Specific criterion for the sum of two closed sets to be closed

Question

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 ?, Bill Johnson posted 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?

Answer 1

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.