Jtdylktuy

Let,
$C(Bbb R):=\
{f:Bbb R to Bbb R| text{ $f$ is continuous and $exists$ a compact set $K$ such that $f(x)=0$, $forall xin K^c$}}.$
Let, $displaystyle g(x)=e^{-x^2}$ for all $xin Bbb R$.
Then prove that there exists a sequence ${f_n}$ in $C(Bbb R)$ such that $f_n to g$ uniformly.

Here $g$ is continuous, so by Weierstress approximation theorem we can say that there always exists a sequence of polynomials ${p_n(x)}$ in $K$ which converges uniformly to $g$ in $K$. But in that case how can I define $f_n(x)$ in such a way that $f_n in C(Bbb R)$ ?

Since $g$ tends to $0$ as $xrightarrowinfty$ you can just take $f_n = gphi_n$
where $phi_n$ is a smooth (for your questions continuous is sufficient) cutoff function such

$phi_n(x)=1$ on $[-n,n]$, $phi_n(x)=0$ on $(-infty,-n-1]cup[n+1,infty)$, and $0le phi_n(x) le 1$ elsewhere.

Let $f_n(x) = begin{cases} g(x), & |x| le n \
g(n)(n+1-|x|), & n<|x| le n+1 \ 0, & text{otherwise}
end{cases}$ (I am using the fact that $g$ is even, but it is clear how this would work for any $g$ that satisfies $g(x) to 0$ as $|x| toinfty$.

Then $|g-f_n|_infty le sup_{|x| ge n} g(x) = g(n)$, hence the convergence is uniform since $lim_{n to infty} g(x) = 0$.