Let $X$ be Banach and $U$ a subspace. Then consider the forward annihilator $$U^{\perp}= \left\{ f \in X': f(u)=0 \ \forall u \in U \right\}$$
where $X'$ is the dual space. Is it true that for any $f \in X'$ there is $g \in U^{\perp}$ such that
$$\inf_{u \in U^{\perp}}\left\lVert u-f \right\rVert= \left\lVert u-g \right\rVert?$$
I only know that such minimizers can be found in Hilbert spaces and on compact spaces. I am not sure about this annihilator space, though.
We can prove this with Hahn-Banach. Fix some arbitrary $f\in X'$, which we can assume isn't in $U^\perp$. For any $u\in U^\perp$ we have \begin{align*} \|f-u\|&=\sup_{\|x\|\leq 1}|f(x)-u(x)| \\ &\geq \sup_{\substack{\|x\|\leq 1 \\ x\in U}}|f(x)-u(x)|\\ & = \sup_{\substack{\|x\|\leq 1 \\ x\in U}}|f(x)|\\ &=\|f|_U\|_U. \end{align*} This implies that $d(f,U^{\perp})\geq \|f\|_U$. Let us consider the restriction $f|_U$ of $f$ to $U$. Define $F$ to be the Hahn-Banach extension of $f|_U$ (so $F\in X^*$ again). We are guaranteed that $\|F\|=\|f|_U\|_U$. On $U$ we have that $g:=f-F=0$. Hence $g\in U^\perp$, and $\|f-g\|=\|f|_U\|$. Thus the distance between $f$ and $U^\perp$ is minimized by $g$.