\documentclass[11pt]{amsart} \pdfoutput=1 \usepackage[usenames,dvipsnames]{xcolor} % \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[letterpaper,margin=1in]{geometry} \usepackage{lmodern} \usepackage{microtype} \usepackage{etoolbox} % \usepackage{lettrine} \usepackage{mathtools} \usepackage{amsmath,amssymb,amsfonts,amsxtra,bm} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{latexsym} \usepackage{nicefrac} % \usepackage{booktabs} \usepackage{makecell,multirow} % \usepackage{colortbl} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{float} \floatstyle{plaintop} \restylefloat{table} \usepackage{footnote, tablefootnote} \usepackage{enumitem} \usepackage{algorithm} \usepackage{algorithmic} \usepackage{comment} \usepackage{xspace} \usepackage{url} \usepackage[most]{tcolorbox} \usepackage[round]{natbib} \usepackage{hyperref} \numberwithin{equation}{section} \makeatletter \definecolor{darkblue}{rgb}{0.0,0.0,0.65} \definecolor{darkred}{rgb}{0.68,0.05,0.0} \definecolor{darkgreen}{rgb}{0.0,0.29,0.29} \definecolor{darkpurple}{rgb}{0.47,0.09,0.29} \renewcommand{\topfraction}{0.85} \renewcommand{\bottomfraction}{0.4} \renewcommand{\textfraction}{0.1} \renewcommand{\floatpagefraction}{0.9} \patchcmd{\section}{\scshape}{\fontfamily{ppl}\bfseries\scshape\selectfont}{}{% \PackageWarning{aiproofs}{Could not bold the section font; amsart may have changed}} \newcommand{\aiauthor}{% \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}% \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany. Email: \texttt{s.sra@tum.de}.}} \makeatother \hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=red!65!black} \setlist{nosep,leftmargin=*} \renewcommand{\bibnumfmt}[1]{{\color{darkblue}\bf\textsc{#1}}.\hskip 6pt} \newenvironment{aiusage}{\section*{AI usage}}{} \newcommand{\aiacknowledgements}[1][]{% \section*{Acknowledgements} \if\relax\detokenize{#1}\relax\else #1\ \fi The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.% } \newcommand{\reals}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\pp}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\bmOne}{\bm{1}} \newcommand{\C}{\mathbb{C}} \newcommand{\Cc}{\mathcal{C}} \newcommand{\Dc}{\mathcal{D}} \newcommand{\Nc}{\mathcal{N}} \newcommand{\Xc}{\mathcal{X}} \newcommand{\Fc}{\mathcal{F}} \newcommand{\symm}{\mathfrak{S}} \newcommand{\ma}{\bm{A}} \newcommand{\mx}{\bm{X}} \newcommand{\my}{\bm{Y}} \newcommand{\mz}{\bm{Z}} \newcommand{\vx}{\bm{x}} \newcommand{\vy}{\bm{y}} \newcommand{\vv}{\bm{v}} \newcommand{\vz}{\bm{z}} \newcommand{\vr}{\bm{r}} \newcommand{\mb}{\bm{B}} \newcommand{\dpp}{\textsc{Dpp}} \newcommand{\etal}{et~al.