I had shown that $f(x)=e^x$ is not tempered distribution.
It is a linear functional of $C_c^{\infty}$ but not on Schwarz space.
Clearly $C_c^{\infty}$ is subspace of S.
I do not have an idea to show rigorously above is not violating Hanh Banach theorem.
Any Help will be appreciated
Indeed, $f$ is a linear functional on the space of compactly supported functions but it is not continuous.
To see this, consider a sequence of functions from the Schwarz space, each supported on the interval $[n,n+1]$, whose all derivatives are at most $1/n$. Then such a sequence goes to 0 in the Schwarz space but when you hit it with $f$ it no longer does so.
As $f$ is discontinuous, it does not have a contonuous extension to the closure of the space of compactly supported functions. This has very little to do with the Hahn-Banach theorem anyway.