Property of linear projections Theorem 15 in Chapter 15 of Peter Lax's functional analysis book says
$X$ is a Banach space, $Y$ and $Z$ are  closed subspaces of $X$ that complement each other $X = Y \oplus Z$, in the sense that every $x\in X$ can be decomposed uniquely as $x = y+z$ where $y\in Y$, $z\in Z$. Denote the two complements of $y$ and $z$ of $x$ by $y = P_Y x$, $z = P_Z x$. Then
1) $P_Y$ and $P_Z$ are linear maps of $X$ on $Y$ and $Z$ respectively.
2) $P_Y^2 = P_Y$, $P_Z^2 = P_Z$, $P_YP_Z = 0$.
3) $P_Y$ and $P_Z$ are continuous.
However, I would like to say that for all $\hat y\in Y$, 
$\|y-x\| \leq \|\hat y - x\|$.
which is a property of projections we already know, but doesn't seem to be built from the definition of projections given by Lax. I'm not too familiar with Banach spaces, so I don't really want to use this property until I know for sure it is implied by the previous properties. 
Can anyone give me some intuition if 1) this property is implied by the definitions, 2) this property may not hold in some special case (Not sure if $X$ is reflexive), or 3) this is implicit in the definition of projection and I'm just reading the text wrong? 
Thanks!
Edit: Just to add clarification, I'm considering any complemented Banach space (infinite dimensional). Not necessarily a Hilbert space, and no indication that it is reflexive.
 A: This property doesn't necessarily hold, and for counterexample you can take $X=\mathbb{R}^2$ as a Banach space.
The above $P_Y,P_Z$ are linear projections, and they have the desired property if and only if they are orthogonal projections.
Assume that $P_Y,P_Z$ are orthogonal projections, in other words $Y$ is orthogonal to $Z$. Then for all $\hat{y}\in Y$, $$\|\hat{y}-x\|^2=\|\hat{y}-y\|^2+\|y-x\|^2\geq\|y-x\|^2$$by Pythagoras. 
Remark: For the term orthogonal to be relevant, we need to assume that $X$ is Hilbert, not only Banach.
A: Let $Y= \operatorname{sp} \{ (1,0)^T \}$, $Z= \operatorname{sp} \{ (1,1)^T \}$. 
Then $P_Y(x) = (x_1-x_2) (1,0)^T$, $P_Z(x) = x_2 (1,1)^T$.
Pick $x=(0,1)^T$, which gives $y=(-1,0)$, $z=(1,1)^T$.
Then
$\|y-x\| = \|z\| = \sqrt{2}$, but with $\hat{y} = (0,0)^T$, we have
$\|\hat{y} -x\| = 1 < \sqrt{2}$.
Addendum:
Here is an example where $P_Y P_Z = P_Z P_Y = 0$:
Let $Y=\operatorname{sp} \{ (1,0)^T \}$, $Z=\operatorname{sp} \{ (0,1)^T \}$, then
$P_Y(x) = (x_1,0)^T$, $P_Z(x) = (0, x_2)^T$.
Let $\|x\|_* = \|(x_1-x_2, x_2)\|_2$.
Choose $x=(1,1)^T$, then $z = P_Z(x) = (0,1)^T$, so
$\|y-x\|_*=\|z\|_* =\sqrt{2}$. If we let $\hat{y} = (0,0)^T \in Y$, then
$\|\hat{y}-x\|_* = 1 < \sqrt{2}$.
