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 $g=0$.
Can real-analytic functions separate functions of moderate growth?
1 Answers
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
$$\int_\mathbb R g(y)\phi_t(y)\ dy=\int_{|y| For the other integral,
$$\Big|\int_{|y| In conclusion, I can choose $R$ so that the integral over $|y|
-
0I 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