Let $X$ be a Banach space and $Y$ a closed subspace of $X$. If $\varphi\in Y^*$, then Hahn-Banach allows us to extend $\varphi$ to a $\tilde\varphi\in X^*$, such that $\|\tilde\varphi\|=\|\varphi\|$. This extension can happen in infinitely many ways, in general.

My question is the following: Can we guarantee the existence of a bounded linear transformation $$ F : Y^*\to X^*, $$ such that $F(\varphi)$ is an extension of $\varphi\in Y^*$ as a bounded linear functional on $X$? (Preferably with $\|F\|=1$.

First unsuccessful attempt: Define $F$ on a (Hamel) basis of $Y^*$, and then linearly extend to the whole of $Y^*$. But this $F$ is not necessarily bounded.

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. If $\varphi\in Y^*$, then Hahn-Banach allows us to extend $\varphi$ to a $\tilde\varphi\in X^*$, such that $\|\tilde\varphi\|=\|\varphi\|$. This extension can happen in infinitely many ways, in general.

My question is the following: Can we guarantee the existence of a bounded linear transformation $$ F : Y^*\to X^*, $$ such that $F(\varphi)$ is an extension of $\varphi\in Y^*$ as a bounded linear functional on $X$? (Preferably with $\|F\|=1$.)

First unsuccessful attempt: Define $F$ on a (Hamel) basis of $Y^*$, and then linearly extend to the whole of $Y^*$. But this $F$ is not necessarily bounded.