by the regularity of Lebesgue measure: $$\int|h_k^{(n)}(x)|dx=\int_K|h_k^{(n)}(x)|dx\le \sup_{x\in\mathbb{R}}|h_k^{(n)}(x)|m(K)\le C_n2^{-2k\epsilon}m(K)<\infty\text{ ($\star$)}$$

The following lemma follows from Theorem 8.22(e) and Theorem 2.52 (a) (Folland)

Lemma: If $f\in \mathcal{C}^{d+1}(\mathbb{R}^d)$ is such that $D^\alpha f\in \mathcal{C}_0(\mathbb{R}^d)$ for all $|\alpha|\le d$ and $D^\alpha f\in L^1(\mathbb{R}^d)$ for all $|\alpha|\le d+1$ then $\check{f}\in L^1(\mathbb{R}^d)$ with $||\check{f}||_{L^1}\le C’\sum_{|\alpha|\le d+1}||D^\alpha f||_{L_1}$ for some $C’\ge 0.$ Hence, by the inversion theorem, $f$ is the fourier transform of $\check{f}.$

We have already shown $L^1$-ness of the functions $h^{(n)}_k.$ Now, $\mathcal{C}_0$-ness is obvious since the functions have compact support. So, by the lemma, there exists functions $u_k\in L^1(\mathbb{R})$ such that $\hat{u}_k=h_k$ for all $k\in\mathbb{N}\cup\{0\}.$

Set $f_k:x\mapsto 2^ku(2^kx).$ Then, using Theorem 7.2(d) (Grandpa Rudin),

$$\hat{f}_k:\xi\mapsto 2^{k}2^{-k}\hat{u}_k(2^{-k}\xi)=\hat{u}_k(2^{-k}\xi)=h_k(2^{-k}\xi)=g_k(\xi)$$

Hence, $g_k$ is the fourier transform of $L^1$ function $f_k.$ Let $C=C’m(K)(C_0+C_1+C_2).$ Then, by $(\star),$ we have

$$||f_k||_{L_1}=\int 2^k|u_k(2^kx)|dx=\int |u_k(y)|dy=||u_k||_{L_1}\le C’\sum_{n=0}^2||h_k^{(n)}||_{L_1}=C2^{-2k\epsilon}$$

Since $\sum_{k=0}^\infty||f_k||_{L_1}\le C\sum_{k=0}^\infty2^{-2k\epsilon}<\infty$ and $L^1$ is complete, $\sum_{k=0}^\infty f_k$ converges to some $f\in L^1(\mathbb{R})$ by Theorem 5.1 (Folland). We claim that $\hat{f}=g.$

For any $\xi\in\mathbb{R},$ we have $$0\le\left|\int f(x)e^{-2\pi i\xi x}dx-\int \bigg(\sum_{k=0}^Nf_k(x)\bigg)e^{-2\pi i\xi x}dx\right|\le \int\left|f(x)-\bigg(\sum_{k=0}^Nf_k(x)\bigg)\right||e^{-2\pi i\xi x}|dx=\left|\left|f-\sum_{k=0}^Nf_k\right|\right|_{L^1}$$

Note that the right hand side approaches $0$ as $N$ tends to infinity and that $$\int \bigg(\sum_{k=0}^Nf_k(x)\bigg)e^{-2\pi i\xi x}dx=\sum_{k=0}^N\int f_k(x)e^{-2\pi i\xi x}dx$$

Hence, by the Squeeze theorem, $$\hat{f}(\xi)=\sum_{k=0}^\infty \hat{f}_k(\xi)=\sum_{k=0}^\infty g_k(\xi)=g(\xi)\text{ for all }\xi\in\mathbb{R}$$

This shows that $g$ is indeed the Fourier transform of an $L^1$ function.