Projection of smallest norm This is related to the question:
Compactness of set of projections
If $X$ is a infinite dimensional Banach space, then any finite dimensional subspace $E$ is complemented, that's an immediate application of Hahn-Banach. The claim I read recently in a book is that we can always find a projection onto $E$ of smallest norm, and that follows by a " compactness argument". The answer is easy in the Hilbert space setting, just take the (norm one) orthogonal projection onto $E$. However, I don't see the argument in general. I thought that the set of projections would be compact, or at least a bounded set of projections would be compact, but that's not the case ( see the answer to the previous question). 
Thanks.
 A: (this proof, if correct, works for all complemented subspaces, finite-dimensional or not)
Let $Y\subset X$ be a finite-dimensional subspace. Let $\mathcal P(Y)$ be the set of bounded projections onto $Y$. The fact that $Y$ is finite-dimensional guarantees that $\mathcal P(Y)\ne\emptyset$. Then $\mathcal P(Y)$ is closed in the strong operator topology. As it is convex, it is also closed in the weak operator topology. 
Let $d=\inf\{\|P\|:\ P\in\mathcal P(Y)\}$. Then there is a sequence $\{P_n\}$ of projections onto $Y$ such that $\|P_n\|\to d$. By cutting $\mathcal P(Y)$ with a closed ball, big enough, we get a weak-operator compact set. Then we may replace $\{P_n\}$ with a weak-operator convergent sequence. Its limit $P$ will belong to $\mathcal P(Y)$, and $\|P\|=d$. 
A: Maybe you could put a kind of weak* topology on the space of operators from $X$ to $E$, that is, there is a base of neighborhoods of the zero operator of the form $N(x,r) = \{T:X \to E | \|T(x)\| \le r\}$ for $x \in E$, $r > 0$.  In fact, this could be regarded as a closed subset of $(X^*)^{\text{dim}(E)}$.  Then the subset of projections will be closed.  And you can intersect this with a large enough unit ball so that it is not empty.
