Let $X$ be a Banach space, $Y \subset X$ a closed subspace. Let $x^* \in X^*$ be such that $\|x^*\| = \sup_{x \in B_X} |x^*(x)| = 1$. Is it then true that $\|x^*|_Y\| = \sup_{y \in B_Y} |x^*(y)| = 1$ too? ($x^*|_Y$ is the restriction of $x^*$ to $Y$)
If we take $y \in B_Y$ then from $\|x^*\| = 1$ it follows $\|x^*|_Y\| \leq 1$. I do not see how to prove it is equal to one.
At the extreme, you can have $Y=\ker x^*$, and so $\|x^*|_Y\|=0$.
And anything in between is possible. Let $X=\mathbb C^2$, put $x^*(a,b)=a$. Then $\|x^*\|=1$. For any $t\in[0,1]$, if you take $Y=\mathbb C\,(t,\sqrt{1-t^2})$, then $$ \|x^*|_Y\|=t. $$