We begin with the simpliest subject of our current theme of study,
namely the
Hilbert transform. It often appears in two forms.
One is as a Fourier multiplier operator: in the case the underlying
space is $ {\mathbb{R}}$,
$ \displaystyle
H(f)(x)=\int_{\mathbb{R}}f^{\wedge}(\xi)\frac{\text{sign}(\xi)}{i}e^{2\pi ix\xi}\thinspace d\xi
$
for $ {f\in L^{2}(\mathbb{R})}$, where
$ \displaystyle
\text{sign}(\xi)=\begin{cases}
1 & \xi>0\\
0 & \xi=0\\
-1 & \xi<0
\end{cases};\ f^{\wedge}(\xi):=\int_{\mathbb{R}}f(x)e^{-2\pi i\xi x}\,dx
$
or in the case the underlying space is a circle $ {\mathbb{T}}$,
$ \displaystyle
H(f)(\theta)=\sum_{n\in\mathbb{Z}}f^{\wedge}(n)\frac{\text{sign}(n)}{i}e^{in\theta}
$
where $ {f\in L^{2}([0,2\pi])}$, and
$ \displaystyle
\text{sign}(n)=\begin{cases}
1 & n>0\\
0 & n=0\\
-1 & n<0
\end{cases};\ f^{\wedge}(n):=\frac{1}{2\pi}\int_{[0,2\pi]}f(x)e^{-inx}\,dx.
$
(Note that the sign function vanishes on a measure zero set on $ {\mathbb{R}}$,
but on a measure 1 set on $ {\mathbb{Z}}$. This ultimately causes the
constant mode of the periodic function jumping out in various identities,
compared to their analogs for functions on $ {\mathbb{R}}$. This is
however not essential.) The other formulation is as a "singular
integral operator'', as will be ellaborated below. It is in the 19th
and earlier 20th centuries that the properties of this operator came
into interest, for its fundamental connection with the following two
questions that arised in Analysis.
The first one is that of convergence, in some appropriate sense, of
the
symmetric partial sum of the Fourier series of an integrable
function $ {f:[0,2\pi]\rightarrow\mathbb{C}}$
$ \displaystyle
S_{N}(f)(\theta):=\sum_{|n|\leq N}f^{\wedge}(n)e^{in\theta}\rightarrow f(\theta)\ ?
$
At that time Lebesgue's integration theory had not been developed.
However, that under the condition of $ {C^{1}}$ regularity of the function,
the pointwise convergence was already clear to Dirichlet, in 1829,
who told in private that he would soon resolve the case when $ {f}$
is continuous. This was disproved in 1873, by du Bois Reymond. Then
after the work of Lebesgue, under the condition of square integrability
of $ {f}$, the $ {L^{2}}$ convergence can be established using elemental
Hilbert space technique, invented during the end of the 19th century.
Later, achieved by Marcel Riesz in 1923, the $ {L^{p}}$ convergence
was shown to be true for all $ {1<p<\infty}$. Nevertheless, the "local''
convergence behavior turns out much more intricate, and the a.e. convergence
for $ {L^{2}}$ functions was eventually resolved by Carleson, in 1966.
The Hilbert transform enters the above question as follows. The symmetric
partial sum can be regarded as the convolution with the
Dirichlet kernel
(this is no incidence: a bounded multiplier must correspond to a distributional
kernel that commutes with translations)
$ \displaystyle
S_{N}(f)(\theta)=f*D_{N}\thinspace(\theta),
$
where
$ \displaystyle
D_{N}(x)=\sum_{|n|\leq N}e^{inx}.
$
The integral operator so defined is actually the orthogonal projection
$ {P_{|n|\leq N}}$ to the space spanned by $ {\{e^{inx}\}_{|n|\leq N}}$,
which is associated with the following multiplier on the intergers
$ \displaystyle
m_{S_{N}}(n)=\chi_{|n|\leq N}=\begin{cases}
1 & |n|\leq N\\
0 & \text{otherwise}
\end{cases}.
$
Denote the multiplier of the Hilber transform on the circle by $ {m_{H}(n):=\frac{1}{i}\text{sign}(n)}$.
And notice that
$ \displaystyle
m_{S_{N}}(n)=\frac{i}{2}\left(m_{H}(n+N)-m_{H}(n-N)\right)+\frac{1}{2}\chi_{\{\pm N\}}.
