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$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Remark 1:

Let \begin{equation*} g(s):=q^T(P+sI)^{-2}q \end{equation*} for real $s$ such that $P+sI$ is invertible. The function $g$ is nonnegative and nonincreasing on the interval $I_P:=(-\la_{\min},\infty)$, and $g(\infty-)=0$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. Moreover, the equation \begin{equation*} g(s)=1 \tag{0}\label{0} \end{equation*} $s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots. So,

• if $q\ne0$, then equation \eqref{0} has only one nonnegative root, so that this root is the largest real root of \eqref{0};

• if $q=0$, then equation \eqref{0} has no real roots.

Let now \begin{equation*} f(z):=\frac12\,z^T Pz+q^T z+r \end{equation*} for $z\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation*} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation*} where \begin{equation*} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation*} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation*} -a^T x=|a|. \tag{3}\label{3} \end{equation*} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation*} x=-(P+tI)^{-1}q, \tag{4}\label{4} \end{equation*} whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the largest root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$.

Case 2: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if and only if $|x|=1$. So, in view of Remark 1, in this case your claim about the minimizer will hold if and only if $q\ne0$ and $|x|=1$. The latter condition will not of course hold in general even if $q\ne0$: e.g., if $n=1$ and $f(z)=z^2+z$, then $x=-1/2$ and $|x|<1$.

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, your claim about the minimizer $x$ will hold if and only if $q\ne0$ and $|x|=1$.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.

$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Remark 1:

Let \begin{equation*} g(s):=q^T(P+sI)^{-2}q \end{equation*} for real $s$ such that $P+sI$ is invertible. The function $g$ is nonnegative and nonincreasing on the interval $I_P:=(-\la_{\min},\infty)$, and $g(\infty-)=0$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. Moreover, the equation \begin{equation*} g(s)=1 \tag{0}\label{0} \end{equation*} $s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots. So,

• if $q\ne0$, then equation \eqref{0} has at most one nonnegative root, and then this root is the largest real root of \eqref{0};

• if $q=0$, then equation \eqref{0} has no real roots.

Let now \begin{equation*} f(z):=\frac12\,z^T Pz+q^T z+r \end{equation*} for $z\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation*} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation*} where \begin{equation*} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation*} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation*} -a^T x=|a|. \tag{3}\label{3} \end{equation*} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation*} x=-(P+tI)^{-1}q, \tag{4}\label{4} \end{equation*} whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the largest root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$.

Case 2: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if and only if $|x|=1$. So, in view of Remark 1, in this case your claim about the minimizer will hold if and only if $q\ne0$ and $|x|=1$. The latter condition will not of course hold in general even if $q\ne0$: e.g., if $n=1$ and $f(z)=z^2+z$, then $x=-1/2$ and $|x|<1$.

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, your claim about the minimizer $x$ will hold if and only if $q\ne0$ and $|x|=1$.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.

$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Remark 1:

Let \begin{equation*} g(s):=q^T(P+sI)^{-2}q \end{equation*} for real $s$ such that $P+sI$ is invertible, that is, for all $s\in I_P:=(-\la_{\min},\infty)$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. The function $g$ is nonnegative, nonincreasing, and $g(\infty-)=0$. Moreover, the equation \begin{equation*} g(s)=1 \tag{0}\label{0} \end{equation*} $s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots.

Let now \begin{equation*} f(z):=\frac12\,z^T Pz+q^T z+r \end{equation*} for $z\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation*} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation*} where \begin{equation*} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation*} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation*} -a^T x=|a|. \tag{3}\label{3} \end{equation*} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation*} x=-(P+tI)^{-1}q, \tag{4}\label{4} \end{equation*} whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the only (and hence the largest) root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$.

Case 2: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if and only if $|x|=1$. However, the equation $g(s)=1$ may not have any real root $s$ at all (e.g. when $q=0$ or, more generally, when $q$ is a short enough nonzero vector orthogonal to the eigenspace of $P$ corresponding to $\la_{\min}$). So, in this case your claim about the minimizer will not hold in general.

