Minimum distance between two subsets of two closed linear subspaces $M,N$ of a Hilbert space $H$ ($H = M \oplus N$)

Question

(The norm below is induced by the inner product defined on $H$)

Let $H$ be a Hilbert space, and let $S$ be unit sphere in $H$ (i.e., $S = \{x \in H : ||x|| = 1\}$). Suppose that $H = M \oplus N$ with $M,N$ being nonzero closed linear subspaces. Show that:

1. there is a minimum distance $d>0$ between the disjoint closed sets $S\cap M$ and $S \cap N$;
2. for any unit vectors $x\in M$ and $y \in N$, $\DeclareMathOperator*{\realpart}{Re}\realpart \langle x,y\rangle \le \alpha \triangleq 1 - {d^2\over 2}$;
3. for any $x \in M$ and $y \in N$, $|\langle x,y \rangle| \le \alpha ||x|| \cdot ||y||$.

In 1, by a minimum distance, I mean $\exists x_M^*\in S\cap M,x_N^* \in S\cap N $ s.t. $\forall x_M \in S\cap M, x_N \in S\cap N$, we have $||x_M^* - x_N^*|| \le ||x_M-x_N||$

Answer 1

Part (1):

The map $\varphi \colon H \times H \to \mathbb{R}$ defined by $$ \varphi(x, y) \colon= \lVert x-y \rVert $$ is continuous and non-negative-valued. So the subset $$ \big\{ \ \lVert x-y \rVert \ \colon \ x \in S \cap M, y \in S \cap N \ \big\} = \big\{ \ \varphi(x, y) \ \colon \ x \in S \cap M, y \in S \cap N \ \big\} $$ of $\mathbb{R}$ is bounded below; let $d$ be the infimum of this set. Then $d \geq 0$ of course. And also $d \leq \varphi(x, y)$ for all $x \in S \cap M$ and for all $y \in S \cap N$.

From here if we can show that $S \cap M$ and $S \cap N$ are compact, then our conclusion will follow. But can we?

Let us assume that $d = 0$.

Then for any $n \in \mathbb{N}$, there are points $x_n \in S \cap M$ and $y_n \in S \cap N$ for which $$ 0 \leq \varphi\left(x_n, y_n\right) < \frac{1}{n}; \tag{1}$$ let us put $$ z_n \colon= x_n - y_n. $$ Then from (1) it is obvious that the sequence $\left(z_n \right)_{n \in \mathbb{N}}$ converges to the zero vector $\mathbf{0}$ in $H$.

In this way we obtain a sequence $\left( x_n \right)_{n \in \mathbb{N}}$ in $S \cap M$ and a sequence $\left( y_n \right)_{n \in \mathbb{N}}$ in $S \cap N$; both these sequences are ultimately sequences in the complete metric space $H$ and if we can show that these sequences are Cauchy, then it will follow that these are convergent to some points $x$ and $y$, respectively, in $H$. Then $$ \lim_{n \to \infty} \varphi\left( x_n, y_n \right) = \lim_{n \to \infty} \left\lVert x_n - y_n \right\rVert = \lVert x-y \rVert = \varphi(x, y). \tag{2}$$ Upon letting $n \to \infty$, from (1) and (2) we obtain $$ d = \varphi(x, y) = \lVert x-y \rVert. \tag{3} $$ Finally, as $S\cap M$ (and $S \cap N$) is a closed set and as $\left( x_n \right)_{n \in \mathbb{N}}$ is a sequence in $S \cap M$ (and $\left( y_n \right)_{n \in \mathbb{N}}$ is a sequence in $S \cap N$), so $x \in S \cap M$ (and $y \in S \cap N$). But as $S \cap M$ and $S \cap N$ are disjoint, so we can also conclude that $x \not\in S \cap N$ and $y \not\in S\cap M$, which implies that $x \neq y$ and hence $\lVert x-y \rVert > 0$, which together with (3) above implies that $d > 0$, as required.

