Lecture 10

The Fourier transform on $\mathcal S$¶

In last lecture we defined the Schwartz class of functions. First, we’ll show that the Fourier transform is well behaved in the Schwarz functions

• $\mathcal S$ is closed under Fourier transform.
• All the properties of Fourier transform works without further assumptions.
• The Fourier transform is invertible.
• The Poisson summation formula.
• The Fourier transform is isometry with $L^2$ norm on $S$.

Secondly, the Schwartz functions serves as “test objects” to define a class of distributions that behave well under Fourier transforms, called “tempered distributions”.

By definition, $f\in \mathcal S$ if and only if $\|f\|_{(k,l)}:=\sup_{x\in \mathbb{R}} |x^k f^{(l)}(x)|<\infty$ for every $k,l \geq 0$. We also let $\|f\|_N:=\max_{k+l = N}\|f\|_{(k,l)}$. The $\| \cdot \|_{(k,l)}$ as well as $\|\cdot\|_N$ are semi-norms on $\mathcal S$.

Lemma. If $f\in L^1(\mathbb{R})$, then $\mathcal Ff$ is bounded.

Proof. $|\mathcal Ff(s)| = |\int_{-\infty}^{+\infty}f(x)e^{-2\pi i x s}dx|\leq \int_{-\infty}^{+\infty}|f(x)|dx = \|f\|_{L^1}$.

Lemma. If $f\in \mathcal S$, then $x^kf^{(l)}(x)$ is $L^1$ for every $(k,l)$.

Proof of above lemma. We show integrability by showing the function is moderate decreasing. $x^kf^{(l)}(x) = \frac{x^k(1+x)^2f^{(l)}(x)} {1+x^2} = \sup|x^k(1+x)^2f^{(l)}(x)|\cdot \frac{1}{1+x^2}$.

Proof of (1). We need to estimate the $(k,l)$-seminorms of $\mathcal Ff(s)$. Using the derivative theorem and multiplication theorem, all functions of the form $s^k\frac{d^l}{ds^l}\mathcal Ff(s)$ is some multiple of Fourier transform of $x^l\frac{d^k}{dx^k}f$. But $x^l\frac{d^k}{dx^k}f$ is $L^1$ because $f\in \mathcal S$, then the previous lemma implies boundedness of Fourier transform. $\square$

Proof of (2). We’ll give two proofs.

First proof. See this discussion.

Second proof. Since $f,g\in \mathcal S$, $f*g$ is moderate decreasing. Indeed, $f*g(x) = \int_{\mathbb{R}}f(x-y)g(y)dy = \int_{|y|\leq \frac{|x|}{2}}f(x-y)g(y)dy+\int_{|y|\geq \frac{|x|}{2}}f(x-y)g(y)dy$. The first term is bounded by $\|g\|_{L^1} \frac{1}{1+(\frac{|x|}{2})^2}$ and second term bounded by $\|f\|_{L^1}\frac{1}{1+(\frac{x}{2})^2}$. By the convolution theorem, $\mathcal F(f*g) = \mathcal Ff \cdot \mathcal Fg$, so $\mathcal F(f*g)\in S$. Then by Fourier inversion, $f*g = \mathcal F^{-1} \mathcal F(f*g) = \mathcal F^{-1}$ of Schwartz function, which is Schwartz. $\square$

Periodization of Schwartz functions¶

The informal derivation of the Fourier transform of rect function in lecture 8 applies to Schwartz functions (even though the Schwartz functions are not compact supported). We make the informal discussion precise as follows.

Periodization Lemma¶

If $f$ is moderate decreasing on $\mathbb{R}$ with $|f(x)|\leq \frac{C}{1+x^2}$, then

(1) $f_T(x) = \sum_{k=-\infty}^{+\infty}f(x - kT)$ converges uniformly.

(2) Let $\widehat{f}_T(n)$ be the Fourier coefficients of $f_T$. Then $\hat{f}_T(n) = \frac{1}{T}\mathcal Ff(\frac{n}{T})$.

(3) $|f_T(x) - f(x)|\leq \frac{C\pi^2}{T^2}$ when $x\in [-\frac{T}{2},\frac{T}{2}]$, this $C$ is the same as the $C$ in moderate decreasing condition of $f$. In particular, this implies $|f_T(x) - f(x)|\to 0$ uniformly on every bounded subset of $\mathbb{R}$.

Proof.