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, in both cases, (i) equation \eqref{0} has at most one root and (ii) if it has a root, then this root is $t=|a|=|Px+q|=\max(0,t)$, for $x$ as above.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.

$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Remark 1:

Let \begin{equation*} g(s):=q^T(P+sI)^{-2}q \end{equation*} for real $s$ such that $P+sI$ is invertible. The function $g$ is nonnegative and nonincreasing on the interval $I_P:=(-\la_{\min},\infty)$, and $g(\infty-)=0$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. Moreover, the equation \begin{equation*} g(s)=1 \tag{0}\label{0} \end{equation*} $s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots. So,

• if $q\ne0$, then equation \eqref{0} has only one nonnegative root, so that this root is the largest real root of \eqref{0};

• if $q=0$, then equation \eqref{0} has no real roots.

Let now \begin{equation*} f(z):=\frac12\,z^T Pz+q^T z+r \end{equation*} for $z\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation*} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation*} where \begin{equation*} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation*} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation*} -a^T x=|a|. \tag{3}\label{3} \end{equation*} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation*} x=-(P+tI)^{-1}q, \tag{4}\label{4} \end{equation*} whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the largest root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$.

Case 2: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if and only if $|x|=1$. So, in view of Remark 1, in this case your claim about the minimizer will hold if and only if $q\ne0$ and $|x|=1$. The latter condition will not of course hold in general even if $q\ne0$: e.g., if $n=1$ and $f(z)=z^2+z$, then $x=-1/2$ and $|x|<1$.

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, your claim about the minimizer $x$ will hold if and only if $q\ne0$ and $|x|=1$.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.

$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Let \begin{equation} g(s):=q^T(P+sI)^{-2}q \end{equation} for real $s$ such that $P+sI$ is invertible, that is, for all real $s$ greater than $-\la_{\min}$, where $\la_{\min}$ is the smallest eigenvalue $\la_{\min}$ of the symmetric positive-definite matrix $P$. The function $g$ is nonnegative, nonincreasing, and $g(\infty-)=0$. Let \begin{equation} f(x):=\frac12\,x^T Px+q^T x+r \end{equation} for $x\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation} where \begin{equation} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation} -a^T x=|a|. \tag{3}\label{3} \end{equation} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation} x=x_t, \quad\text{where } x_s:=-(P+sI)^{-1}q, \end{equation} whence $g(t)=x^T x=1$ and hence \begin{equation} f(x)=h(t), \end{equation} where \begin{equation} \begin{aligned} h(s)&:=f(x_s)=\frac12\,q^T (P+sI)^{-1} P(P+sI)^{-1}q-q^T (P+sI)^{-1}q+r \\ &:=-\frac12\,q^T (P+sI)^{-1}q-\frac s2\,q^T (P+sI)^{-2}q+r \\ &:=-\frac12\,q^T (P+sI)^{-1}q-\frac s2\,g(s)+r. \end{aligned} \end{equation} Recall that the continuous function $g$ is nonincreasing. So, the set $S:=\{s\in\R\colon g(s)=g(t)\}=\{s\in\R\colon g(s)=1\}$ is of the form $[t_1,t_2]\cap(-\la_{\min},\infty)$ for some real $t_1,t_2$ such that $t_2\ge t$. So, for all $s$ in the interior of the set $S$ we have \begin{equation} \begin{aligned} h'(s)&=\frac12\,q^T (P+sI)^{-2}q-\frac 12\,g(t)=0. \end{aligned} \end{equation} So, $f(x_s)=h(s)=h(t)=f(x)$ for all $s\in S$. Therefore and because $x=x_t$ is the unique minimizer of $f$ on $B$, we conclude that $x_s=x_t=x$ for all $s\in S$. In particular, $x_{t_2}=x$. So, in Case 2, the minimizer of $f$ on $B$ is \begin{equation} x=x_{t_2}=-(P+t_2I)^{-1}q, \end{equation} and $t_2>0$ is the largest real root $s$ of the equation \begin{equation} g(s)[=q^T(P+sI)^{-2}q]=1, \end{equation} as desired.