\@\xspace} \newcommand{\half}{\tfrac{1}{2}} \newcommand{\kron}{\otimes} \newcommand{\inv}[1]{{#1}^{-1}} \newcommand{\nlsum}{\sum\nolimits} \newcommand{\nlprod}{\prod\nolimits} \newcommand{\ip}[2]{\langle {#1},\, {#2} \rangle} \newcommand{\iplr}[2]{\left\langle {#1},\, {#2}\right\rangle} \newcommand{\set}[1]{\left\lbrace#1\right\rbrace} \newcommand{\norm}[1]{\|{#1}\|} \newcommand{\bignorm}[1]{\left\lVert #1\right\rVert} \newcommand{\pnorm}[2]{\|#1\|_{#2}} \newcommand{\enorm}[1]{\bigl\|#1\bigr\|_{2}} \newcommand{\frob}[1]{\|#1\|_{\text{F}}} \newcommand{\fdel}[2]{\frac{\partial {#1}}{\partial {#2}}} \newcommand{\what}{\widehat} \newcommand{\ld}{\ell{}d} \newcommand{\gm}{\#} \newcommand{\nf}[2]{\nicefrac{#1}{#2}} \newcommand{\pfrac}[2]{\left(\tfrac{#1}{#2}\right)} \def\sint{\begingroup\textstyle\int\endgroup} % \DeclareMathOperator{\trace}{tr} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Diag}{Diag} \DeclareMathOperator{\real}{Re} \DeclareMathOperator{\imag}{Im} \DeclareMathOperator{\per}{per} \DeclareMathOperator{\Vol}{Vol} \DeclareMathOperator{\Imm}{Imm} \DeclareMathOperator{\mdet}{det} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\qjsd}{QJSD} \DeclareMathOperator{\qjrd}{QJRD} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator*{\argmax}{argmax} \newtheorem{theorem}{Theorem}[section] \newtheorem{conj}[theorem]{Conjecture} \newtheorem{prop}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corr}[theorem]{Corollary} \newtheorem{rmk}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{defn}[theorem]{Definition} \newtheorem{prob}[theorem]{Problem} \definecolor{cdarkred}{RGB}{180,70,30} % \definecolor{cdarkgreen}{RGB}{0,130,0} % \definecolor{azure}{RGB}{240,255,255} \definecolor{cmidgreen}{RGB}{201,219,216} \definecolor{clightgreen}{RGB}{200,255,200} \definecolor{clightgray}{RGB}{240,240,240} \definecolor{cmygreen}{RGB}{10,120,0} \definecolor{cmyblue}{RGB}{50,50,255} \definecolor{sblue}{RGB}{100,100,255} \definecolor{cmypurple}{RGB}{150,50,180} \definecolor{cmyred}{RGB}{238, 64, 0} \definecolor{cdarkblue}{RGB}{30,75,170} \definecolor{csharpred}{RGB}{255,0,0} \newcommand{\sred}[1]{{\color{csharpred} #1}} \newcommand{\darkred}[1]{{\color{cdarkred} #1}} \newcommand{\bred}[1]{\textbf{\sred{#1}}} \newcommand{\darkblue}[1]{{\color{cdarkblue} #1}} \newcommand{\bblue}[1]{\textbf{\darkblue{#1}}} \newcommand{\darkbred}[1]{\textbf{\color{cdarkred} #1}} \newcommand{\darkgreen}[1]{{\color{cmygreen} #1}} \newtcolorbox{dashedbox}[1][]{ nobeforeafter, colback=yellow!10!white, boxrule=0pt, enhanced jigsaw, top=2pt, left=2pt, right=2pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \newtcbox{\tcbmathbox}[1][]{ nobeforeafter, colback=yellow!30!white, boxrule=0pt, enhanced jigsaw, top=0pt, left=1pt, right=1pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \newcommand{\Hc}{\mathcal H} \newcommand{\Pc}{\mathcal P} \newcommand{\Ec}{\mathcal E} \newcommand{\eps}{\varepsilon} \newcommand{\one}{\mathbf 1} \DeclareMathOperator{\Ran}{Ran} \DeclareMathOperator{\Ker}{Ker} \allowdisplaybreaks \hypersetup{ pdftitle={On Lin's Arithmetic-Geometric Means Conjecture for Singular Values}, pdfauthor={Suvrit Sra} } \makeatletter \newcommand{\preprintslink}[2]{% \thanks{First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.