Lecture 11

Duality formula¶

The Fourier inverse transform is given by $\mathcal F^{-1} f(x) = \int_{-\infty}^{+\infty}f(y)e^{2\pi i yx}dy$, so $\mathcal F^{-1}f(-x) = \int_{-\infty}^{+\infty}f(y)e^{-2\pi i y x}dy = \mathcal Ff(x)$. If we define the operator ^- by $f\mapsto f^-$, where $\mathcal f^-$ is determined by $f^-(x) = f(-x)$, the we obtain $(\mathcal F^{-1}f)^- = \mathcal Ff$.

Distribution theory¶

In previous lectures we introduced the Fourier transform and its basic properties. In particular, we showed that there exists a particular class $\mathcal S$ of smoothly rapid decreasing functions, or Schwartz functions, on which the Fourier transform is nicely behaved, in particular, the Fourier transform is bijective (inversion theorem) and perserves $L^2$ norms (Pluncherel identity). Of course, it is far from enough to only restrict Fourier transform to Schwartz functions since most examples in real application are not Schwartz functions. How about $L^1$ functions? The Riemann-Lebesgue lemma (see problem set 5) and our previous discussion on Fourier series shows that the regularity of $f$ corresponds to decay of its Fourier transform. “Taking derivatives $\frac{d}{dx}$” and “multiplication by $x$” are Fourier transform of each other, so they should be in some sense equivalent. However, it’s quite annoying that it is not always possible to take derivatives if the function is not smooth, but it is always OK to multiply by $x$. So in order to have a complete theory (closed under Fourier transforms and inversion works), the functions has to be infinitely regular and infinitly fast decreasing, then we go back to Schwartz functions. In order to develop a sufficiently general functions, one has to consider objects to which “taking derivatives” is as easy as “multiplication by $x$”. Of course, functions does not fit into our scope. Our next plan is to generalize theory of Fourier transforms to more general objects, called tempered distributions, and show that everything still works. More importantly, we can make precise the Fourier transform of δ measure, which plays an essential rule in electronic enginnering.

Test functions¶

Definition¶

A test function on $\mathbb{R}$ is a smooth function with compact support.

Lemma. There exists a non-negative smooth function with compact support on $\mathbb{R}$.

Proof. $\phi(x) = \begin{cases} e^{-(\frac{1}{1-|x|^2})}, -1<x<1 \\ 0, |x|\geq 1\end{cases}$.

You are going to check this in problem set 5.

We have the following fundamental lemma:

Lemma. If $f,g$ are continuous functions on $\mathbb{R}$ such that $\int f\varphi = \int g\varphi$ for every $\varphi \in \mathscr{C}_c^{\infty}(\mathbb{R})$, then $f = g$.

Proof. Suppose for some $x_0$, $f(x_0)\neq g(x_0)$. WLOG assume $f(x_0) - g(x_0) > 0$. Since $f,g$ are continuous, there is neighbourhood of $x_0$, say $(x_0 - \delta,x_0 + \delta)$ such that $f(x) - g(x)\geq \epsilon$ in the neighbourhood. Let $\varphi(x) = j_\delta(x-x_0)$, then $\int_{x_0 - \delta}^{x_0 + \delta} (f - g)\varphi \geq \epsilon$, contradicts to the assumption that $f-g$ annihilates all test functions.

Topology on $\mathscr{C}_c^{\infty}(\mathbb{R})$¶

On $\mathscr{C}^{\infty}(\mathbb{R})$ there is a collection of semi-norms given by $\|f\|_{k,K}:=\sup_{x\in K}|\frac{d^k}{dx^k}f(x)|$. They are semi norms because $\|f\|_{k,K} = 0$ may not imply $f\equiv 0$.

