Define H(a) to be the space of holomorphic functions $f(z)$ on $S\_a:=\{z:|\Re z| < a\}$ with
$$ ||f||^2\_a:=\int\_{S\_a} |f(x+iy)|^2 (1+|x+iy|)^{100} dx \, dy < \infty. $$
Two questions:


1. Is H(2) dense in H(1)?
2. Are the entire functions dense in H(1)?


Is there a nice kernel to convolve things with?


This occurs in the following form: When building an equality of integral transforms, I need to use holomorphy of the test function out to some large domain (wide vertical strip), but the final equality converges absolutely even when the test function is holomorphic on a much smaller domain. I want to know if I can relax the assumptions on the test function. (Awfully sure the answer is yes, but why?)


One note: I mean the natural inclusion of H(2) into H(1), so no composing with conformal maps.



density of holomorphic functions in vertical strips
It's possible I'm over-thinking this:
If $f \in H(1)$, let
$$ f\_\delta(z) = \frac{1}{\sqrt{\pi\delta}} \int\_{(0)} f(w) \exp ((w-z)^2/\delta) dw, $$
then
$$ f\_\delta(x+iy) = \frac{i\exp (x^2/\delta)}{\sqrt{\pi\delta}} \int\_{-\infty}^\infty f(i(t+y)) e^{-2itx/\delta} \exp (-t^2/\delta) dt, $$
so $f\_\delta$ is entire (integral converges rapidly for all $z$), and $f \in H(b)$ for all $b>1$, by interchanging integrals in the norm and using
$$ (1+|(x+iy)-it|) \le (1+|t|)(1+|x+iy|). $$
Also, on $S\_a$,
$$ f\_\delta(z) = \frac{1}{\sqrt{\pi\delta}} \int\_{(0)} f(z+w) \exp (w^2/\delta) dw, $$
by contour shifting.
Then $f\_\delta \to f$ as $\delta \to 0$, by the usual approximation to the identity argument.


This works if we change the norm to
$$ ||f||^2\_a:=\sup\_{|\sigma|<a} \int\_{-\infty}^\infty |f(\sigma+it)|^2 (1+|t|)^{100} dt. $$