}} \makeatother \begin{document} \title[On Lin's Arithmetic--Geometric Means Conjecture for Singular Values]{On Lin's Arithmetic--Geometric Means Conjecture\\ for Singular Values} \aiauthor \begin{abstract} Let $A$ and $B$ be positive semidefinite, and let $0\le t\le 1$. \citet{lin2016drury} conjectured that \begin{equation*} \lambda_j((1-t)A+tB) \ge \sigma_j^{1/2}(A^{2(1-t)}B^{2t}), \qquad j=1,\ldots,n. \end{equation*} The case $t=\frac12$ is Drury's theorem, which answered a long-standing question of Bhatia and Kittaneh. Lin remarked that Drury's method does not seem to reach other values of $t$. We prove his conjecture for every $t\in[0,1]$. The proof builds on Drury's argument, but replaces his block-matrix step (unique to the midpoint $t=\frac12$) by an eigenvalue count based on an inertia identity and a one-variable inequality. \end{abstract} \preprintslink{linconj}{8 October 2026} \maketitle \vspace*{-8pt} \section{Introduction}\label{sec:intro} Finding matrix analogues of scalar inequalities has long been a theme of matrix analysis. Many works study analogues of the arithmetic--geometric mean inequality; see, e.g., the books~\citep{bhatia1997matrix,bhatia2007psd,zhan2002book}. Some of the most useful of these inequalities compare singular values $\sigma(\cdot)$ and eigenvalues $\lambda(\cdot)$ of matrices. A fundamental example of such an inequality is \begin{equation}\label{eq:drury} \sigma_j^{1/2}(AB) \le \tfrac12\lambda_j(A+B),\qquad A, B \succeq 0, \qquad j=1,\ldots,n, \end{equation} which was raised as a question by~\citet{bhatia2000notes} and proved twelve years later by~\citet{drury2012question}. \citet{bhatia1990singular} proved the weaker inequality $\sigma_j(A^{1/2}B^{1/2})\le\frac12\lambda_j(A+B)$. This inequality was generalized by~\citet{ando1995matrix} to $\sigma_j(A^{1-t}B^{t})\le\lambda_j((1-t)A+tB)$ for $t\in [0,1]$. Revisiting Drury's proof, \citet[Conjecture~5.2]{lin2016drury} conjectured the corresponding weighted form of~\eqref{eq:drury}, and wrote that Drury's method \emph{``does not seem to lead to a solution.''} To our knowledge no proof has appeared since, and Theorem~\ref{thm:main} supplies one. \begin{theorem}[Lin's conjecture]\label{thm:main} Let $A,B\succeq 0$ and let $0\le t\le 1$. Then, for every $j=1,\ldots,n$, \begin{equation}\label{eq:1} \sigma_j^{1/2}(A^{2(1-t)}B^{2t}) \le \lambda_j((1-t)A+tB). \end{equation} \end{theorem} More broadly, both Ando's inequality and Theorem~\ref{thm:main} are members of the family \begin{equation}\label{eq:q-family} \sigma_j^{q}(A^{(1-t)/q}B^{t/q})\le\lambda_j((1-t)A+tB), \qquad j=1,\ldots,n, \end{equation} indexed by $q>0$. A brief calculation shows that if inequality~\eqref{eq:q-family} holds for an exponent $q$, then it also holds for every larger exponent, so the statements become stronger as $q$ decreases. The cases $q\ge1$ therefore follow from Ando's inequality, % while Lin's conjecture is the case $q=1/2$, outside that range. Theorem~\ref{thm:main} gives~\eqref{eq:q-family} for every $q\ge1/2$, and another brief calculation shows that at $t=1/2$ the exponent $1/2$ is the best. % \section{Proof of Theorem~\ref{thm:main}}\label{sec:proof} \noindent\emph{\textbf{Proof summary.