$
Then, in terms of operator,
$ \displaystyle
S_{N}(f)(\theta)=\frac{ie^{-iN\theta}H(e^{iN\theta}f)-ie^{iN\theta}H(e^{-iN\theta}f)}{2}+\frac{P_{\pm N}}{2}.
$
Hence the study of the boundedness properties of the family of operators
$ {S_{N}}$ reduces to the boundedness property of operators $ {H}$ and
$ {P_{\pm N}}$.
The second question is concerned with a different summation method
of Fourier series, and its connection to complex analysis. Here for
each $ {\theta\in[0,2\pi]}$, the Fourier series is called
Abel summable
to $ {f(\theta)}$ if
$ \displaystyle
\sum_{n\in\mathbb{Z}}f^{\wedge}(n)r^{|n|}e^{in\theta}\rightarrow f(\theta)
$
as $ {r\rightarrow1}$. Similar to the symmetric partial sum, Abel summation
can also be written as a convolution of $ {f}$ with a kernel. This time
it is the
Poisson kernel for the unit disk
$ \displaystyle
f*\mathcal{P}_{r}\thinspace(\theta),
$
where
$ \displaystyle
\mathcal{P}_{r}(\theta)=\sum_{n\in\mathbb{Z}}r^{|n|}e^{in\theta}=\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}.
$
We remark that if for a.e. $ {\theta\in[0,2\pi]}$, a Fourier series
is Abel summable and moreover, $ {f^{\wedge}(n)=0}$ for all $ {n<0}$,
then $ {f}$ , originally defined on the unit circle, can be extended
to a holomorphic function inside the unit disk, by
$ \displaystyle
F(z)=\sum_{n\in\mathbb{Z}}f^{\wedge}(n)z^{n}.
$
It is of interest that when the a.e. convergence holds. In sharp contrast
to the symmetric partial sum method, it turns out that the result
holds even for integrable functions. This is to be explained by the
fact that the family of Poisson kernels, indexed in $ {r}$, forms an
approximations to the identity, while it is not the case for
the Dirichlet kernels.
Now we ask the converse, that is, given a holomorphic function $ {F}$
in the unit disk, then under what conditions $ {F}$ will have radial
limit on the circle, i.e.
$ \displaystyle
\lim_{r\rightarrow1^{-}}F(re^{i\theta})\ \text{ exists for a.e. $\theta\in [0,2\pi]$?}
$
Suppose $ {F}$ has the power series expansion $ {\sum_{n=0}^{\infty}a_{n}z^{n}}$.
Now by Parseval's identity
$ \displaystyle
\sum_{n=0}^{\infty}|a_{n}|^{2}r^{2n}=\frac{1}{2\pi}\int_{[0,2\pi]}|F(re^{i\theta})|^{2}d\theta,\ 0\leq r<1.
$
Thus, assume that
so that the series $ {\sum_{n=0}^{\infty}|a_{n}|^{2}}$ converges to
the above quantity, and thus has a corresponding $ {L^{2}}$ function
$ {F_{0}(e^{i\theta}).}$ Since $ {L^{2}([0,2\pi])\subset L^{1}([0,2\pi])}$,
by the above paragraph we have the limit $ {\lim_{r\rightarrow1^{-}}F(re^{i\theta})}$
exists for a.e. $ {\theta}$. This result was obtained by Fatou.
