The upper plane #

Denote $\bH$ as the open upper half-plane in $\C \simeq \R^2$. $\dH$ denotes $\R$ in $\C$.

$L^2(\dH \to \C, \lambda)$ is a Hilbert space.

Fourier transform #

For $f \in L^1(\R \to \C)$, the Fourier transform is $\what f: \R \to \C$. $$ \what f(\omega) = \F[f] (\omega) := \int _ {\R} f(x) \exp(-i \omega x)\dd x = \pd{f(x)}{e^{i\omega x}} $$

• We have $\what f(\omega) \to 0$ as $\omega \to \pm \infty$ for $f \in \L^1(\R \to \C)$.
  • The proof is similar to the proof for Fourier series, using a density argument.
• And $\what f$ is bounded. $\n{\what f} _ \infty \le \n{f} _ 1$. $$ |f(\omega)| \le \int _ {\R} |f(x)| \dd x = \n{f} _ 1 $$

The Fourier transform $\f: L^1 \to L^\infty$ is a continuous linear map.

$\what f$ is uniformly continuous for $f \in L^1$ #

For $f \in L^1(\R \to \C)$, $\what f(\omega)$ is uniformly continuous on $\R$.

• Consider indicator $f(x) = 1 _ {[b, c]}(x)$. $\what f$ is uniformly continuous. $$ \what f(\omega) = \int _ {b}^c e^{-i\omega x} = \begin{cases}i(e^{-2\pi c \omega} - e^{-2 \pi b \omega})/\omega & \text { if } \omega \neq 0 \\ c-b & \text { if } \omega=0\end{cases} $$
• Therefore $\what f$ is uniformly continuous for $f \in \square^1(\R \to \C, \lambda)$.
• There exists a sequence of $\p{f _ n} \subseteq {\square}^1(\R \to \C, \lambda)$. Where $f _ n \rightrightarrows _ {\R}f$.
• Since $\n{\what f _ n - \what f} _ {\infty} \le \n{f _ n - f} _ 1$, $\what f _ n \rightrightarrows _ {\R} \what f$.
• Since uniform limit of uniformly continuous functions is uniformly continuous. $\what f(\omega)$ is uniformly continuous.

Derivative of Fourier transform #

Suppose $f \in L^1(\R \to \C)$. If $g(x) = (-ix) f(x) \in L^1$.

Then $\what f(\omega) \in C^1$ and $(\what f)'(\omega)=\what g(\omega)$.

• The exchange of integral and differentiation is justified by dominated convergence.

Fourier transform of derivative #

Suppose $f \in L^1(\R \to \C)$ and $f \in C^1$, $f' \in L^1$. Then $(f')\phat (\omega) = i\omega \what f(\omega)$.

• The given conditions guarantees $f(x) \to 0$ as $x \to \pm \infty$.
  • Since $f, f' \in L^1$, For $\epsilon > 0$ there exists $a \in \R$ where $$ \int _ {a}^\infty \abs{f'(x)} \dd x < \epsilon \land \abs{f(a)} < \epsilon $$
  • Now for $b > a$, and $f' \in C$, we have $$ \abs{f(b)} = \abs {\int _ {a}^b f'(x) \dd x + f(a)} \le \int _ {a}^\infty \abs{f'(x)} \dd x + \abs{f(a)} < 2 \epsilon $$
• Now since $f' \in C$, we can apply integration by parts: $$ \what{(f')}(\omega) = \int _ \R f'(x) \exp(-i \omega x) \dd x = i\omega \int _ \R f(x) \exp (-i\omega x) \dd x = i \omega \what f(\omega) $$

Fourier transform: linear transforms #

Suppose $f \in L^1(\R \to \C)$.

• $\f[f(x - b)] = \int f(x - b) e^{-i\omega x} \dd x = e^{-ib \omega} \what f(\omega)$.
• $\f[e^{ibx} f(x)] = \what f(\omega - b)$.
• $\f[f(bx)] = \frac{1}{\abs{b}}\what f(\frac{t}{b})$.
• $\f[\overline{f(-x)}] = \overline {\what f(\omega)}$. $$ \F [\overline{f(-x)}] (\omega) = \int _ \R \overline{f(-x)} e^{-i \omega x} \dd x = \overline{\int _ \R f(x) e^{-i \omega x}\dd x} = \overline{\what f(\omega)} $$

Integral of a function times a Fourier transform #

Suppose $f, g \in L^1(\R)$, then $$ \begin{aligned} \int _ \R \what f(x) g(x) \dd x & = \int _ \R g(x) \int _ {\R} f(y) e^{-iyx} \dd y \dd x = \iint f(y) g(x) e^{ixy} \dd x \times \dd y = \int _ {\R} f(x) \what g(x) \dd x \end{aligned} $$ The change of order follows from Fubini's theorem.

