If every $x\in X$ is uniquely $x=y+z$ then $\|z\|+\|y\|\leq C\|x\|$

Question

Problem Statement

Let $(X,\|\cdot \|)$ be a Banach space and $Y$, $Z$ closed subspaces of $X$. If every $x \in X$ can be uniquely represented as $x=y+z$ for $y \in Y$ and $z \in Z$ then show that there exists $c$ such that $\|y\|\leq C\|x\|$ and $\|z\| \leq C\|x\|$

Attempt

If $L:X \rightarrow Y$ given by $L(x)=L(y+z)=y,$ then we will be done if we can show that $L$ is a bounded linear operator. Alternatively, by adding the two inequalities, we can show that the norm $\|\cdot\|$ is equivalent to $\| \cdot \|_{Y\times Z}$, where $\|(y,z)\|=\|y\|+\|z\|$. I attempted the first approach via the closed graph theorem but I am unable to make any headway.

Answer 1

Let $W:= Y \oplus Z$, which is a Banach space when endowed with the norm $\|(y,z)\|= \|y\|+\|z\|$.

Define a map from $W$ to $X$ by sending $(y,z) \mapsto y+z$. Note that this map is injective and surjective by the given conditions. Moreover this map is bounded because clearly $\|y+z\|\leq \|y\|+\|z\|$. By the open mapping theorem, the inverse map is bounded as well, so that $\|y\|+\|z\|\leq C\|x\|$.