Vrftsjtryk

I have this issue that I'm kind of clueless about, it is peripheral to what I typically do. I will state all the assumptions meticulously, even though I suspect they are not all needed. It is problem 3.1.1 in Stengers 1993 "Numerical Methods Based on Sinc and Analytic Functions".

Let $d>0$ and define $$D_d= {win mathbb{C} : lvert Im w rvert < d}$$ and for $0<epsilon<1$ $$D_d(varepsilon)= {win mathbb{C} : lvert Re w rvert < varepsilon lvert, ; lvert Im w rvert < d (1-varepsilon) }$$
Let $1leq p <infty$ and define
$$N_p(f,D_d) = left ( lim_{varepsilonto 0} int_{partial D_d(varepsilon)} lvert f(z) rvert^p lvert dz rvert right ) ^{1/p}$$

Define $H^p(D_d)$ the family of functions that are analytic and has $N_p(f,D_d)<infty $.

Let $0<delta < d$ and $h>0$. Define
$$D_d(n,delta)= left{win mathbb{C} : lvert Re w rvert < left(n+frac12right) h , ; lvert Im w rvert < delta right}$$

Let
$$
E(n,delta,f)(z) = frac{sin(pi z / h)}{2 pi i} int_{partial D(n,delta)} frac{f(zeta)}{(zeta-z) sin ( pi zeta /h)} d zeta.
$$

I would like to show that
$$
E(n,delta,f)(zeta) = f(zeta)- sin(pi zeta/h) sum_{k=-n}^n frac{(-1)^k f(kh)}{pi( zeta-kh)/h}
$$

We have
$$
frac{E(n,delta,f)(zeta)}{sin(pizeta/h)} = frac{1}{2 pi i} int_{partial D(n,delta)} frac{f(w)}{(w-zeta) sin ( pi w /h)} ,mathrm{d}w.
$$
The RHS is almost the Cauchy integral formula for $dfrac{f(zeta)}{sin(pizeta/h)}$, except for the fact that $sin(pi w/h)$ has simple zeroes at $w=kh$, $k=-n,-n+1,dots,n-1,n$ lying inside $D(n,delta)$ contributing to the integral (and if $zeta$ happens to be one of these points you need to work a bit more for this double pole, for example, take limit over different $zeta$s).

So by the homology form of Cauchy's theorem we just need to prove
$$
frac{1}{2pi i}int_{partial B_epsilon(kh)} frac{f(w)}{(w-zeta) sin ( pi w /h)} ,mathrm{d}w=frac{(-1)^k f(kh)}{pi( kh-zeta)/h}
$$
where $0<epsilon<h/2$ is chosen so that $zetanotinoverline{B_epsilon(kh)}$. But near $w=kh$, we have a holomorphic $g$ vanishing at $kh$ such that
$$
sin(pi w/h)=(-1)^kfrac{pi}{h}(w-kh)cdot(1+g(w)).
$$
So
begin{align*}
&frac1{2pi i}int_{partial B_epsilon(kh)} frac{f(w)}{(w-zeta) sin ( pi w /h)} ,mathrm{d}w\
&=frac1{2pi i}
int_{partial B_epsilon(kh)} frac{(-1)^k dfrac{f(w)}{(w-zeta)(1+g(w))pi/h}}{w-kh} ,mathrm{d}w\
&=left.(-1)^k frac{f(w)}{(w-zeta)(1+g(w))pi/h}rightrvert_{w=kh}
end{align*}
and the result follows.

Addendum

• The function $g(z)$ is easily constructed from the Taylor series expansion of $sin(pi z/h)$ at $z=kh$
  begin{align*}
  sin(pi z/h)&=(-1)^ksin(pi(z-kh)/h)\
  &=(-1)^ksum_{j=0}^inftyfrac{(-1)^j}{(2j+1)!}left[frac{pi(z-kh)}{h}right]^{2j+1}\
  &=(-1)^kfrac{pi(z-kh)}{h}times\
  &quadleft{1+sum_{j=1}^inftyfrac{(-1)^j}{(2j+1)!}left[frac{pi(z-kh)}{h}right]^{2j}right}
  end{align*}
  so $g(z)=sum_{j=1}^inftyfrac{(-1)^j}{(2j+1)!}left[frac{pi(z-kh)}{h}right]^{2j}$.




• When $zeta=ell h$, we can note that $$int_{partial D(n,delta)} frac{f(w)}{(w-zeta) sin ( pi w /h)},mathrm{d}w$$ is finite, multiplying by $sin(pizeta/h)/(2pi i)=sin(ellpi)/(2pi i)=0$ thus gives you $0$. This is also the limit as $zeta'tozeta$ of
  begin{align*}
  E(n,delta,f)(zeta')&=f(zeta')- sum_{k=-n}^n frac{(-1)^k f(kh)sin(pizeta'/h)}{pi( zeta'-kh)/h}\
  &to f(ell h)- sum_{k=-n}^n (-1)^k f(kh)color{red}{underbrace{lim_{zeta'toell h}frac{sin(pi zeta'/h)}{pi( zeta'-kh)/h}}_{begin{cases}(-1)^ell&text{ if }k=ell\0&text{otherwise}end{cases}}}\
  &= 0,
  end{align*}
  which is morally how we should make sense of the formula.