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


Added:


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?



Integral kernel of form $e^{-<x,y>^2}$
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 $x=0$.


