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## Integral kernel of form $e^{-<x,y>^2}$

Let $K(x,y): \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ given by $K(x,y) = e^{-< x,y>^2}$ where $<\cdot,\cdot>$ denote the canonical inner product. Define integral operator $T:C(\mathbb{R}^n) \to C(\mathbb{R}^n)$ by $(Tf)(x) = \int_{\mathbb{R}^n}K(x,y)f(y) dy$ whenever this integration makes sense. (Here, $C(\mathbb{R}^n)$ is a space of continuous functions.)

Now consider a positive real analytic function $f(x)$ which decays exponentially in $< x,x>$. For such $f$, $Tf$ clearly makes sense. The question in my mind is, would $Tf$ be real analytic?

I think since the convergence is not only uniform, but also monotone, it's plausible to expect $Tf$ being analytic. But I have no clue how am I going to show this.

Another guess that I have is $f(x)$ smooth with some decaying condition also would imply analyticity of $Tf$, though I wouldn't need that much.

I'll appreciate any kind of comments on the problem. Thanks!

Junehyuk Jung

an alternate question will be this:

Let $0 \leq f_1 (x) \leq f_2 (x) \leq \cdots$ be the sequence of real analytic function defined on $\mathbb{R}^n$, which converges to some bounded function $f(x)$. Will $f(x)$ be real analytic?

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If $f(y)$ decays as $e^{-c|y|^2}$, the integral makes sense for complex $x$ with small imaginary part (more precisely, if $x=u+iv$, the inequality $|v|^2<c$ is sufficient for convergence) and defines an analytic function whose trace on $\mathbb R^n$ is $Tf$. The smoothness of $f$ is irrelevant. – fedja Dec 9 2010 at 4:41
Thanks for the comment! I've just verified that your advise works. – Junehyuk Jung Dec 9 2010 at 5:20

The answer to your second question is No. Take $$f_n(x)=\sqrt{1+x^2}-\sqrt{n^{-2}+x^2}.$$ You have $0\le f_n\le f_{n+1}$ and the limit $f$ is $\sqrt{1+x^2}-|x|$, which is not analytic at $x=0$.