The space of holomorphic functions on the unit disk satisfying the
bound (
1) is called the
Hardy space $ {H^2(\mathbb{D})}$,
with norm $ {\|F\|_{H^{2}(\mathbb{D})}}$ defined to be
the quantity on the LHS of (
1). It can be identified with
the closed subspace $ {\ell^{2}(\mathbb{Z}_{+})}$ of the Hilbert space
$ {\ell^{2}(\mathbb{Z})}$, or equivalently the space $ {\mathcal{S}}$
of square integrable functions on the cirlce whose Fourier coefficients
vanish for $ {n<0}$. In view of this, define the orthogonal projection
from $ {L^{2}([0,2\pi])}$ to $ {\mathcal{S}}$, and write in the corresponding
element in $ {H^{2}(\mathbb{D})}$
$ \displaystyle \begin{array}{rcl}
P(f)(z): & = & \sum_{n=0}^{\infty}a_{n}z^{n}.\\
& = & \sum_{n=0}^{\infty}\left(\frac{1}{2\pi}\int_{[0,2\pi]}f(e^{i\theta})e^{-in\theta}\thinspace d\theta\right)z^{n}\\
& = & \frac{1}{2\pi}\int_{[0,2\pi]}f(e^{i\theta})\sum_{n=0}^{\infty}\left(e^{-i\theta}z\right)^{n}\thinspace d\theta\\
& = & \frac{1}{2\pi i}\int_{S^{1}}\frac{f(\zeta)}{\zeta-z}\thinspace d\zeta,
\end{array} $
which turns out to be the
Cauchy integral on the circle. On
the other hand, notice that
For this reason, $ {H(f)}$ is often called
conjugate to $ {f}$,
and hence can be also identified the imaginary part of the
Cauchy kernel
of the unit circle
$ \displaystyle
\frac{1}{i(\zeta-z)}d\zeta=\frac{ie^{i\psi}d\psi}{i(e^{i\psi}-re^{i\theta})}=\mathcal{P}_{r}(\theta-\psi)d\psi+\frac{d\psi}{2}+i\mathcal{Q}_{r}(\theta-\psi)d\psi.
$
Here $ {Q_{r}(\theta)}$ is known as the
conjugate Poisson kernel.
In explicit form,
$ \displaystyle
\mathcal{Q}_{r}(\theta-\psi)=\frac{r\sin(\theta-\psi)}{1-2r\cos\theta+r^{2}}\rightarrow\frac{1}{2}\cot(\frac{\theta-\psi}{2})
$
as $ {r\rightarrow1^{-}}$. Unlike Poisson's kernel, it is not absolutely integrable,
thus called "singular'', near the diagonal $ {\theta=\psi}$, but
note that $ {\cot(x)}$ is an odd function on $ {[-\pi,\pi]}$, which is
one manifestation of the {cancellation property} for such singular
integral operators.
Now we turn to the analogous results of the above for functions on
the real line, where the unit disk is replaced by the upper half space
$ {\mathbb{R}_{+}^{2}:=\{(x,y)\in\mathbb{R}^{2}:y>0\}}$. Starting with
a function $ {F_{0}^{\wedge}\in L^{2}(0,\infty)}$, define its
Laplace transform
to be
which can be viewed as an "integral version'' of Abel summation
of the Fourier series $ {\sum_{n\geq0}f^{\wedge}(n)}$. (The analogue
for $ {\sum_{n\in\mathbb{Z}}f^{\wedge}(n)}$ will be given by $ {\int_{\mathbb{R}}f^{\wedge}(\xi)e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi}$.)
Since for any $ {\delta>0}$, if $ {y\geq\delta}$, the integral has the
uniform bound
$ \displaystyle
\int_{[0,\infty)}|F_{0}^{\wedge}(\xi)e^{2\pi i\xi z}|\thinspace d\xi\leq\left(\int_{[0,\infty)}|F_{0}^{\wedge}(\xi)|^{2}\thinspace d\xi\right)^{1/2}\left(\int_{[0,\infty)}e^{-4\pi\xi\delta}\thinspace d\xi\right)^{1/2},
$
so $ {F}$ is in fact holomorphic in $ {\mathbb{R}_{+}^{2}}$. Plancherel's
identity implies that
$ \displaystyle
\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\int_{[0,\infty)}|F_{0}^{\wedge}(\xi)|^{2}e^{-4\pi\xi y}\thinspace d\xi\leq\|F_{0}^{\wedge}\|_{L^{2}(0,\infty)}^{2}
$
and by the monotone convergence theorem
$ \displaystyle
\sup_{y>0}\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\|F_{0}^{\wedge}\|_{L^{2}(0,\infty)}^{2}.
$
As before, the
Hardy space $ {H^{2}(\mathbb{R}_{+}^{2})}$ is
defined to be the space of holomorphic functions on $ {\mathbb{R}_{+}^{2}}$
satisfying the condition
$ \displaystyle
\sup_{y>0}\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx<\infty,
$
and equipped with the norm $ {\|F\|_{H^{2}(\mathbb{R}_{+}^{2})}}$ to
be the above quantity.
Given such a function $ {F}$ with suitable decaying
property, via a contour integration argument, its
inverse Laplace transform
satisfies, for any $ {y_{1},y_{2}>0}$.
