What will be the support of the convolution of two test functions. If $g\in C^{\infty}_c$ defined on $\Bbb R^n$ and K is the support of function $g$. I want to find the support of $g_\epsilon$. Where $g_\epsilon$ is regularization of $g$.

Regularization of $g$ is defined as:
$g_\epsilon$:=  $g*\omega_\epsilon$ (convolution of g and a test function) i.e. $g_\epsilon(x
)$=$\int_{\Bbb R^n}g(x-y).\omega_\epsilon(y)dy $
I think, this process is also known as mollification. 
Here 
$ \omega_\epsilon (x)= 
\begin{cases}
\frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}},  &\text{for |x|< $\epsilon$ } \\
0, & \text{otherwise}  \\
\end{cases} $ 

I only know that support of convolution of two compactly supported functions is again a compact set. And, as support of $\omega_\epsilon $ is a ball $B(0,\epsilon)$ and support of g is a compact set K(given) then support of their convolution should be the intersection of the support of $g(x-y)$ and $\omega_\epsilon(y)$. 
Because for non zero function integration remains non zero... 
 A: Reminder: The sum of two compact sets, e.g. K an the closed ball $M := B_\varepsilon(0)$ is compact again, as it is the continuous image of a
compact set $K \times M$ in the product domain. Therefore, if both supports are compact you can omit the closure on the right hand side. 
A: Definition: Let $f:\mathbb{R}^N\to\mathbb{R}$ be a function. Consider the family $\omega_i$ of all open sets on $\mathbb{R}^N$ such that for each $i$, $f=0$ a.e. in $\omega_i$, then $f=0$ a.e. on $\omega=\cup\omega_i$ and we define the support of $f$ by $$\operatorname{supp}(f)=\mathbb{R}^N\setminus\omega$$
Suppose that $f,g\in C^\infty(\mathbb{R})\cap L^1(\mathbb{R})$. We can prove that $\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$.
Take any $x\in\mathbb{R}$ and note that $$(f\ast g)(x)=\int_\mathbb{R}f(x-y)g(y)=\int_{(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)}f(x-y)g(y)$$
For $x\notin \operatorname{supp}(f)+\operatorname{supp}(g)$, we have that $(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)=\emptyset$ and then $(f\ast g)(x)=0$. It follows that if $x\in (\operatorname{supp}(f) + \operatorname{supp}(g))^c$ then $$(f\ast g)(x)=0$$ 
Therefore $$\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$$
Now let's prove that the equality is not necessary. Consider the functions in $\mathbb{R}$ defined by $f(x)=x\chi_{[-1,1]}(x)$ and $g(x)=\chi_{[-2,2]}(x)$, where $\chi_A$ is the characteristic function of the set $A$. 
Note that in this case $\overline{\operatorname{supp}(f)+\operatorname{supp}( g)}=[-3,3]$, however, the interval $(-1,1)$ does not belong to the support of $f\star g$ (which is equal to $[-3,-1]\cup [1,3]$).
Remark 1: The proof I gave here can be found in Brezis's book of Functional Analysis.
Remark 2: I don't know a expliclty characterization of the support of the convolution, but by the given formula, you can see that if the two functions has compact support, then does the convolution.
Update: I have corrected some errors in the text.