}} The proof has four steps. The first two follow Drury's argument~\citep{drury2012question} exposited in \citet[Sections~2--4]{lin2016drury}. Write $a=1-t$ and $b=t$. In Step~1 we use homogeneity of~\eqref{eq:1} to reduce the task to proving $\lambda_j(aA+bB)\ge 1$. Step~2 reduces this task to an eigenvalue inequality for $aA+bB_0$, where $B_0 \preceq B$ is a rank-$j$ matrix. Drury's proof of that inequality works only at $t=1/2$, so we introduce Steps~3 and~4 to replace it. Step~3 counts eigenvalues using an inertia identity, which reduces the inequality to the sign of a quadratic form on a subspace; Step~4 settles this sign using an elementary scalar inequality. \subsection{Rescaling.} Using $a$ and $b$, inequality~\eqref{eq:1} reads \begin{equation}\label{eq:lin-target} \lambda_j(aA+bB)\ge\sigma_j^{1/2}(A^{2a}B^{2b}), \qquad j=1,\ldots,n. \end{equation} If $t=0$ or $t=1$ the inequality is trivial; so we assume $t\in(0,1)$. Also, it suffices to consider $A, B \succ 0$ (the $\succeq 0$ cases follow by continuity). We can also assume $a\le b$ (otherwise flip the roles of $A$ and $B$ and note $\sigma(AB)=\sigma(BA)$), so that $b \ge 1/2$. Fix $j\in\{1,\ldots,n\}$. If $\sigma_j(A^{2a}B^{2b})=0$, then~\eqref{eq:lin-target} is trivial; otherwise by homogeneity we may scale~\eqref{eq:lin-target} so that $\sigma_j(A^{2a}B^{2b})=1$, and reduce the task to proving that $\lambda_j(aA+bB)\ge1$. \subsection{Rank reduction.} By Weyl's monotonicity principle, lowering $B$ in the L\"owner order can only lower $\lambda_j(aA+bB)$. Following Drury, we lower $B$ to a matrix $B_0$ of rank~$j$ by keeping only the part of $B$ that produces the $j$ largest singular values of $A^{2a}B^{2b}$, and resetting those singular values to~$1$. More precisely, let $T=A^{2a}B^{4b}A^{2a}=(A^{2a}B^{2b})(A^{2a}B^{2b})^*$. Since $\lambda(T)=\sigma^2(A^{2a}B^{2b})$, the rescaling step tells us $\lambda_j(T)=1$. Choose an orthonormal basis $v_1,\ldots,v_n$ with $Tv_k=\lambda_k(T)v_k$, and let $E$ be the orthogonal projection onto $\text{span}(v_1,\ldots,v_j)$. Since $T\succeq0$ and $\lambda_k(T)\ge\lambda_j(T)=1$ for $k\le j$, \[ T-E=\sum_{k\le j}(\lambda_k(T)-1)v_kv_k^*+\sum_{k>j}\lambda_k(T)v_kv_k^*\succeq0. \] Thus, $T \succeq E$. Congruence with $A^{-2a}$ then yields \begin{equation}\label{eq:B0-def} B^{4b}=A^{-2a}TA^{-2a}\quad\succeq\quad B_0^{4b} :=A^{-2a}EA^{-2a}. \end{equation} Because $b\ge1/2$, the exponent $1/(4b)$ is at most $1$, so by L\"owner--Heinz on~\eqref{eq:B0-def} we obtain $B\succeq B_0$. Hence by Weyl $\lambda_j(aA+bB)\ge\lambda_j(aA+bB_0)$. \vskip4pt \noindent\emph{\textbf{Why such a choice of $B_0$?}} The point of this choice is that $B_0$ is determined by $A$ and by its own support projection $P$. Let $V=A^{2a}B_0^{2b}$. By~\eqref{eq:B0-def}, $VV^*=A^{2a}B_0^{4b}A^{2a}=E$, so $V$ is a partial isometry with final projection $E$. Since $A^{2a}$ is invertible, $\Ker V=\Ker B_0^{2b}=\Ran(I-P)$, so the initial projection of $V$ is $P$. In other words, \begin{equation}\label{eq:initial-projection} B_0^{2b}A^{4a}B_0^{2b}=P. \end{equation} The rank of $P$ is $j$, because $B_0^{4b}=A^{-2a}EA^{-2a}$ has rank $j$. Let $S=PA^{4a}P|_{\Ran P}$ be the compression of $A^{4a}$ to $\Ran P$, which is positive definite because $A$ is. Write $B_0=PXP$ with $X\succ0$ on $\Ran P$, and let $Y=X^{2b}$. Compressing~\eqref{eq:initial-projection} to $\Ran P$ gives $YSY=I_{\Ran P}$, and multiplying on the left and on the right by $Y^{-1}$ gives $S=Y^{-2}$. Since $Y\succ0$, uniqueness of positive square roots gives $Y=S^{-1/2}$, and therefore $X=S^{-1/(4b)}$. Thus \begin{equation}\label{eq:S-B0} B_0=PS^{-1/(4b)}P. \end{equation} It remains to prove \begin{equation}\label{eq:weighted-drury} \lambda_j(aA+bB_0)\ge1. \end{equation} Since~\eqref{eq:S-B0} expresses $B_0$ through $A$ and $P$ alone, we prove~\eqref{eq:weighted-drury} for every $A\succ0$ and every orthogonal projection $P$ of rank~$j$, with $B_0$ defined by~\eqref{eq:S-B0}. At $t=1/2$ the matrix $B_0=P(PA^2P)^{-1/2}P$ is Drury's matrix $B_1$~\citep[Section~2]{lin2016drury}, and~\eqref{eq:weighted-drury} is the inequality $\lambda_j(A+B_1)\ge2$ that Drury proves by an eigenvalue estimate for $2\times2$ block matrices. That estimate uses that $B_0$ is the inverse square root of $PA^2P$ on $\Ran P$, and we see no analogue of it for the fractional power $S^{-1/(4b)}$. \subsection{Eigenvalue count.} Inequality~\eqref{eq:weighted-drury} says that $aA+bB_0-I$ has at least $j$ nonnegative eigenvalues. This matrix is $H+bB_0$ with $H=aA-I$: a Hermitian matrix, in general indefinite, plus a positive semidefinite perturbation carried by the $j$-dimensional space $\Ran P$. The following inertia identity transfers such an eigenvalue count to a matrix that acts only on the space carrying the perturbation. It is the finite-dimensional form of the Birman--Schwinger principle~\citep{schwinger1961bound}. \begin{lemma}[Inertia identity]\label{lem:inertia} Let $H=H^*$ be invertible on a finite-dimensional Hilbert space $\Hc$. Let $J\colon\Pc\to\Hc$ be an isometric inclusion, and let $W\succ0$ on $\Pc$. Set \[ M=H+JWJ^*, \qquad G=W^{-1}+J^*H^{-1}J. \] Then \[ n_{\ge 0}(M)=n_+(H)+n_{\le 0}(G), \] where $n_{\ge0}$ counts eigenvalues $\ge0$, $n_+$ counts eigenvalues $>0$, and $n_{\le0}$ counts eigenvalues $\le0$, each with multiplicity. \end{lemma} \begin{proof} Consider the Hermitian matrix \[ \mathcal L= \begin{pmatrix} H & J \\ J^* & -W^{-1} \end{pmatrix} \quad\text{on }\Hc\oplus\Pc. \] The Schur complements in $\mathcal L$ of its blocks $-W^{-1}$ and $H$ are \[ H+JWJ^*=M \qquad\text{and}\qquad {-W^{-1}}-J^*H^{-1}J=-G. \] By inertia additivity for Schur complements~\citep{haynsworth1968determination}, $\mathcal L$ therefore has the inertia of both $M\oplus(-W^{-1})$ and $H\oplus(-G)$. As $-W^{-1}\prec0$ and $H$ is invertible, counting nonnegative eigenvalues gives $n_{\ge0}(M)=n_{\ge0}(\mathcal L)=n_+(H)+n_{\le0}(G)$. \end{proof} To apply Lemma~\ref{lem:inertia}, $H=aA-I$ must be invertible, that is, $1/a$ must not be an eigenvalue of $A$. By continuity we may assume so: for fixed $P$, the left side of~\eqref{eq:weighted-drury} depends continuously