$ \displaystyle
\int_{\mathbb{R}}F(x+iy_{1})e^{-2\pi i\xi(x+iy_{1})}\thinspace dx=\int_{\mathbb{R}}F(x+iy_{2})e^{-2\pi i\xi(x+iy_{2})}\thinspace dx.
$
In other words, there is a
deck transformation rule
$ \displaystyle
F_{y_{1}}^{\wedge}(\xi)e^{2\pi y_{1}\xi}=F_{y_{2}}^{\wedge}(\xi)e^{2\pi y_{2}\xi}.
$
A limiting argument shows this it is true for all $ {F\in H^{2}(\mathbb{R}_{+}^{2})}$.
Hence the function $ {F_{0}^{\wedge}(\xi):=F_{y}^{\wedge}(\xi)e^{2\pi y\xi}}$
is well defined, and by Plancherel's identity
$ \displaystyle
\int_{\mathbb{R}}|F(x+iy)|^{2}\thinspace dx=\int_{\mathbb{R}}|F_{0}^{\wedge}(\xi)|^{2}e^{-4\pi y\xi}\thinspace dx.
$
Thus,
$ \displaystyle
\|F\|_{H^{2}(\mathbb{R}_{+}^{2})}=\int_{\mathbb{R}}|F_{0}^{\wedge}(\xi)|^{2}\thinspace dx
$
and $ {H^{2}(\mathbb{R}_{+}^{2})}$ can be identified with the Hilbert
space $ {L^{2}(0,\infty)}$ via (
3).
It then follows that $ {F}$ has boundary values $ {F_{0}}$, the inverse
Fourier transform of $ {F_{0}^{\wedge}}$, in the $ {L^{2}}$ sense. It
also holds a.e., once we have the
Poisson integral representation
$ \displaystyle
F(x+iy)=\int_{[0,\infty)}F_{0}^{\wedge}(\xi)e^{2\pi i\xi z}\thinspace d\xi=\int_{\mathbb{R}}F_{0}(x-t)\mathcal{P}_{y}(t)\thinspace dt
$
or more generally,
$ \displaystyle
\int_{\mathbb{R}}f^{\wedge}(\xi)e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi=\int_{\mathbb{R}}f(x-t)\mathcal{P}_{y}(t)\thinspace dt
$
where $ {\mathcal{P}_{y}(x)=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}}}$ is
the
Poisson kernel for the upper half space. To see this, define
the
Cauchy kernel for $ {y>0}$
$ \displaystyle
C(x+iy)=\int_{\mathbb{R}}e^{2\pi i|\xi|z}\thinspace d\xi=\int_{\mathbb{R}}2\chi_{[0,\infty)}e^{2\pi i\xi z}\thinspace d\xi=\frac{i}{\pi z}.
$
Then
$ \displaystyle \begin{array}{rcl}
\mathcal{P}_{y}(x) & = & \int_{\mathbb{R}}e^{-2\pi|\xi|y}e^{2\pi ix\xi}\thinspace d\xi\\
& = & \int_{[0,\infty)}e^{-2\pi\xi y}e^{2\pi ix\xi}\thinspace d\xi+\int_{[0,\infty)}e^{-2\pi\xi y}e^{-2\pi ix\xi}\thinspace d\xi\\
& = & \frac{1}{2}(C(x+iy)+C(-x+iy))\\
& = & \frac{1}{\pi}\frac{y}{x^{2}+y^{2}}.
