let $V$ be the vector space of all sequences of real numbers which are eventually $0$, that is $ V=\{s=(a_1,a_2, . . .) | \exists N$ such that $a_N = a_{N+1} = · · · = 0\}$.
We know the natural linear map $V → V''$ is not an isomorphism, where $V''$ is the double dual of $V$, but I am not sure why. Could someone explain this to me, thanks a lot!
We have seen in your last question, that $V'$ can be identified with the space $\def\R{\mathbf R}\def\N{\mathbf N}\R^\N$ of all real sequences, via: $$ (b_n) \mapsto \sum_n a_n b_n, \quad a \in \p, b \in V $$ We will construct an element of $V''$, that is a linear map on $\R^\N$, that does not lie in the image of the natural map. Let $c \subseteq \R^\N$ denote the subspace of converging sequences, and let $\ell \in (\p)'$ be a linear map such that $\ell\colon c \ni (a_n) \mapsto \lim a_n$ (such a linear map exists by the axiom of choice: choose any basis of $c$, extend to a basis of $\R^\N$, define $\ell$ by linear extension). Suppose $\ell$ were represented by some $b \in V$. Let $e^k \in c$ denote the sequence $(\delta_{kn})_n$, then $$ \ell(e^k) = \lim_n \delta_{kn} = 0 $$ Hence $$ b_k = b(e^k) = \ell(e^k) = 0$$ So $b =0$, but $\ell\ne 0$. That is $\ell \in V''\setminus V$.