Is infimum achieved? Suppose $V$ is a Banach space, and $V_0$ is a closed subspace of $V$. I know that the following quantity is well-defined for every $w\in V\sim V_0$:
$$\inf\{\|w+v_0\|~|~v_0\in V_0\}$$
Is this infimum ever achieved by a particular $v_0$? If so, why?
 A: No. This is true if $V_0$ is a finite-dimensional subspace, or if $V$ is a Hilbert space, but not in general. The following is a counter-example :
Define
$$
V := c_0 := \{(x_n) \in \ell^{\infty} : \lim x_n = 0\}
$$
with the induced (supremum) norm from $\ell^{\infty}$. And let
$$
V_0 = \{ (x_n) \in c_0 : \sum 2^{-n} x_n = 0 \}
$$
I claim that $V_0$ does not have this "best approximation property".


*

*Let $y = (y_n) \in V\setminus V_0$, and set $\lambda := \sum 2^{-n} y_n \neq 0$. Then consider the elements
$$
x^1 := y - \frac{2}{1}\lambda (1,0,0,0, \ldots )
$$
$$
x^2 := y - \frac{4}{3}\lambda (1,1,0,0,0, \ldots )
$$
$$
x^3 = y - \frac{8}{7} \lambda (1,1,1,0,0, \ldots )
$$
and so on, then
$$
\|y - x^n\| = \left ( 1 - 2^{-n}\right )^{-1} |\lambda | \to |\lambda |
$$
Hence, $d(y, V_0) \leq |\lambda |$

*However, for any $x = (x_n) \in V_0$, then
$$
|\lambda | = \left | \sum 2^{-n} (y_n - x_n) \right | \leq \sum 2^{-n} | y_n - x_n |
$$
But, since $y\neq x$, and $y_n, x_n \to 0$, there is $n_0 \in \mathbb{N}$ such that
$$
|y_n - x_n | < \|y-x\| \quad\forall n\geq n_0
$$
Hence,
$$
| \lambda | < \sum 2^{-n} \|y-x\| < \|y-x\|
$$
Hence, there is no $x\in V_0$ such that $\|y-x\| = | \lambda |$.


There is also an interesting paper by R.R. Phelps (see this), which connects this best approximation property to a dual "unique hahn-banach extension" property, which might be worth a look.