Convolution #

Suppose $f, g \in \L(\R \to \C)$. The convolution of $f$ and $g$ is denoted as $f * g$, and is the function defined by $$ (f * g)(x) = \int _ \R f(y) g(x - y) \dd y $$

• If $f, g \in L^1(\R)$, $f * g$ is defined almost everywhere, and $$ \n{f * g} _ 1 \le \int _ {\R} \int _ {\R} |f(y)||g(x - y)| \dd x \dd y \le \n{f} _ 1 \n{g} _ 1 $$ following Tonelli's Theorem.
• $(f * g)(x) = (g * f)(x)$ when $(f * g)$ is defined at $x$.
  • This follows from a linear change of variable.

$\L^p$-norm of convolution #

Suppose $f \in L^1(\R \to \C, \sigma)$ and $g \in L^p(\R \to \C, \sigma)$. For $p \in [1, \infty]$ and $q = p'$. $$ \n{f * g} _ p = \sup _ {h \in \L^{q}, \n{h} _ q = 1} \abs{\pd{f * g}{h}} = \sup _ {h \in \L^q, \n{h} _ q = 1} \abs{\int _ {\R} (f * g)(x)h(x) \dd x} $$

• $f * g$ is finite almost surely.
  • Apply the procedure used in proving the $\L^p$-norm of cyclic convolution to $|f| * |g|$.
• $\n{f * g} _ p = \n{f} _ 1 \n{g} _ p$.
  • Apply the procedure again on $f * g$.

Fourier transform on $L^1$: convolution #

Suppose $f, g \in L^1(\R \to \C, \sigma)$. Then $\f[f * g] (\omega) = \what f(\omega) \what g(\omega)$. $$ \begin{aligned} \int _ {\R} (f * g)(x) e^{-ix\omega} \dd x &= \int _ \R \p{\int _ {\R} f(y) g(x - y) \dd y} e^{-ix \omega} \dd x\\ &= \int _ \R\p{\int _ \R g(x - y) e^{-ix\omega} \dd x}f(y) \dd y \\ & = \int _ \R \what g(\omega) e^{-iy\omega} f(y) \dd y = \what g(\omega) \what f(\omega) \end{aligned} $$

Poisson kernel on $\R$ #

For $y > 0$, define Poisson kernel $P _ y: \R \to (0, \infty)$ as $$ P _ y(x) = \frac{1}{\pi} \frac{y}{x^2 + y^2} $$

• The kernel integrates to one: $$ \begin{aligned} \int _ \R P _ y(x) \dd x & = \int _ \R \frac{1}{\pi} \frac{y}{x^2 + y^2} \dd x = \left .\frac{1}{\pi} \arctan\p{\frac{x}{y}} \right | _ {-\infty}^\infty = 1 \end{aligned} $$
• The kernel concentrates to zero as $y \downarrow 0$. That is $$ \lim _ {y \downarrow 0} \int _ {\R - B(0, \delta)} P _ y(x) \dd x = 0 $$
  • Notice the monotonicity, and the fact that $f(y) = {1}/\p{2\pi y}$.

Similarly, for $f \in L^p(\R \to \C)$ where $p \in [1, \infty]$. Define $\P _ yf := f * P _ y$.

• This is well defined, since $P _ y \in L^q$.

Poisson convergence theorems on $\R$ #

Suppose $f \in UC,B(\R \to \C)$. Then $\P _ y f\rightrightarrows _ {\R} f$ as $y \downarrow 0$.

• A modification of the proof on $\dD$ gives the result.

Suppose $f \in L^p(\R \to \C)$, $p \in [1, \infty)$. Then $\P _ y f \to f$ in $L^p$ as $y \downarrow 0$.

Fourier inversion formula #

Suppose $f, \what f \in L^1(\R \to \C)$ then $$ f(x) = \F^{-1}[\what f(\omega)] := \frac{1}{2\pi}\int _ \R \what f(\omega) e^{ix \omega} \dd \omega $$

