Suppose $(X,\Vert\cdot\Vert_X)$ is a Banach space, $Y$ is a subspace of $X$ such that $Y^\perp = \{0\}$, and suppose $Y$ is endowed with its own norm $\Vert\cdot\Vert_Y$ such that $B_Y \subsetneq B_X \cap Y$, so that in particular $X^* \subset Y^*$. Let $M$ be a subspace of $X^*$ and $x \in X$ is such that $B_{Y^*} \cap M \subset x^{-1}(-1,1)$ (where $x$ is understood as a linear functional on $X^*$), can we then assert that there exists $y \in Y$ such that $\Vert y\Vert_Y \leq 1$ and $y|_M = x|_M$?
Thanks for any input.
You can lift $x|_M$ to $Y^{**}$ but not necessarily to $Y.$ For a counterexample consider the embedding of $Y=C[0,1]$ in $X=L^1[0,1]$ with $M=X^*$ and $x=\tfrac 12 1_{[0,1/2]}.$ Functionals in $B_{Y^*}\cap X^*$ correspond to bounded functions with $L^1$ norm at most $1,$ so are in $x^{-1}(-1,1),$ but $x$ is not equal to anything in $Y.$