\end{array} $
Notice that for $ {y>0}$
$ \displaystyle
\frac{i}{\pi z}=\mathcal{P}_{y}(x)+i\mathcal{Q}_{y}(x)
$
where $ {\mathcal{Q}_{y}(x)=\frac{1}{\pi}\frac{x}{x^{2}+y^{2}}}$ is
the
conjugate Poisson kernel. We would like to know what happens
when $ {y\rightarrow0}$. This is the clearest in terms of multipliers. We have,
corresponding to the above,
$ \displaystyle
2\chi_{[0,\infty)}e^{-2\pi\xi y}=e^{-2\pi|\xi|y}+\frac{1}{i}\text{sign}(\xi)e^{-2\pi|\xi|y}
$
and as $ {y\rightarrow0}$,
$ \displaystyle
2\chi_{[0,\infty)}(\xi)=1+\text{sign}(\xi)=1+im_{H}(\xi),
$
where $ {m_{H}(\xi)=\frac{1}{i}\text{sign}(\xi)}$ is the Fourier multiplier
of the Hilbert transform on the real line. One thus has
$ \displaystyle
H(f)(x)=\lim_{y\rightarrow0}\int_{\mathbb{R}}f(x-t)\mathcal{Q}_{y}(t)\thinspace dt
$
in $ {L^{2}}$. We shall now focus on distributional property of $ {\mathcal{Q}_{y}(t)}$
as $ {y\rightarrow0}$. Assume $ {\varphi}$ is a test function. For any $ {\epsilon>0}$,
$ \displaystyle \begin{array}{rcl}
& & \int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\
& = & \int_{|t|<\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt+\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\
& = & \int_{|t|<\epsilon}(\varphi(t)-\varphi(0))\mathcal{Q}_{y}(t)\thinspace dt+\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt\\
& =: & A+B
\end{array} $
Here the {cancellation property} of $ {\mathcal{Q}_{y}(x)}$ plays a
role. It is clear that $ {A=O(\epsilon)}$, and so
$ \displaystyle
\int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt=\lim_{\epsilon\rightarrow0}\int_{|t|\geq\epsilon}\varphi(t)\mathcal{Q}_{y}(t)\thinspace dt.
$
We now wish to take $ {y\rightarrow0}$. When $ {|t|\geq\epsilon}$ it is clear
that $ {\mathcal{Q}_{y}(t)\rightarrow\frac{1}{\pi t}}$ as $ {y\rightarrow0}$. This amounts
to estimate
$ \displaystyle
\frac{1}{\pi}\int_{|t|\geq\epsilon}\varphi(t)\thinspace\frac{dt}{t}-\int_{\mathbb{R}}\varphi(t)\mathcal{Q}_{\epsilon}(t)\thinspace dt=\int_{\mathbb{R}}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt,
$
where
$ \displaystyle
\Delta_{\epsilon}(t)=\begin{cases}
-\mathcal{Q}_{\epsilon}(t) & \text{if }|t|<\epsilon\\
\frac{1}{t}-\mathcal{Q}_{\epsilon}(t) & \text{if }|t|\geq\epsilon
\end{cases},
$
which again satisfies a cancellation property. It's then not hard
to see that
$ \displaystyle \begin{array}{rcl}
& & \left|\int_{\mathbb{R}}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt\right|\\
& = & \left|\int_{|t|<\epsilon}(\varphi(t)-\varphi(0))\Delta_{\epsilon}(t)\thinspace dt\right|+\left|\int_{\epsilon\leq|t|<1}(\varphi(t)-\varphi(0))\Delta_{\epsilon}(t)\thinspace dt\right|+\left|\int_{|t|\geq1}\varphi(t)\Delta_{\epsilon}(t)\thinspace dt\right|\\
& = & O(\epsilon+\epsilon\log\epsilon+\epsilon).
\end{array} $
We conclude that $ {\mathcal{Q}_{y}(x)\rightarrow\frac{1}{\pi}\text{p.v.}(\frac{1}{x})}$
in distribution as $ {y\rightarrow0}$, and the Hilbert transform ca be written
as a singular integral operator
$ \displaystyle
H(f)=\frac{1}{\pi}\text{p.v.}(\frac{1}{x})*f.
$
Moreover, since $ {\mathcal{P}_{y}(x)\rightarrow\frac{1}{\pi}\delta_{0}}$ in
distributon as $ {y\rightarrow0}$, denoting the distributional limit of $ {\frac{1}{z}}$
to be $ {\frac{1}{x+i0}}$ as $ {y\rightarrow0}$, we have the following distributional
identity for kernels
$ \displaystyle
\frac{1}{x+i0}=\text{p.v.}(\frac{1}{x})-i\pi\delta
$
corresponding to identity for multipliers above.
Again, one can define the orthogonal projection $ {P}$ from $ {L^{2}(\mathbb{R})}$
to the closed subspace that contains functions whose Fourier transforms
are supported on $ {[0,\infty)}$. It is realized by the
Cauchy integral
$ \displaystyle
F(z)=\int_{\mathbb{R}}\frac{f(t)}{t-z}\thinspace dt=\int_{[0,\infty)}f^{\wedge}(\xi)e^{2\pi i\xi z}\thinspace d\xi,\ z\in\mathbb{R}_{+}^{2}
$
which maps $ {L^{2}(\mathbb{R})}$ onto $ {H^{2}(\mathbb{R}_{+}^{2})}$.