(1) Note that $f_T$ is periodic function, so we only need to consider $x$ in its period, say, $x\in [\frac{T}{2},\frac{T}{2}]$. $|f(x - kT)|\leq \frac{C}{1+(x-kT)^2} = \frac{C}{1+T^2(\frac{x}{T}-k)^2}\leq \frac{C}{1+T^2(k-\frac{1}{2})^2}$. Note that $\sum_{k=-\infty}^{+\infty}\frac{C}{1+T^2(k - \frac{1}{2})^2}$ converges, so $f_T$ converges uniformly by Wererstrass $M$-test.

(2)

(3) $|f_T(x) - f(x)| = |\sum_{k\neq 0,k\in \mathbb{Z}}f(x - kT)| \leq \sum_{k \in \mathbb{Z}, k\neq 0}\frac{C}{1+(x - kT)^2}\leq \frac{C}{T^2}\sum_{k \neq 0,k\in \mathbb{Z}} \frac{1}{\frac{1}{T^2}+(\frac{x}{T}-k)^2} \leq \frac{C}{T^2}\cdot 2\sum_{k\geq 1}\frac{1}{(k - \frac{1}{2})^2}$. To verify the the last inequality, observe that $-\frac{T}{2}\leq x\leq \frac{T}{2}\implies -\frac{1}{2}\leq \frac{x}{T}\leq \frac{1}{2} \implies (\frac{x}{T}-k)^2 \geq (k - \frac{1}{2})^2$ for every $k\neq 0$.

Now we summerize three important consequences of the periodization lemma.

The Poisson summation formula¶

When $f$ is in $\mathcal S$, the derivatives of $f$ are moderate decreasing so $f_T$ is smooth. Then the Fourier series of $f_T$ converges uniformly on $[-\frac{T}{2},\frac{T}{2}]$, then we obtain

The Inversion theorem¶

We know that with Fourier coefficients we can obtain the function by taking Fourier series.

$f_T(x) = \sum_{n = -\infty}^{+\infty}\hat{f_T}(n)e^{\frac{2\pi}{T}i n x} = \sum_{n = -\infty}^{+\infty}\frac{1}{T}\mathcal Ff(\frac{n}{T})e^{2\pi i\frac{n}{T}x}$.

For every $x\in \mathbb{R}$, when $T$ is sufficiently large, $x$ will be in $[-\frac{T}{2},\frac{T}{2}]$. Then (3) of periodization lemma implies $f_T(x)\to f(x)$. For $\epsilon > 0$, choose $T,N$ such that $|f_T(x) - f(x)|<\epsilon$ and $|\sum_{n = -N}^N \frac{1}{T}\mathcal Ff(\frac{n}{T})e^{2\pi i \frac{n}{T}x} - \int_{-\frac{N}{T}}^{+\frac{N}{T}}\mathcal Ff(s)e^{2\pi i x s}ds|<\epsilon$. If $\mathcal Ff$ is $L^1$ (this is ensured when $f\in \mathcal S$), then one can further enlarge $N$ to make $|\int_{-\infty}^{+\infty}Ff(s)e^{2\pi i s x}ds - \int_{-\frac{N}{T}}^{\frac{N}{T}}\mathcal Ff(s)e^{2\pi i s x}ds|<\epsilon$. It follows that by letting $T\to \infty$, we get

The Plancherel identity¶

Apply the Parsval identity to $f_T$ we get $\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}|f_T(x)|^2dx = \sum_{n = -\infty}^{+\infty}|\widehat{f_T}(n)|^2$. By the periodization lemma we get $\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}|f_T(x)|^2dx = \sum_{n = -\infty}^{+\infty}|\frac{1}{T}\mathcal Ff(\frac{n}{T})|^2 \implies \int_{-\frac{T}{2}}^{\frac{T}{2}}|f_T(x)|^2 dx = \sum_{n = -\infty}^{+\infty}|\mathcal Ff(\frac{T}{n})|^2 \cdot \frac{1}{T}$. The RHS, as a Riemann sum of , goes to $\int_{-\infty}^{+\infty}|\mathcal Ff(s)|^2ds$ when $T\to \infty$; the LHS goes to $\int_{-\infty}^{+\infty}|f(x)|^2 dx$ by uniform convergence and previous argument. We have derived

More generally, the Plancherel theorem is true for all $L^2$ functions $f$ on $\mathbb{R}$ such that $\mathcal Ff$ is $L^2$. This is because $\mathcal S$ is dense in $L^2$.

Appendix: The inverse Fourier transform of $\mathrm{sinc}$¶

Proof. (1) change of variable $y = \frac{x}{a}$, note that here one needs to pay attention to the sign of $a$.

(2) Integration by parts twice.