07/06/2014, 02:11 PM
Hm.
Define the Weierstrass transform:
\( \mathcal{W}\{f\}(x) = \frac{1}{\sqrt{4\pi}}\int_{-\infty}^\infty f(t) \; e^{-\frac{(x-t)^2}{4}} \; dt \)
And it's inverse
\( \mathcal{W}^{-1}\{f\}(x)_{x0} = \frac{1}{\sqrt{4\pi}} \int_{x_0-i\infty}^{x_0+i\infty} f(t)e^{\frac{(x-t)^2}{4}}\;dt \)
Where \( x_0 \) is in the strip of convergence of the forward transform.
This transform has the nice property that \( \mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n \)
Now let \( f(x) \) be a function that is exponentially bounded (i.e. \( |f(x+yi)| < Ce^{a|y|} \) on some strip \( a < \mathfrak{R}(z) < b \). First, we have that \( \mathcal{W}^{-1}\{f\}(x)_{c} \) is convergent for all \( x \in \mathbb{R} \) if \( a < c < b \). To show this, let \( t =a+bi \). Then we have
\( |f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < |Ce^{\frac{(x-(a+bi))^2}{4}+\alpha |b|}| \)
For sufficiently large \( |b| \)
With some simple algebra, we can rearrange that to
\( |f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < Ce^{\alpha |b| +\frac{1}{4}(a^2-b^2-2ax+x^2)} \)
This is \( o(e^{-|b|}) \) as \( b \rightarrow \pm\infty \) since \( \alpha|b| - \frac{b^2+c}{4} < -|b| \) for sufficiently large \( |b| \). Thus the integral converges as the integrand will decay quickly on the line of integration.
Now let \( a_n \) be an arbitrary sequence of complex numbers and
\( Ge(a_n,x) = \sum_{k=0}^{\infty} a_k x^k \)
\( He(a_n,x) = \sum_{k=0}^{\infty} a_k H_k(x) \)
If \( Ge(a_n,x) \) is exponentially bounded on some strip, we can take it's inverse Weierstrass transform. If \( Ge(a_n,x) \) is also entire, we can exchange the integral and sum and use the identity \( \mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n \) to get
\( \mathcal{W}^{-1}\{ Ge(a_n,t)\}(x)_{c} = \sum_{k=0}^{\infty} a_k H_k(\frac{x}{2}) \)
Which is just \( He(a_n,x) \) Which also means that \( \sum_{k=0}^{\infty} a_k H_k(\frac{x}{2}) \) converges for all \( x \in \mathbb{R} \)
So if we let \( h_n \) be the nth coefficient in the Hermite series for some function \( f(x) \), we get that
\( -h_0iz + \sum_{n=0}^{iz-1} f(n) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(z) \).
converges for all \( x \in \mathbb{R} \) if
\( g(z) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) z^k \)
defines an entire function that is exponentially bounded on some strip \( a < \mathfrak{R}(z) < b \)
It would probably be easy to weaken the growth bounds to something like \( |f(x+iy)|<Ce^{\alpha y^2} \) for \( 0<a<\frac{1}{4} \), but the integral would most likely no longer converge for all real x.
Define the Weierstrass transform:
\( \mathcal{W}\{f\}(x) = \frac{1}{\sqrt{4\pi}}\int_{-\infty}^\infty f(t) \; e^{-\frac{(x-t)^2}{4}} \; dt \)
And it's inverse
\( \mathcal{W}^{-1}\{f\}(x)_{x0} = \frac{1}{\sqrt{4\pi}} \int_{x_0-i\infty}^{x_0+i\infty} f(t)e^{\frac{(x-t)^2}{4}}\;dt \)
Where \( x_0 \) is in the strip of convergence of the forward transform.
This transform has the nice property that \( \mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n \)
Now let \( f(x) \) be a function that is exponentially bounded (i.e. \( |f(x+yi)| < Ce^{a|y|} \) on some strip \( a < \mathfrak{R}(z) < b \). First, we have that \( \mathcal{W}^{-1}\{f\}(x)_{c} \) is convergent for all \( x \in \mathbb{R} \) if \( a < c < b \). To show this, let \( t =a+bi \). Then we have
\( |f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < |Ce^{\frac{(x-(a+bi))^2}{4}+\alpha |b|}| \)
For sufficiently large \( |b| \)
With some simple algebra, we can rearrange that to
\( |f(a+bi)e^{\frac{(x-(a+bi))^2}{4}}| < Ce^{\alpha |b| +\frac{1}{4}(a^2-b^2-2ax+x^2)} \)
This is \( o(e^{-|b|}) \) as \( b \rightarrow \pm\infty \) since \( \alpha|b| - \frac{b^2+c}{4} < -|b| \) for sufficiently large \( |b| \). Thus the integral converges as the integrand will decay quickly on the line of integration.
Now let \( a_n \) be an arbitrary sequence of complex numbers and
\( Ge(a_n,x) = \sum_{k=0}^{\infty} a_k x^k \)
\( He(a_n,x) = \sum_{k=0}^{\infty} a_k H_k(x) \)
If \( Ge(a_n,x) \) is exponentially bounded on some strip, we can take it's inverse Weierstrass transform. If \( Ge(a_n,x) \) is also entire, we can exchange the integral and sum and use the identity \( \mathcal{W}\{H_n(\frac{x}{2})\}(t) = t^n \) to get
\( \mathcal{W}^{-1}\{ Ge(a_n,t)\}(x)_{c} = \sum_{k=0}^{\infty} a_k H_k(\frac{x}{2}) \)
Which is just \( He(a_n,x) \) Which also means that \( \sum_{k=0}^{\infty} a_k H_k(\frac{x}{2}) \) converges for all \( x \in \mathbb{R} \)
So if we let \( h_n \) be the nth coefficient in the Hermite series for some function \( f(x) \), we get that
\( -h_0iz + \sum_{n=0}^{iz-1} f(n) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) H_k(z) \).
converges for all \( x \in \mathbb{R} \) if
\( g(z) = \sum_{k=1}^{\infty} \left(\sum_{n=1}^{\infty} h_n {n \choose k-1} \frac{B_{n-k+1}}{k} (-2i)^{n-k}\right) z^k \)
defines an entire function that is exponentially bounded on some strip \( a < \mathfrak{R}(z) < b \)
It would probably be easy to weaken the growth bounds to something like \( |f(x+iy)|<Ce^{\alpha y^2} \) for \( 0<a<\frac{1}{4} \), but the integral would most likely no longer converge for all real x.