Part (2):

For any points $x \in S \cap M$ and $y \in S\cap N$, we have $$ \lVert x - y \rVert \geq d > 0. $$

So if $x$ is a unit vector in $M$ and $y$ is a unit vector in $N$, then $x \in S \cap M$ and $y \in S \cap N$ and hence we have $$ \Re \langle x, y \rangle = \frac{1}{4} \left( \lVert x + y \rVert^2 - \lVert x-y\rVert^2 \right) \leq \frac{1}{4} \left( \left( \lVert x \rVert + \lVert y \rVert \right)^2 - d^2 \right) = \frac{1}{4} \left( (1 + 1)^2 - d^2 \right) = 1 - \frac{d^2}{4}. $$

Part (3):

For any $x \in M$ and $y \in N$, we have $x \neq \mathbf{0}$ and $y \neq \mathbf{0}$ and so $\lVert x \rVert > 0$ and $\lVert y \rVert > 0$, which implies that $\frac{1}{\lVert x \rVert}x \in S\cap M$ and $\frac{1}{\lVert y \rVert} y \in S\cap N$, and therefore we obtain $$ \left\langle \frac{1}{\lVert x \rVert}x, \frac{1}{\lVert y \rVert} y \right\rangle \leq 1 - \frac{d^2}{4}, $$ and so $$ \langle x, y \rangle \leq \left( 1 - \frac{d^2}{4} \right) \lVert x \rVert \, \lVert y \rVert. $$

Hope my explanations are clear enough. If not, please feel free to ask for clarification wherever you need to!

Answer 2

Proof of (1):

Let $d = \inf\limits_{x_M\in S \cap M \\x_N \in S\cap N}||x_M - x_N||$, then $\exists \{x_{M_n}\}_{n = 1}^\infty \in S\cap M$ and $ \{x_{N_n}\}_{n = 1}^\infty\in S\cap N$ s.t. $||x_{M_n} - x_{N_n}||\xrightarrow{n\to \infty}{} d$ . Since both $S \cap M$ and $S\cap N$ are compact, we can find convergent subsequences $\{x_{M_{n_p}}\}_{p = 1}^\infty$ and $\{x_{N_{n_p}}\}_{p = 1}^\infty$. Suppose $x_{M_{n_p}}\xrightarrow{p\to \infty}{}x_{M^*}\in S \cap M$ and $x_{N_{n_p}}\xrightarrow{p\to \infty}{}x_{N^*}\in S\cap N$. By continuity of norm, we know $||x_{M^*} - x_{N^*} ||= d$. Since $S\cap M$ and $S\cap N$ are disjoint, we have $x_{M^*} \ne x_{N^*}$, showing that $d > 0$.

Proof of (2):

Since $x \in M $ and $y \in N$ are unit, we have $x \in S \cap M$ and $y \in S\cap N $. Thus, $\begin{align} d^2 &\le ||x-y||^2 \\ &=||x||^2 + ||y||^2 - 2\DeclareMathOperator*{\realpart}{Re}\realpart\langle x,y \rangle \\ &= 2 - 2\realpart\langle x,y \rangle\end{align}$.

$\Rightarrow \realpart\langle x,y \rangle \le 1- {d^2\over 2}$

Proof of (3):

If $x=0$ or $y = 0$, $|\langle x,y \rangle| \le \alpha ||x|| \cdot ||y||$ surely holds.

Otherwise, it is equivalent to prove $|\langle x,y \rangle| \le \alpha$ where $x \in S\cap M$ and $y \in S \cap N$.

$\begin{align} d^2 &\le ||x-y||^2 \\ & = ||x||^2 - 2 \realpart\langle x,y\rangle + ||y||^2 \\ & = 2 - 2 \realpart\langle x,y\rangle\\ & \le 2 - 2|\langle x,y \rangle|\end{align}$.

$\Rightarrow |\langle x,y \rangle| \le 1 - {d^2\over 2}$.

$\DeclareMathOperator*{\QED}{Q.E.D.} \QED$