Let $Y$ and $Z$ be two closed subspaces of a Banach space $X$ with $Y\cap Z=\{0\}$.


I know that $Y+Z$ is a closed subspace of $X$ $\iff \exists \alpha > 0:\quad \lVert y\rVert \le \alpha\lVert y+z\rVert \forall y∈Y,\forall z∈Z$.


However, reading this question [A criterion for the sum of two closed sets to be closed ?](https://mathoverflow.net/questions/88413/a-criterion-for-the-sum-of-two-closed-sets-to-be-closed), Bill Johnson [posted](https://mathoverflow.net/a/88438) that: the standard equivalence to the sum being closed is that the unit spheres of $Y$ and $Z$ are a positive distance apart i.e. $\exists r>0\quad \lVert y−z\rVert ≥r\quad \forall y\in Y\,\forall z\in Z\quad s.t.\quad \lVert y\rVert=\lVert z\rVert=1$.


Could anybody provide me with a proof or rather a reference to where I can see the proof of this equivalence?



Specific criterion for the sum of two closed sets to be closed
Let $\alpha>1$ such that
$$
\frac1{\alpha-1}=d(S(Y),S(Z)).
$$
Then for all $0\ne y\in Y$, $0\ne z\in Z$ we have
\begin{multline\*}
\frac1{\alpha-1}\le\left\lVert \frac{y}{\lVert y\rVert}-\frac{z}{\lVert z\rVert}
\right\rVert
\le \left\lVert \frac{y}{\lVert y\rVert}-\frac{z}{\lVert y\rVert}
\right\rVert
+
\left\lVert\frac{z}{\lVert y\rVert}-
\frac{z}{\lVert z\rVert}
\right\rVert
\\
=\frac{\lVert y-z\rVert}{\lVert y\rVert}
+\frac1{\lVert y\rVert}\Bigl\lvert\lVert z\rVert-\lVert y\rVert\Bigr\rvert.
\end{multline\*}
Replacing $z$ with $-z$ we get
$$
\lVert y\rVert
\le (\alpha-1) \lVert y+z\rVert +\Bigl\lvert \lVert y\rVert-\lVert z\rVert\Bigr\rvert
\le \alpha\lVert y+z\rVert,
$$
where in the last step we use the inverted triangle inequality.


The converse direction is trivial.