• Therefore $\f$ and $\f^{-1}$ are related by a scaling and time reversal.
• Observe: $$ \begin{aligned} \F^{-1} [e^{-y |\omega|}] (x) & = \frac{1}{2\pi}\int _ \R e^{-y|\omega|} e^{i \omega x} \dd \omega\\ &= \frac{1}{\pi} \re \int _ {0}^\infty e^{-y \omega} e^{i \omega x} \dd \omega = \frac{1}{\pi}\re \frac{1}{ix - y} = \frac{1}{\pi} \frac{y}{x^2 + y^2} = P _ y(x) \end{aligned} $$
• Further observe: $$ \begin{aligned} \frac{1}{2\pi}\int _ \R \what f(\omega) e^{-y|\omega|} e^{i x \omega} \dd \omega & = \frac{1}{2\pi} \int _ \R \p{\int _ \R f(z) e^{-i z \omega}\dd z} e^{-y |\omega|} e^{ix \omega} \dd \omega\\ &= \int _ \R f(z)\F^{-1}[e^{-y|\omega|}] (x - z) \dd z\\ & = f(x) * \F^{-1}[e^{-y |\omega|}] (x) = f(x) * P _ y(x) = \P _ yf(x) \end{aligned} $$
• As $y \downarrow 0$, $\text{LHS}\to \f^{-1}[\what f(\omega)]$ pointwise.
  • By dominated convergence, which requires $\what f(\omega) \in L^1$.
• As $y \downarrow 0$, $\text{RHS} \to f(x)$ in $\L^p$ norm.
• Therefore $\f^{-1}[\what f(\omega)] = f(x)$ due to uniqueness of convergence. And we are done.
• At last, we can verify that $\f[P _ y(x)] (\omega) = e^{-y|\omega|}$.
  • This is interesting, as directly computing the result is difficult.
  • $\f[e^{-y|\omega|}] = P _ y(x)$. Therefore $e^{-y|\omega|} = \f[P _ y(x)]$. By applying $\f$ on both sides.

Since $f(x) = \f[\what f] (-x)$, and that $\what f \in L^1$, $f$ must be essentially uniformly continuous and bounded.

The inversion $\f^{-1}: L^1 \to L^\infty$ is also a continuous linear map.

Fourier transform on $L^1$: injective map #

Suppose $f(x) \in L^1$ and $\what f(\omega) = 0$, then $f = 0$.

• Since $f, \what f \in L^1$, $f = \f^{-1}[\what f] = 0$.

Therefore $\f, \f^{-1}$ are injective linear map on $L^1$.

Fourier transform on $L^2$: Plancherel's Theorem #

Suppose $f \in L^1, L^2(\R \to \C)$, and $\what f \in L^1$, then $$ \begin{aligned} \n{f} _ 2^2 &= \int _ {\R} f(x) \overline{f(x)} \dd x = \int _ {\R} f(-x) \overline{f(-x)} \dd x \\ &= \int _ \R 2\pi\what{\what f}(x) \overline {f(-x)} \dd x = 2\pi\int _ \R \what f(x) \overline{\what f(x)} \dd x = 2 \pi \norm{\what f(x)} _ {2}^2 \end{aligned} $$ Suppose $f \in L^1, L^2(\R \to \C)$, then $\n{f} _ 2^2 = 2\pi \n{\what f} _ 2^2$ is also true.

• Consider $f * P _ y$, $(f * P _ y) \phat \in L^1$. Thus $\n{f * P _ y} _ 2 = \sqrt{2\pi} \n{(f * P _ y)\phat} _ 2$.
• As $y \downarrow 1$, both LHS and RHS converges to $\n{f} _ 2$. TODO

Since $L^1$ is dense in $L^2$, this allows us to extend continuous maps $\f$ and $\f^{-1}$ to $L^2$.

• Notice that $\what f$ notation denotes the integral, and $\f$ notation is the transform.

Schwartz functions #

The Schwartz functions $\S(\R^n \to \C)$ is a subset of $\E(\R^n\to \C)$. $$ \S(\R^n):= \left \{ \phi \in C^\infty(\R^n): \forall k \in \N^n,\forall m \in \N, \exists C _ {m, k} \ge 0, \forall t: |t|^{m}\left|D^{k} \phi(t)\right| \leq C _ {m,k} \right \} $$ Or equivalently, all partial derivatives of $\phi$ decrease to zero faster than any power of $1/|t|$.

• $\S$ is a vector space over $\C$.
• All functions in $\S$ are bounded and uniformly continuous.
• $\S$ is a subspace of $L^p(\R^n )$, for all $p \in [1, \infty]$.
• $\phi \in \S$ then $D^k \phi \in \S$ for any index $k \in \N^n$.
• $\D$ is a proper subspace of $\S$.
• $\D$ is dense in $\S$. for any $\phi \in \S$ there exists a sequence $\phi _ \nu(t) \in \D$ that converge in $\S$ to $\phi(t)$.

Fourier transform on $\S(\R)$ #

Suppose $\phi \in \S(\R \to \C)$, then $\f[\phi] \in \S(\R)$ as well.

• By repeated differentiation of $\f[\phi]$ for all $k \in \N$: $$ D^{k} \f[\phi] (\omega)=\f [(-ix)^{k} \phi(x)] (\omega) $$
  • $\f[\phi] \in C^\infty(\R)$, and all derivatives are bounded.
• By repeated differentiation of $\phi$ and compute the Fourier transform, for all $k \in \N$: $$ \f [D^{k} \phi] (\xi)=(i\xi)^{k} \f[\phi] (\xi) $$
  • $\f[\phi] \in \S(\R)$ since the RHS must be bounded.

Therefore $\f: \S \to \S$ is a linear isomorphism.