Lecture 17

The Fourier-Laplace transform¶

Let $g$ be a timelimted smooth function, say $\text{supp} g\subset [-R,R]$. Then the Fourier transform of $g$ can be extended to a complex valued function $\mathcal Fg(z):=\int_{-\infty}^{+\infty}g(t)e^{-2\pi i z t}dt$. Let $z = x+iy$, then $\mathcal Fg(z) = \int_{-R}^{R}g(t)e^{-2\pi i tx}e^{2\pi yt}dt$. Moreover, $\partial_{\bar z}\mathcal Fg(z) = \int_{-R}^R g(t)\partial_{\bar z}e^{-2\pi i t z}dt = 0$ because $e^{-2\pi i t z}$ is holomorphic. This implies $\mathcal Fg(z)$ is a holomorphic function on the whole complex plane, we call such function an entire function.

Another way to see that $\mathcal F g$ is entire is to use power series. From complex analysis we know that $e^{z} = \sum_{k\geq 0}\frac{z^n}{k!}$ and the series converges unformly on compact sets of $\mathbb{C}$, so $e^{-2\pi i zt} = \sum_{k\geq 0 }\frac{(-2\pi i z t)^k}{k!} = \sum_{k\geq 0}(-2\pi i)^kt^k\frac{1}{k!}z^k$. If $g$ is supported on $[-R,R]$, then $\int_{-R}^{R}g(t)e^{-2\pi i t z}dt =\int_{-R}^R \sum_{k\geq 0}\frac{(-2\pi i z)^k}{k!} t^k g(t)dt \leq \sup|g|\cdot\sum_{k\geq 0} 2R^{k+1}(-2\pi i)^k\frac{z^k}{k!}$. The Cauchy-Hadamard criterion implies that the series on the right hand side has convergence radius ∞ and defines an entire function. To justify the switch of integral and summation, let $s_n = \sum_{k=0}^n\frac{(-2\pi i z)^k}{k!} t^k$, then $|s_n|\leq \sum_{k\geq 0}\frac{(2\pi|t||z|)^k}{k!} = e^{2\pi |z||t|}$ which is continuous and thus integrable on $[-R,R]$ (as a function of $t$) and dominate $|s_n|$, then apply dominated convergence theorem.

We’ll consider the power series expansion in the proof of central limit theorem in later lecture.

Paley-Wiener theorem¶

Theorem¶

Let $U(z)$ be an entire function with growth condition

(eqn:plcondition) for every $k\geq 0$. Then $U$ is the Fourier-Laplace theorem of a smooth function supported on $[-R,R]$.

Proof. The idea to find $u$ is Fourier inversion. Let $U(x)$ be the restriction of $U$ on the real line. Then the growth condition eqn:plcondition implies that $U(x)$ is a Schwartz function. To see this, first the inequality implies $U(x)$ is rapidly decreasing, then by Cauchy estimate, the derivative of a holomorphic function can be estimated by its value. Then $u(t) = \int_{\mathbb{R}_x}U(x)e^{2\pi i x t}dx$ is a Schwartz function. To show that $U$ is the Fourier Laplace transform of $u$, we’ll show that $u$ is supported in $[-R,R]$. Then $\mathcal Fu(z) = U(z)$ because they are entire functions but coincide on the real line.

We claim that $u(t) = \int_{-\infty}^{+\infty}U(x)e^{2\pi i x t}dx = 0$ when $|t|>R$. For $r>0$ and $b\in \mathbb{R}$ (note: we are not assuming $b>0$ even though the picture looks like $b>0$), let $I_1,I_2,I_3,I_4$ be path integrals as shown in the picture, they depend on $r$ though not explicitly written. Since $U(z)$ is entire function, the Cauchy integral formula (or Morera theorem) implies $I_1+I_2+I_3+I_4 = 0$. Note that $\lim_{r\to \infty}I_1 = u(t)$. First, we show that $I_2$ and $I_4$ go to zero when $r\to \infty$, so we get $u(t) = -\lim_{r\to \infty}I_3$.

