Concave-modulus intersection of Hilbert balls

For a nonnegative function $\delta$ with $\delta (0)=0$, consider the finite weighted inequality ${\sum }_{i,j}{a}_i{a}_j[\delta (\mathrm{\parallel }{v}_i\mathrm{\parallel }{)}^2+\delta (\mathrm{\parallel }{v}_j\mathrm{\parallel }{)}^2-\delta (\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }{)}^2]\ge 0$ for Hilbert vectors $v_i$ and nonnegative scalars $a_i$. The function $\delta (t)=t$ satisfies it, since the sum is $2\mathrm{\parallel }{\sum }_i{a}_i{v}_i{\mathrm{\parallel }}^2$. Each truncated function ${\delta }_A(t)=\min \{t,A\}$, $A>0$, also satisfies it. Indeed, for the subsum with BOTH $\mathrm{\parallel }{v}_i\mathrm{\parallel },\mathrm{\parallel }{v}_j\mathrm{\parallel }<A$, replacing the negative truncated-distance square by the negative full-distance square can only decrease the sum, and this subsum is then $2\mathrm{\parallel }{\sum }_{\mathrm{\parallel }{v}_i\mathrm{\parallel }<A}{a}_i{v}_i{\mathrm{\parallel }}^2\ge 0$. Every remaining summand is separately nonnegative, because at least one of its two positive squares is $A^2$ while its negative square is at most $A^2$. The property is closed under pointwise limits, and under taking $\delta =({\sum }_{k=1}^N{c}_k{\delta }_k^2{)}^{1\mathrm{/}2}$ with $c_k\ge 0$, since the weighted expression is linear in ${\delta }^2$.

The function $\eta (t)=\omega (\sqrt{t}{)}^2$ is nondecreasing, continuous, zero at zero and concave. Here is a proof that covers nonsmooth moduli. For integers $n\ge 1$, interpolate $\omega$ on grids $0,1\mathrm{/}n,2\mathrm{/}n,\dots ,n$ and extend constantly after $n$; these concave nondecreasing polygonal functions converge pointwise to $\omega$. On each affine piece ${\omega }_n(r)=a+br$, its slope $b$ is nonnegative and its intercept $a$ is nonnegative because concavity and ${\omega }_n(0)=0$ imply ${\omega }_n(r)\ge r{\omega }_n^{\mathrm{\prime }}(r)$. Thus ${\omega }_n(\sqrt{t}{)}^2=a^2+2ab\sqrt{t}+b^{2}t$ is concave on that piece. At a junction its one-sided derivatives are ${\omega }_n(r)b\mathrm{/}r$, so the slope drops when $b$ drops. Continuity at zero then proves global concavity for ${\eta }_n$, and passing to the pointwise limit proves it for $\eta$. Interpolate $\eta$ on the same type of grids and flatten the final slope to zero. Every such polygonal ${\eta }_n$ is a finite sum ${\sum }_k{c}_k\min \{t,t_k\}$ with $c_k\ge 0$: take $c_k$ to be the decrease of slope at the breakpoint $t_k$. Hence $\sqrt{{\eta }_n(t^2)}=({\sum }_k{c}_k\min \{t,\sqrt{t_k}{\}}^2{)}^{1\mathrm{/}2}$ has step 1.1's property. Passing to the limit gives that property for $\omega (t)=\sqrt{\eta (t^2)}$.