Case 2: $t=0$, that is, $x=-P^{-1}q$. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(0)\le1$. However, the equation $g(s)=1$ may not have any real root $s$ at all (e.g. when $q=0$ or, more generally, when $q$ is a short enough nonzero vector orthogonal to the eigenspace of $P$ corresponding to $\la_{\min}$). So, in this case your claim about the minimizer will not hold in general.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and uisng the spherical symmetry of $B$.

$\newcommand\R{\Bbb R}\newcommand\la{\lambda}$Remark 1:

Let \begin{equation*} g(s):=q^T(P+sI)^{-2}q \end{equation*} for real $s$ such that $P+sI$ is invertible, that is, for all $s\in I_P:=(-\la_{\min},\infty)$, where $\la_{\min}>0$ is the smallest eigenvalue of the symmetric positive-definite matrix $P$. The function $g$ is nonnegative, nonincreasing, and $g(\infty-)=0$. Moreover, the equation \begin{equation*} g(s)=1 \tag{0}\label{0} \end{equation*} $s\in I_P$ has at most one root, because (i) if $q\ne0$, then $g'(s)=-2q^T(P+sI)^{-3}q<0$ for all $s\in I_P$ and hence $g$ is strictly decreasing and (ii) if $q=0$, then $g(s)=0$ for all $s\in I_P$ and hence equation \eqref{0} has no roots.

Let now \begin{equation*} f(z):=\frac12\,z^T Pz+q^T z+r \end{equation*} for $z\in\R^n$. The function $f$ is strictly convex. So, $f$ attains its minimum on the closed unit ball $B$ at a unique point $x\in B$. As you noted, we have \begin{equation*} 0\le\min_{y\in B}a^T(y-x)=-|a|-a^T x, \tag{1}\label{1} \end{equation*} where \begin{equation*} a:=\nabla f(x)=Px+q \tag{2}\label{2} \end{equation*} and $|\cdot|$ is the Euclidean norm. Since $x\in B$, we have $-a^T x\le|a|$. So, by \eqref{1}, \begin{equation*} -a^T x=|a|. \tag{3}\label{3} \end{equation*} So, for $$t:=|a|,$$ we have one of the following two cases:

Case 1: $t>0$. Then, since $x\in B$, \eqref{3} and \eqref{2} imply $x=-a/|a|=-\frac1t\,(Px+q)$, so that $|x|=1$ and
\begin{equation*} x=-(P+tI)^{-1}q, \tag{4}\label{4} \end{equation*} whence $g(t)=x^T x=1$ and therefore, by Remark 1, $t>0$ is the only (and hence the largest) root $s$ of equation \eqref{0} -- as desired, because, obviously, here $t=\max(0,t)$.

Case 2: $t=0$, that is, $a=0$, that is, $x=-P^{-1}q$ -- so that \eqref{4} still holds. In this case $1\ge x^T x=q^TP^{-2}q$, so that $g(t)=g(0)\le1$. So, in this case, $t$ is a root of \eqref{0} if and only if $|x|=1$. However, the equation $g(s)=1$ may not have any real root $s$ at all (e.g. when $q=0$ or, more generally, when $q$ is a short enough nonzero vector orthogonal to the eigenspace of $P$ corresponding to $\la_{\min}$). So, in this case your claim about the minimizer will not hold in general.

We conclude that, in both cases, \eqref{4} holds for the unique minimizer $x$ of $f$. Moreover, in both cases, (i) equation \eqref{0} has at most one root and (ii) if it has a root, then this root is $t=|a|=|Px+q|=\max(0,t)$, for $x$ as above.

These conclusions can also be obtained a bit more elementarily, by first diagonalizing $P$ and using the spherical symmetry of $B$.