When $b>0$, $|I_2|\leq \int_{0}^b |U(r+iy)||e^{2\pi i t(r+iy)}|dy \leq \sup_{y\in [0,b]}|U(r+ib)|\int_{0}^b e^{-2\pi t y}dy$. The growth condition on $|U(z)|$ implies $\sup_{y\in [0,b]}|U(r+ib)|\leq C_k \frac{e^{2\pi R|b|}}{(1+\sqrt{r^2+b^2})^k}$. So $|I_2|\leq const\cdot \frac{e^{2\pi R|b|}}{(1+\sqrt{r^2+b^2})^k}\to 0$ when $r\to \infty$. The same result holds when $b<0$ and for $I_4$ by the same argument.

We have showed that for every $b\in \mathbb{R}$, $u(t) = -\lim_{r\to \infty}I_3 = \int_{-\infty}^{\infty}U(x+ib)e^{2\pi t(x+ib)}dx$, so $|u(t)|\leq C_2 e^{2\pi Rb - 2\pi t b} \int_{-\infty}^{+\infty} \frac{1}{(1+|x|)^2}dx$ for every $b$. When $b>0$, $2\pi Rb - 2\pi t b<0$ when $R-t<0$, i.e. $t>R$, this implies $u(t)=0$ because the non-negative number which is less than any negative exponential is 0. Similarly, when $b<0$, we $2\pi Rb - 2\pi tb<0$ when $t<-R$, which implies $u(t) = 0$ when $t<-R$. The proof is complete.

Distribution with compact support¶

The Paley-Wiener theorem can be generalized to distirbutions. But

Let $T$ be a distribution. We say $T$ is zero or zero mass at $x$ if for some open neighbourhood $U$ of $x$, $(T,\varphi) = 0$ for every $\varphi\in \mathscr{C}_c^{\infty}(U)$. As the case of functions, the support of $T$ is the set where $T$ is not zero, which is automatically closed. The set $\mathscr{E}'$ denote the distributions with compact support.

One important classes of distribution with compact support are compact supported measures.

Topology of $\mathscr{E}'$¶

Recall that a distribution is a coninuous linear functional on $\mathscr{C}_c^{\infty}(\mathbb R)$. A natural question is the dual of $\mathscr{C}^{\infty}(\mathbb{R})$. For a distribution $T$, $(T,f)$ may not be defined when $f\in \mathscr{C}^\infty(\mathbb{R})$ even when $T$ is a continuous function. Not suprisingly this can be done when $T$ has compact support.

Let $T$ be a distribution supported on $K$, where $K$ is a compact set. Let $\psi \in \mathscr{C}_c^{\infty}(R)$ which is $\equiv 1$ on $K$. For $f\in \mathscr{C}^{\infty}(\mathbb{R})$, we write $f = \psi f + (1-\psi)f$. The $(T,\psi f)$ is defined because $\psi f$ has compact support, the other part $(1-\psi f)$, though may not have compact support, is gurenteed to support outside the support of $T$ so it’s meaningful to assign zero. So we define $(T,f):=(T,\psi f)$. To show that it is well-defined, let $\psi'\in \mathscr{C}_c^{\infty}(\mathbb{R})$ satisfying also $\psi' \equiv 1$ on $K$. Then $(T,\psi f) - (T,\psi' f) = (T,(\psi - \psi')f) = 0$ because $\psi - \psi'$ is supported outside $K$. Moreover we see that $|(T,f)| = |(T,\psi f)|\leq C_K \sum_{0\leq l\leq k}\sup_{x\in K}|(f\psi)^{(l)}(x)|\leq C_K' \sum_{l\leq k}\sup_{x\in K}|f^{(l)}(x)|$ for every $f\in \mathscr{C}^{\infty}(\mathbb{R})$.

