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 $S_a$, $$ f_\delta(z) = \frac{1}{\sqrt{\pi\delta}} \int_{(0)} f(z+w) \exp (z^2/\delta) dw, $$ 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 $$ ||f||^2_a:=\sup_{|\sigma|<a} \int_{-\infty}^\infty |f(\sigma+it)|^2 (1+|t|)^{100} dt. $$

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 $$ ||f||^2_a:=\sup_{|\sigma|<a} \int_{-\infty}^\infty |f(\sigma+it)|^2 (1+|t|)^{100} dt. $$

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 $$ (1+|t+y|) \le (1+|t|)(1+|y|). $$ Also, on $S_a$, $$ f_\delta(z) = \frac{1}{\sqrt{\pi\delta}} \int_{(0)} f(z+w) \exp z^2 dw, $$ by contour shifting. Then $f_\delta \to f$ as $\delta \to 0$, by the usual approximation to the identity argument.

Maybe the norm should really be $$ ||f||^2_a:=\sup_{|\sigma|<a} \int_{-\infty}^\infty |f(\sigma+it)|^2 (1+|t|)^{100} dt \, ? $$

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 $S_a$, $$ f_\delta(z) = \frac{1}{\sqrt{\pi\delta}} \int_{(0)} f(z+w) \exp (z^2/\delta) dw, $$ 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 $$ ||f||^2_a:=\sup_{|\sigma|<a} \int_{-\infty}^\infty |f(\sigma+it)|^2 (1+|t|)^{100} dt. $$