Convergence of a sequence of projections

Question

Let $C \subset \mathbb{R}^n$ be a compact, convex set, and $P \in \mathbb{R}^{n \times n}$ be a positive definite matrix ($P \succ 0$).

Consider the projection $\Pi_P: \mathbb{R}^n \rightarrow C$ defined as $$ \Pi_P(v) := \arg\min_{x \in C} \left\| x-v \right\|_P $$ for all $v \in \mathbb{R}^n$.

I am wondering if, for all $v \in \mathbb{R}^n$, it holds that $$ \lim_{\epsilon \rightarrow 0^{+}} \,\Pi_{P+ \epsilon I}\left( -\left( P+ \epsilon I \right)^{-1} v \right) = \Pi_P( -P^{-1} v) $$ where $I$ denotes the identity matrix.

Comments. It can be shown that $ \lim_{\epsilon \rightarrow 0^{+}} \,\Pi_{P}\left( -\left( P+ \epsilon I \right)^{-1} v \right) = \Pi_P( -P^{-1} v) $. I have shown the claim for the simpler case $P = p I$, $p>0$.

Answer 1

(I use $-v$ instead of $v$ in my answer to save some minus signs.)

The unique projections are characterized by the variational inequalities \begin{align*} \big\langle \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - c , \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - (P + \varepsilon \, I)^{-1} v \big\rangle_{P + \varepsilon \, I} &\le 0 \qquad\forall c \in C ,\\ \big\langle \Pi_{P} P^{-1} v - c , \Pi_{P} P^{-1} v - P^{-1} v \big\rangle_P &\le 0 \qquad\forall c \in C . \end{align*} Now, we use $c = \Pi_{P} P^{-1} v$ in the first inequality and $c = \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v$ in the second one.

Adding the resulting inequalities and switching to the Euclidean scalar product yields \begin{equation*} \big\langle \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - \Pi_{P} P^{-1} v , (P + \varepsilon \, I) \, \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - P \, \Pi_{P} P^{-1} v \big\rangle \le 0 . \end{equation*} Now, \begin{align*} &\big\lVert \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - \Pi_{P} P^{-1} v \big\rVert_P^2 \\ &\qquad+ \varepsilon \, \big\langle \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - \Pi_{P} P^{-1} v , \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v \big\rangle \le 0 . \end{align*} follows. Since the projections belong to $C$ and since $C$ is bounded, we obtain \begin{equation*} \big\lVert \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v - \Pi_{P} P^{-1} v \big\rVert_P \le \varepsilon \, \big\lVert \Pi_{P + \varepsilon \, I} (P + \varepsilon \, I)^{-1} v \big\rVert_{P^{-1}} \le \varepsilon \, L , \end{equation*} where $L$ depends on the diameter of $C$ and on the eigenvalues of $P$.

Answer 2

I asssume when you write $\|\cdot\|_P$, you're talking about the norm corresponding to the inner product $\langle x, y \rangle_P = x^T P y$. Thus $\Pi_R(-R^{-1} v)$ is the $x \in C$ that minimizes $$F_R(x) = \|x + R^{-1} v\|_R^2 - v^T R^{-1} v = x^T R x + 2 x^T v $$

Since $F_P(x)$ is strictly convex, it has a unique minimum on the compact convex set $C$, let's say at $x_0$. For any $\delta > 0$ there is $\eta > 0$ such that $F_P(x) > F_P(x_0) + \eta$ for $\|x - x_0\| > \delta$. As $\epsilon \to 0+$, $F_{P+\epsilon I} \to F_P$ uniformly on $C$, so for sufficiently small $\epsilon$ we will have $F_{P+\epsilon I}(x) > F_{P}(x_0) + \eta/2$ for $\|x - x_0\| > \delta$ while $F_{P+\epsilon I}(x_0) < F_P(x_0) + \eta/2$, and thus $\|\Pi_{P+\epsilon I}(-(P+\epsilon I)^{-1} v) - \Pi_P(-P^{-1} v)\| \le \delta$. Since this is true for all $\delta > 0$, we conclude that $$\lim_{\epsilon \to 0+} \Pi_{P+\epsilon I}(-(P+\epsilon I)^{-1} v) = \Pi_P(-P^{-1} v)$$