Note that the topology of $\mathscr{C}^{\infty}(\mathbb{R})$ is determined by the semi-norms $\|\cdot\|_{l,K}$, and $f_n\to f$ if and only if for every compact $K$ and every $l$, $\|f_n - f\|_{l,K}\to 0$. Then for $T\in \mathscr{E}'$, $|(T,f_n)-(T,f)|\leq C_{K_{T}}\cdot \sum_{l\leq k}\|f_n - f\|_{l,K_T}\to 0$. If we denote by $\mathscr{E}$ the space $\mathscr{C}^{\infty}(\mathbb{R})$ together with its topology, then $\mathscr{E}'$ are continuous linear functionals on $\mathscr{E}$, which justifies the notation.

Action of distribution depending on parameters¶

To verify the Cauchy-Riemann Fourier-Laplace transform of distribution we need to “differenciate under action of distributions”. These nice properties are consequence of continunity of $T$, as we’ll see.

(=lem:diff)

Lemma¶

Let $\phi(x,y)\in \mathscr{C}^{\infty}(X\times Y)$, where $X,Y$ are open sets on $\mathbb{R}$. Let $T\in \mathscr{E}'$. Then

(1) $(T(y),\phi(x,y))$ defines a smooth function on $X$.

(2) $\partial^{(k)}_x(T(y),\phi(x,y)) = (T(y),\partial^{k}_x(\phi(x,y)))$.

Proof.

Let $g(x) = (T(y),\phi(x,y))$. First we show that $g(x)$ is continuous. Note that if $\phi(x,y)\in \mathscr{C}^{\infty}(X,Y)$, then $\phi(x+h,y)\to \phi(x,y)$ in $\mathscr{C}^{\infty}(Y)$ when $h\to 0$. Since $T$ is compact support, $T$ is continuous on $\mathscr{C}^\infty(Y)$, this implies $g(x+h) = (T(y),\phi(x+h,y))\to (T(y),\phi(x,y)) = g(x)$ when $h\to 0$.

Then we consider derivatives using this observation. Taylor expansion of $\phi(x,y)$ near $x$ implies $\phi(x+h,y) = \phi(x,y)+h\partial_x\phi(x,y) + \frac{h^2}{2}\partial_{xx}\phi(x+\theta h,y)$ for $|\theta|<1$. Then $g(x+h) = g(x)+h (T,\partial_x\phi)+\frac{h^2}{2}(T,\partial_{xx}\phi(x+\theta h,y))$.

$|\frac{g(x+h)-g(x)}{h}-(T,\partial_x\phi)| = \frac{h}{2}|(T(y),\partial_{xx}\phi(x+\theta h),y)|\leq \frac{h}{2}(|T(y),\partial_{xx}\phi(x,y)|+\epsilon)\to 0$ when $h\to 0$. The ε comes continunity of $(T,\partial_{xx}(x,y)$ when $h$ is small.

Iterating the above argument shows that $g$ is smooth and its derivative is given by $g^{(k)}(x) = (T(y),\partial_x^{(k)}\phi(x,y))$.

Obtaining smooth function from distribution¶

In problem set 7, we proved the following result

The second bullet is a consequence of $\mathscr{C}^\infty$ convergence of Riemann sum, still by a continunity argument. To verify this, we must show for every $g\in \mathscr{C}_c^{\infty}(\mathbb{R})$, $(\phi*T,g) = (T,\phi^-*g) = \int (T*\phi)(x)g(x)dx$. Let $R_h = \sum_{k\in \mathbb{Z}}\phi^-(x-kh) g(kh)h$ be the Riemann sum of $\phi^-*g$. $(T,R_h) = \sum_{k\in \mathbb{Z}}h g(kh)(T,\phi^-(x - kh)) = \sum_{k\in \mathbb{Z}}hg(kh)(T*\phi)(kh)$. Since $R_h\to \phi^-*g$ in $\mathscr{C}_c^\infty(\mathbb{R})$ and $T$ is a distribution, the LHS $\to$ $(T,\phi^-*g) = (\phi*T,g)$ when $h\to 0$. But observe that the RHS, as a Riemann sum of $\int (T*\phi)(x)g(x)dx$, goes to $\int (T*\phi)(x)g(x)dx$. This finishes the proof. Note that actually the summation $\sum_{k\in \mathbb{Z}}$ is actually a finite sum because $\phi,\psi,g$ are compactly supported.