on $A\succ0$, and $1/a$ is an eigenvalue of $A+\eps I$ for only finitely many $\eps>0$. Let $m$ be the number of eigenvalues of $A$ greater than $1/a$, so that $n_+(H)=m$. Apply Lemma~\ref{lem:inertia} with $\Hc=\mathbb C^n$, $\Pc=\Ran P$, $J$ the inclusion map, and $W=bS^{-1/(4b)}\succ0$. Then $JWJ^*=bB_0$, so $M=aA+bB_0-I$, and since $(aA-I)^{-1}=-(I-aA)^{-1}$, \begin{equation}\label{eq:count} n_{\ge0}(aA+bB_0-I)=m+n_{\le0}(G), \qquad G=b^{-1}S^{1/(4b)}-P(I-aA)^{-1}P\big|_{\Ran P}. \end{equation} Thus~\eqref{eq:weighted-drury} holds as soon as $G$ has at least $j-m$ nonpositive eigenvalues. It is enough to find a subspace $U\subseteq\Ran P$ with $\dim U\ge j-m$ on which the quadratic form of $G$ is nonpositive. Indeed, for such a subspace $n_{\le0}(G)\ge\dim U$: otherwise the span $N$ of the eigenvectors of $G$ for eigenvalues $\le0$ would have $\dim N<\dim U$, so $U$ would contain a nonzero vector $v\perp N$, and $\ip{Gv}{v}>0$. The two terms of $G$ indicate where to look. The first is positive definite. The second, $-P(I-aA)^{-1}P$, is negative on vectors built from eigenvectors of $A$ with eigenvalues $x<1/a$, on which $(I-aA)^{-1}$ acts by the positive numbers $1/(1-ax)$. Let $\Fc$ be the span of the eigenvectors of $A$ with eigenvalues in $[0,1/a)$, and let $U=\Ran P\cap\Fc$. Because $1/a$ is not an eigenvalue of $A$, $\Fc$ has dimension $n-m$, and since $\Ran P$ has dimension $j$, \[ \dim U\ge j+(n-m)-n=j-m. \] It therefore remains to prove \begin{equation}\label{eq:form} \ip{Gu}{u}\le0\qquad\text{for every }u\in U. \end{equation} \subsection{Scalar inequality.} By homogeneity it suffices to prove~\eqref{eq:form} for a unit vector $u\in U$. Let $\mu$ be the spectral measure of $A$ associated with $u$, that is, the probability measure on $\mathbb R$ with $\ip{f(A)u}{u}=\int f\,d\mu$ for every function $f$ on the spectrum of $A$. Since $u\in\Fc$, the measure $\mu$ is supported in $[0,1/a)$. Since $u\in\Ran P$, \[ \ip{Su}{u}=\ip{A^{4a}u}{u}=\sint x^{4a}\,d\mu(x) \qquad\text{and}\qquad \ip{P(I-aA)^{-1}Pu}{u}=\sint\frac{d\mu(x)}{1-ax}. \] Put $\gamma=1/(4b)$; because $b\ge1/2$, we have $0<\gamma\le1$. Concavity of $x\mapsto x^\gamma$ and the spectral measure of $S$ associated with $u$ give $\ip{S^\gamma u}{u}\le\ip{Su}{u}^\gamma$. By~\eqref{eq:count}, therefore, \begin{equation}\label{eq:form-bound} \ip{Gu}{u} \le b^{-1}\left(\sint x^{4a}\,d\mu(x)\right)^{\gamma} -\sint\frac{d\mu(x)}{1-ax}. \end{equation} So~\eqref{eq:form} follows once the right side of~\eqref{eq:form-bound} is shown to be nonpositive for every probability measure $\mu$ on $[0,1/a)$. Lemma~\ref{lem:scalar-resolvent} proves this bound by integrating an inequality in one real variable; its proof is in Appendix~\ref{app:scalar}. % \begin{lemma}[Scalar inequality]\label{lem:scalar-resolvent} Let $01$, \[ g''(y)=4a^{1-4a}\,\frac{(1-y^{-1})^{4a}(2y-1-4a)}{y^2(y-1)^2}, \] so $g$ is concave on $[1,y_*]$ and convex on $[y_*,\infty)$, where $y_*=(1+4a)/2$ (convex throughout if $y_*\le1$). Since $2-(1+4a)(1-a)=4a^2-3a+1>0$, we have $y_*