Can you project unto closed subspaces of normed spaces that are not necessarily pre-Hilbert? I'm working through some notes for my signal processing class and they introduce the whole notion of pre-Hilbert spaces (inner product spaces) essentially only in order to be able to project elements onto closed subspaces.
My question is: suppose $V$ is a general normed space (not necessarily derived from an inner product) and $U \subset V$ is a complete subset, can we not use the same proof to show that $\forall v \in V, \exists u \in U$ such that $||u-v||$ is minimal? In that case, the only thing the inner product does is to make the projection orthogonal, right?
 A: You can always project onto any subspace at all. Namely, given any $W\leq V$, you can choose an algebraic basis $B_1$ of $W$ and extend it to a basis $B=B_1\sqcup B_2$ of $V$. Then for any $v\in V, v=\sum_{b_1\in B_1} a_{b_1}b_1+\sum_{b_2\in B_2}a_{b_2}b_2$, you can put $F(v)=\sum_{b_1\in B_1} a_{b_1}b_1$, which gives you a projection $V\to W$, and any projection is of this form.
If $V$ is a topological vector space, e.g. a normed space, then such a projection will typically not be continuous, however, and $F(v)$ may not minimize the distance from $v$ to $W$ (even if $F$ happens to be continuous with respect to some given metric). In general, there may be no continuous projection at all, not even if $W$ is a closed (or complete, when it makes sense) subspace of $V$.
If $V$ is an arbitrary normed space, and you try to mimic the proof you mention, it does does not work: given any $v\in V$, even if you choose a sequence $w_n\in W$ such that $\lvert v-w_n\rvert<1/n+\inf_{w\in W}\lvert v-w\rvert$, the sequence $w_n$ need not be Cauchy. For example, if $V$ is the plane with the $L^\infty$ norm and $W$ is spanned by $(0,1)$, then for $v=(1,0)$, you can choose $w_n=(0,(-1)^n)$ and this is clearly not Cauchy, even though $\lvert v-w_n\rvert=1=\inf_{w\in W}\lvert v-w\rvert$. In this case, a distance-minimizing projection does exist, but it is not unique.