Fourier-Laplace transform of distributions with compact support¶

Alternative definition of Fourier transform for compactly supported distributions¶

Recall that when we defined the Fourier transform for distributions, we tried a function $(\mathcal F T_f,g) = \int_y \int_x f(x)e^{-2\pi i xy}dx dy$ and applied Fubini theorem to conclude that $(\mathcal FT_f,g) = (T,\mathcal Fg)$ and defined the Fourier transform to be $(\mathcal FT,g) = (T,\mathcal Fg)$. But what if we do not use Fubini theorem, and then $(\mathcal FT_f,g) = ((T_f(x),e^{-2\pi i x y}),g(y))$, then try to define $\mathcal FT$ be the function $y\mapsto (T(x),e^{-2\pi i xy})$? Unfortunately this would not be successful for tempered distribution $T$ because $e^{-2\pi i xy}$ is not compacted supported and this function can be infinity. But if $T$ has compact support, then $T$ acts on all smooth functions and this approach works! Moreover, by the continunity argument, the function $(T(x),e^{-2\pi i x y})$ is a smooth function on $y$.

Definition. Let $T$ be a distribution with compact support. The Fourier transform of $T$ is a smooth function given by $\mathcal FT(x) = (T(y),e^{-2\pi i xy})$.

Of course, we need to show the new definition of Fourier transform as a smooth function concide (in the sense of distribution) as the original definition. By the identity principle of distributions, it sufficies to check that their actions on $\mathscr{C}_c^{\infty}$ are the same, i.e. $\int_{y}(T(y),e^{-2\pi i xy})g(x)dx = (T,\mathcal Fg)$. This follows from the same argument of proving $T*\phi = \phi*T$ by writing $\mathcal Fg = \int g(y)e^{-2\pi i x y}$ as a Riemann sum $\sum_{k\in \mathbb{Z}}g(kh)e^{-2\pi i xkh}h$.

The Fourier-Laplace transform¶

Now we’re ready to define the Fourier-Laplace transform of compactly supported distributions.

By the “differentiation under integral sign” lemma, $\partial_{\bar{z}}\mathcal FT(z) = (T(t),\partial_{\bar{z}}e^{-2\pi i t z}) = 0$, and $\mathcal FT(z) = 0$.

Example¶

The Fourier-Laplace transform of $\delta_a$ is $\mathcal F\delta_a(z) = (\delta_a(t),e^{-2\pi i zt}) = e^{-2\pi i a z}$. In particular, $\mathcal F(\frac{\delta_a + \delta_{-a}}{2}) = \cos(2\pi a z)$, $\mathcal F(-\delta_a + \delta_{-a}) = 2i\sin(2\pi a z)$.

Let $\chi_{[-a,a]}$ be the step function. Then $\mathcal F\chi_{[-a,a]}(z) = (\chi_{-a,a}(t),e^{-2\pi i t z}) = \int_{-a}^a e^{-2\pi i t z}dt = \frac{1}{-2\pi i z}(e^{-2\pi i az}-e^{2\pi i az}) =\frac{-2i \sin(2\pi a z)}{-2\pi i z} = \frac{\sin(2\pi a z)}{\pi z}$. This is an entire function with power series $\frac{1}{z}\cdot (z -\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots) = \sum_{n = 0}^{+\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}$.