what is the relation of smooth compact supported funtions and real analytic function? What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.
 A: If by "test function" you mean a smooth function with compact support (as opposed to a Schwarz function), then the answer is no. Consider an analytic function $f \in C_c^\infty(\mathbb R^n)$ and a point $x$ outside of its support. Because $f$ is analytic, it must equal its own Taylor series developed at $x$, i.e. $0$, and since the radius of convergence is infinite, $f$ must be the zero function.
The upshot of this is that analytic functions are very rigid objects. Loosely speaking, perturbing an analytic function in a single point causes global changes to "ripple" out from the perturbation in order to preserve analyticity.
A: A test function is a smooth and compactly supported function, while a real analytic function is smooth and given by a power series uniformly converging at every point.
Perhaps I will only give a hint as an answer to your second question: A test function must be equal to zero everywhere outside some compact set. If it were also analytic, then what would the coefficients of its power series have to be?
A: Assume $f$ is a real-analytic test function which is not identically zero; this will lead to a contradiction. Let $K$ denote the support of $f$. The idea (also present in OP's attempt) is to expand $f$ into a power series centered at some boundary point $x_0$ of the support; find that the coefficients are all zero, obtain a contradiction. Key steps:


*

*A nonempty bounded set must have nonempty boundary (because the only sets with empty boundary are $\varnothing$ and $\mathbb R^n$). 

*Let $x_0$ be a boundary point of $K$. Note that $x_0\in K$ and $x_0$ is also   a limit point of the complement of $K$.  

*All derivatives of $f$ are identically zero on the complement of $K$ 

*They are also continuous on all of $\mathbb R^n$. By continuity they are equal to zero at $x_0$.  

*The power series of $f$ centered at $x_0$ is $0+0(x-x_0)+\dots$

*Therefore, $f$ is equal to zero  in some neighborhood of $x_0$. But this contradicts the definition of support of $K$.

A: I am trying to prove that if some test function $f$ in $\Bbb R^n$ is real analytic then it must be equivalent to zero function. Please correct me if there is any problem.

Proof: 
Let us define a test function,  
$ f_\epsilon (x)= 
\begin{cases}
\frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}},  &\text{for |x|< $\epsilon$ } \\
0, & \text{everywhere else}  \\
\end{cases} $  
$\color{red}{\text{Edited in: This was a false start}}$. You should work with a general test function, not  a particular one. That is, simply say: let $f:\mathbb R^n\to \mathbb R$ be a test function. By definition of a test function, $f$ has compact support. The goal is to show that if $f$ is real-analytic, then the support is  empty. 

We know, this function has derivatives of all orders and also has a compact support.
Now, Let us suppose that it is also a real Analytic function. 
If this function is a real analytic then it must has a Taylor series expansion about any point $x_0$ in its domain which will converge to $f_\epsilon (x)$ for a nbdd. of $x_0$. 
Now in particular take Taylor expansion about $\pm\epsilon$
$f_\epsilon (x) \approx  0 + 0x + 0x^2 + . . .$
The Taylor series expansion does in fact converge to the function $f_\epsilon^˜ (x) = 0$.
We can see that $f_\epsilon^˜ (x)$ and $f_\epsilon (x)$ are two different functions for |x|< $\epsilon$. 
But they must be equal since $f_\epsilon (x)$ is real analytic, which is only possible when $f_\epsilon (x) \equiv 0$.



*

*Is it enough to prove it. Personally I don't feel that its sufficient... 

*If I need to elaborate it by writing its Taylor series then calculations of derivatives is again a problem for me. I don't know how to write derivatives of $f_\epsilon (x)$. And in above proof, the way I approximate its Taylor series at $\epsilon$ looks quite tricky... isn't it? Please give me your suggestions and feel free to edit this proof. 


Thanks you so much. 