First suppose the index set is finite and nonempty. If $\omega$ is identically zero, the pairwise hypothesis makes all the $y_i$ equal and their common value works. Otherwise $\omega (t)>0$ for every $t>0$: if $\omega (t_{\ast })=0$ with $t_{\ast }>0$, monotonicity makes it zero on $[0,t_{\ast }]$, and concavity bounds every subsequent slope by zero, forcing it to be identically zero. If some $x_i=x$, then $y=y_i$ works directly by the pairwise hypothesis. In the remaining case all $r_i=\mathrm{\parallel }{x}_i-x\mathrm{\parallel }$ and $b_i=\omega (r_i)$ are positive. In the finite-dimensional affine span of the $y_i$, minimize $h(y)={\max }_i\mathrm{\parallel }y-y_i{\mathrm{\parallel }}^2\mathrm{/}{b}_i^2$. A minimum exists because this continuous function tends to infinity with $\mathrm{\parallel }y\mathrm{\parallel }$ in that span; its perpendicular projection from $H_2$ never increases any distance. Let $y$ be a minimizer and let $J$ be its active set. The origin belongs to the convex hull of $(y-y_i)\mathrm{/}{b}_i^2$, $i\in J$: otherwise that finite convex hull has a nonzero nearest point $z$, whose variational inequality gives $⟨z,(y-y_i)\mathrm{/}{b}_i^2⟩\ge \mathrm{\parallel }z{\mathrm{\parallel }}^2>0$ for every active $i$. Moving a sufficiently small distance in direction $-z$ strictly decreases all active functions and keeps the finitely many inactive ones below the old maximum, a contradiction.

Choose nonnegative coefficients ${\beta }_i$, summing to one over $J$, with ${\sum }_{i\in J}{\beta }_i(y-y_i)\mathrm{/}{b}_i^2=0$, and normalize $a_i=({\beta }_i\mathrm{/}{b}_i^2)\mathrm{/}({\sum }_{j\in J}{\beta }_j\mathrm{/}{b}_j^2)$. Then $\sum a_i=1$ and $y=\sum a_i{y}_i$. Write $h(y)={\lambda }^2$; every active index has $\mathrm{\parallel }y-y_i{\mathrm{\parallel }}^2={\lambda }^2{b}_i^2$. The Hilbert variance identity gives ${\sum }_{i,j\in J}{a}_i{a}_j\mathrm{\parallel }{y}_i-y_j{\mathrm{\parallel }}^2=2{\sum }_{i\in J}{a}_i\mathrm{\parallel }{y}_i-y{\mathrm{\parallel }}^2=2{\lambda }^2{\sum }_i{a}_i{b}_i^2$. Apply the pairwise hypothesis and step 2.1's weighted inequality to $v_i=x_i-x$. They bound that same sum above by ${\sum }_{i,j}{a}_i{a}_j\omega (\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }{)}^2\le 2{\sum }_i{a}_i\omega (\mathrm{\parallel }{v}_i\mathrm{\parallel }{)}^2=2{\sum }_i{a}_i{b}_i^2$. The latter scalar is positive, so $\lambda \le 1$. The minimizer therefore belongs to EVERY target ball, proving finite intersection.

Now let $I$ be arbitrary and nonempty, and fix $i_0\in I$. For every finite $J\subset I$ containing $i_0$, the set $C_J={\bigcap }_{i\in J}\bar{B}(y_i,b_i)$ is nonempty by step 4.1, closed and convex. Let $p_J$ be its nearest point to zero by [F1]. Since $C_J$ lies in the fixed ball with center $y_{i_0}$ and radius $b_{i_0}$, the numbers $\mathrm{\parallel }{p}_J{\mathrm{\parallel }}^2$ have a finite supremum $D$. By [A1], for each integer $n\ge 1$, choose finite $J_n$ with $\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2>D-1\mathrm{/}n$ and replace them by their cumulative unions, preserving this inequality. They are increasing. For $m\ge n$, [F1] gives $\mathrm{\parallel }{p}_{J_m}-p_{J_n}{\mathrm{\parallel }}^2\le \mathrm{\parallel }{p}_{J_m}{\mathrm{\parallel }}^2-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\le D-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2$, so they form a Cauchy sequence and converge to a point $p\in H_2$.

For any fixed $i\in I$, let $q_n=p_{J_n\cup \{i\}}$. Again [F1] gives $\mathrm{\parallel }{q}_n-p_{J_n}{\mathrm{\parallel }}^2\le \mathrm{\parallel }{q}_n{\mathrm{\parallel }}^2-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\le D-\mathrm{\parallel }{p}_{J_n}{\mathrm{\parallel }}^2\to 0$. Hence $q_n\to p$; since every $q_n$ lies in the closed ball $\bar{B}(y_i,b_i)$, so does $p$. This works for each $i$ without enumerating $I$, and proves the required full intersection. If $I$ is empty, zero belongs to its intersection by convention. This proves all cases.