$\newcommand{\*}{{\ast}}$The following question came to my mind when I was reading "5.10 Weak Convergence" (in normed spaces and their duals) from David G. Luenberger's "Optimization by Vector Space Methods" (1969).
Suppose $V$ is any normed linear space that is not necessarily finite-dimensional, not necessarily complete, and not necessarily hosting an inner product. Suppose $V^{\*}$ is the normed dual of $V$, and $V^{\* \*}$ is the normed dual of $V^{\*}$. Suppose $y_1^{\*}$ and $y_2^{\*}$ are any elements in $V^{\*}$ which act identically on $V$ in the "pointwise sense". That is, $y_1^{\*}(x) = y_2^{\*}(x)$ for all elements $x$ in $V$. Does it follow that $z^{\* \*}(y_1^{\*}) = z^{\* \*}(y_2^{\*})$ for all elements $z^{\* \*}$ in $V^{\* \*}$? Does it also follow that $y_1^{\*} = y_2^{\*}$?
My hunch is that both answers are "yes", since the hypothesis of the question seems to fully characterize $y_1^{\*}$ and $y_2^{\*}$ as identical "functions" on $V$. I am not so sure however since $V^{\* \*}$ is generally "larger" than $V$. Would appreciate some help in form of a proof if the answer(s) is (are) "yes", or counterexamples if the answer(s) is (are) "no".
Thanks.