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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)| 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 $F_t(x)=-i(x+i)f_t(x)$?

1 Answers 1

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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| 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| where $M(R)={\rm max}_{|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| 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 $x$, $f(x)=0$.

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    I realize it is not clear what I am asserting I am proving. The statement is: Assume $f$ is continuous and of moderate growth and that $\int f(y)\phi^t(x+y)\ dy=0$ for all $x$ and $t$ sufficiently large. Then $f=0$.2011-11-10