Orthogonal projection of a point centrally-symmetric closed convex subset of $\mathbb R^n$ never expands the coordinates of the point

Question

Let $C$ be a closed convex subset of $\mathbb R^n$ which is symmetric about the standard coordinate axes. For example, think of $C$ as the unit-ball for an $\ell_p$-norm, for some $p \in [1,\infty]$. Let $x=(x_1,\ldots,x_n) \in \mathbb R^n$ and $y = (y_1,\ldots,y_n) \in C$ be the orthogonal projection of $x$ onto $C$.

Question. Is it true that $|y_i| \le |x_i|$ for all $i$ ?

The above claim "appears true", at least when one draws a diagram (of course this is itself not a proof).

My attempted proof

Fix an index $i \in [n]$. If $y_i = x_i$, there is nothing to show. Otherwise, suppose $x_i \ge 0$. Then because $x_i$ is symmetric about the coordinate axes, we must have $y_i \ge 0$ (see details further below). Also, we must $x-te_i \in C$ for sufficiently small positive $t$ (otherwise $y_i=x_i$). Here $e_i$ is the $i$th standard basis vector in $\mathbb R^n$. Now, by the Kolmogorov characterization of projections, we have

$$ 0 \ge (x-y)^\top (x-te_i-x) = -tx_i+ty_i, $$ that is, $0 \le y_i \le x_i$ as claimed. Similarly, if $x_i \le 0$, then use $x+te_i$ instead of $x-te_i$ to get $x_i \le y_i \le 0$. The result would then follow.

An omitted detail. Suppose $x_i$ and $y_i$ have opposite signs, i.e $x_i y_i < 0$, and let $z$ be obtained from $y$ by flipping the sign $i$th coordinate. Because $C$ is symmetric about the $i$th coordinate axis, it must contain $z$. Then, $$ \|z-x\|^2 - |y-x\|^2 = (-y_i-x_i)^2 - (y_i-x_i)^2 = (y_i+x_i)^2 - (y_i-x_i)^2 = 4x_i y_i < 0, $$ which contradicts the fact that $y$ is the point of $C$ which is closest to $x$.

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However, I'm not 100% about my proof above. Thanks in advance.

Answer 1

Fix $j \in \{1, ..., n\}$. As I understand your symmetry condition, for any $y \in C$, we have $y^{j, 0} \in C$ which is defined via $y^{j, 0}_i := y_i$ for $i \neq j$ and $y^{j, 0}_j = 0$. Then by the Pythagorean theorem, if $y$ is the projection of $x$ onto $C$, we get $$ \|y^{j, 0} - x \|^2 \geq \|y - y^{j, 0}\|^2 + \|x-y\|^2, $$ and solving this leads to $x_j y_j \geq y_j^2$ and thus the claim.