Can real-analytic functions separate functions of moderate growth? Let $g$ be a continuous function of moderate growth. I want to prove that if $\int_\mathbb R g\cdot f=0$ for all $f$ in a family of (real-analytic, polynomially-decaying) functions described below, then $g=0$.
The family of functions is made up of those $f$ that are the restriction to $\mathbb R$ of a function holomorphic in a fixed strip $|{\rm Im}(z)|<a$ and satisfy $f(x+iy)=O\big((1+x^2)^{-t}\big)$. Obviously, $t$ must be sufficiently large to cancel the moderate growth of $g$. (Holomorphy is needed to guarantee that the $f$ have exponentially decaying Fourier transforms, which is necessary for my larger project.)
This type of result is well-known when the $f$ have compact support (e.g., Theorem 1.2.5 in Hormander's "The Analysis of Linear Partial Differential Operators I"), and there are results in similar situations. For example, if $f=O\big((1+|x|)^{-\alpha}\big)$ for $\alpha>1$, and $\int_\mathbb R f=1$, then the family $f_\epsilon(x)=\epsilon^{-1}f(x/\epsilon)$ works for a wide class of $g$ (Theorem 8.15 in Folland's "Real Analysis"). Unfortunately, if $f$ is analytic in a strip, each $f_\epsilon$ will be analytic in a strip that goes to zero with $\epsilon$, and I think I need them to be analytic in a fixed strip (plus, I'm not sure this method will work for moderately-growing $g$).
I've been trying to prove this using the subfamily (as $t\rightarrow\infty$)
$$f_t(x)={\Gamma(t)\over\sqrt{\pi}\Gamma(t-1/2)}{a^{2t-1}\over (a^2+x^2)^t}$$
but I am currently unable to obtain a sufficient bound on $\int_{|x|>R}f_t\cdot g$ to deduce that $\lim_{t\rightarrow\infty}\int_{|x|>R}f_t\cdot g=0$ (assuming that happens!). I get stuck pretty quickly: by moderate growth, $g(x)\le C(a^2+x^2)^N$, for some $C$, $N$, so 
$$\int_{|x|>R}f_t\cdot g\le C\int_{|x|>R}{\Gamma(t)\over\sqrt{\pi}\Gamma(t-1/2)}{a^{2t-1}\over (a^2+x^2)^{t-N}}$$
I'd thought I'd be able to calculate this, but my attempts so far have failed...
To summarize, my specific questions are: 
(1) Is it false that $\int_\mathbb R f\cdot g=0$ implies $g=0$, for $f$ in the above family?
(2) Is there a reference to a proof of this or a related fact that I might adapt into a proof?
(3) Is there a bound on $\int_{|x|>R}f_t\cdot g$ from which I can deduce that $\lim_{t\rightarrow\infty}\int_{|x|>R}f_t\cdot g=0$?
(3') Can you replace $f_t$ by $F_t(x)=-i(x+i)f_t(x)$?
 A: I was able to answer my question in the affirmative (turns out, I just needed to use a different bound on my function of moderate growth). I'm copying my write-up, which has a few notational differences with my question: I use $\phi_t$ instead of $f_t$ for the above family, and $f$ is the function of moderate growth instead of $g$. 
Since $f$ has moderate growth, for $|x|$ sufficiently large, $f(x)=O\big(|x|^N\big)$. Assume $t\gg N$. Fix $x$. Then
$$f(x)=\int_\mathbb R f(x)\phi_t(y-x)\ dy=\int_\mathbb R (f(x)-f(y))\phi_t(y-x)\ dy+\int_\mathbb R f(y)\phi_t(y-x)\ dy$$
The second integral vanishes by assumption. Change variables $y\rightarrow y+x$ and set $g(y)=f(x)-f(y+x)$. Note that $g$ is still of moderate growth and $g(0)=0$. Take $R>0$ and split the first integral into two pieces
$$\int_\mathbb R g(y)\phi_t(y)\ dy=\int_{|y|<R} g(y)\phi_t(y)\ dy+\int_{|y|\ge R} g(y)\phi_t(y)\ dy$$
We show the integral over $|y|\ge R$ goes to zero. Take $C_R$ such that for $|y|\ge R$, $|g(y)/y^N|\le C_R$.
$$\Big|\int_{|y|\ge R}\phi_t(y)g(y)\ dy\Big|\le C_R\int_{|y|\ge R}\phi_t(y)|y|^N\ dy\le C_R\int_\mathbb R\phi_t(y)|y|^N\ dy$$
$$=C_R{a^{N}\Gamma(N/2+1/2)\Gamma(t-N/2-1/2)\over \sqrt{\pi}\Gamma(t-1/2)}$$
The asymptotic expression of the Beta function shows that this is asymptotic to
$$C_R {a^N\Gamma(N/2+1/2)\over \sqrt{\pi}t^{N/2}}$$
and thus goes to zero as $t\rightarrow\infty$. (Note that $C_R$ may blow up as $R$ gets small.) (Also note that if you replaced $|y|^N$ with $(a^2+y^2)^N$, the asymptotic expression would not go to zero, which is where I was running into problems.)
For the other integral,
$$\Big|\int_{|y|<R}\phi_t(y)g(y)\ dy\Big|\le M(R)\int_{|y|<R}\phi_t(y)\ dy\le M(R)\int_\mathbb R\phi_t(y)\ dy=M(R)$$
where $M(R)={\rm max}_{|x|<R}|g(x)|$. Since $g$ is continuous and $g(0)=0$, $M(R)\rightarrow 0$ as $R\rightarrow 0$.
In conclusion, I can choose $R$ so that the integral over $|y|<R$ is as small as I want, and then I can choose $t$ so that the integral over $|y|\ge R$ is as small as I want. So for any $x$, $f(x)=0$.