Let $\mathscr{C}_K^{\infty}(\mathbb{R})$ consists of $f\in \mathscr{C}_c^{\infty}(\mathbb{R})$ with $supp(f)$ actually $\subset K$. A sequence $(f_n)_{n = 1}^{\infty}$ converges to $f$ in $\mathscr{C}_K^{\infty}(\mathbb{R})$ if and only if $\|f_n - f\|_{k,K}\to 0$ for every $k$.

A function $f$ is in $\mathscr{C}_c^{\infty}(\mathbb{R})$ if and only if there exists some $K$ such that $f\in \mathscr{C}_K^{\infty}(\mathbb{R})$. A sequence $(f_n)_{n = 1}^{\infty}$ converges to $f$ in $\mathscr{C}_c^{\infty}(\mathbb{R})$ if and only if the whole sequence is in $\mathscr{C}_K^{\infty}(\mathbb{R})$ for some $K$ and $f_n \to f$ in $\mathscr{C}_K^{\infty}(\mathbb{R})$.

Distributions¶

Motivation & Example¶

Let $D = \sum_{k = 0}^N f_k \frac{d}{dx^k}$ be a differential operator of order $N$ (we assume $f_k$ to be $L^1$). For test function φ, let $T(\varphi) := \int \sum_{k = 0}^N f_k(x) \frac{d^k}{dx^k}\varphi(x) dx$. Then $T$ is a continuous linear functional on $\mathcal D$ with $|T(\varphi)|\leq \max_{0\leq k\leq N}\{\|f_k\|_{L^1}\} \cdot \sum_{0\leq k\leq N}\|\varphi\|_{k,K}$.

Definition¶

A distribution is a linear functional $T:\mathcal D \to \mathbb{C}$ such that for every compact set $K\subset \mathbb{R}$, there exists a constant $C> 0$ and $N\in \mathbb{Z}_+$ ($C,N$ depend on $K$) such that for every $f\in \mathcal{D}$ with $supp(f)\subset K$,

If there exists an $N$ which holds for every $K$, we call the distribution $T$ of order $N$. For example, a differential operator of order $N$ is a distribution of order $N$.

An alternative definition of distribution is a linear functional that is continuous on $\mathcal D$, note that here to be continuous means that for every sequence $(f_n)_{n = 1}^{\infty}$ such that $f_n\to f$ in $\mathcal D$, $T(f_n)\to T(f)$.

Proof of equivalence of definitions¶

We call the first definition $A$ and second definition $B$. Let $T$ be a linear functional on $\mathcal D$.

$A\implies B:$ Suppose $T$ satisfies definition $A$. Let $K$ be the uniform support of $(f_n)$, and $C,k$ be the constants coming from $K$ and definition $A$. Then $|T(f_n) - T(f)| =|T(f_n - f)|\leq C\sum_{0\leq k\leq N}\|f_n - f\|_{k,K}$. Since $\|f_n - f\|_{k,K}\to 0$ for every $k$, in particular this implies $\sum_{0\leq k\leq N}\|f_n - f\|_{k,K}\to 0$.

$B\implies A:$ Prove by contradiction. Suppose $T$ satisfies definition $B$ but not definition $A$. Then there exists a compact $K$ such that every $C,N$ there exists $f$ with $|T(f)|> C\sum_{0\leq k\leq N}\|f\|_{k,K}$. Take $C = N = j$ for integers $j>0$, we obtain a sequence $(f_j)_{j = 1}^\infty$ such that $|T(f_j)|>j\sum_{0\leq k \leq j}\|f_j\|_{k,K}$. By changing $f_j$ to $\frac{f_j}{|T(f_j)|}$ if necessary, we can assume $T(f_j)= 1$ for every $j$. Then $\|f_j\|_{k,K}< \frac{1}{j}$ for every $j$ and every $k\leq j$, this implies $f_j\to 0$ in $\mathcal D$. Since $T$ satisfies definition $B$, $T(f_j) \to T(0) = 0$. But with our choice $T(f_j) = 1$ for every $j$, this is a contradiction.