functional analysis complementary subspace Let $Y$ and $Z$ be closed subspaces in a Banach space $X$. Show that each $x \in X$ has a unique decomposition $x = y + z$, $y\in Y$, $z\in Z$ iff $Y + Z = X$ and $Y\cap Z = \{0\}$. Show in this case that there is a constant $\alpha>0$ such that $ǁyǁ + ǁzǁ \leq\alphaǁxǁ$ for every $x \in X$
 A: Hint: Assume that $Y\cap Z=\{0\}$. If $z+y=x=z'+y'$ s.t. $z,z'\in Z$, $y,y'\in Y$ then $z-z'=y'-y$.
For the other direction, if $y\in Y\cap Z$ then $0+0=y+(-y)$.
A: I think that the constant part requires the Open Mapping Theorem. 
Consider the space $Y\oplus Z$ with the norm $\|y\oplus z\|_1=\|y\|+\|z\|$. It is easy to see that it is a Banach space. Then we define the map $T:Y\oplus Z\to X$ given by $T(y\oplus z)=y+z$. This map is clearly linear and bijective. Moreover,
$$
\|T(y\oplus z)\|=\|y+z\|\leq\|y\|+\|z\|=\|y\oplus z\|_1,
$$
so $T$ is bounded. By the Open Mapping Theorem $T$ is open, which means that $T^{-1}$ is bounded. So, given $x\in X$ with $x=y+z$, $y\in Y$, $z\in Z$, there exists $\alpha=\|T^{-1}\|>0$ such that
$$
\|y\|+\|z\|=\|y\oplus z\|_1=\|T^{-1}(x)\|_1\leq\alpha\|x\|.
$$
For the sake of completeness, here is a proof for the first part of the question. Note that both sides of the implication have the assertion $X=Y+Z$, so what we have to prove is 
$$
\mbox{unique decomposition }\iff\ Y\cap Z=\{0\}.
$$
So assume first that $X=Y+Z$ with unique decomposition, and let $w\in Y\cap Z$. By the decomposition, $w=y+z$, $y\in Y$, $z\in Z$. But as $w\in Y$, we get $z=w-y\in Y$. So $w=(y+z)+0$ is another decomposition of $w$; by the uniqueness, $z=0$. A similar argument shows that $y=0$, and thus $w=0+0=0$. This shows that $Y\cap Z=\{0\}$. 
Conversely, suppose that $Y\cap Z=\{0\}$. If $x=y_1+z_1=y_2+z_2$ with $y_1,y_2\in Y$, $z_1,z_2\in Z$, then we have $y_1-y_2=z_2-z_1\in Z$, so $y_1-y_2\in Y\cap Z$ and $y_1=y_2$. A similar argument shows that $z_1=z_2$. So the decomposition is unique. 
