Squared distance is strictly convex along geodesics in a hadamard manifold

The distance formula and the absence of cut points. By [F2], for every $w\in T_{p}M$ the segment $t\mapsto {\exp }_p(tw)$, $t\in [0,1]$, is the unique affinely parametrized geodesic segment from $p$ to ${\exp }_p(w)$, and it minimizes; hence $d_g(p,{\exp }_p(w))=\mathrm{\mid }w{\mathrm{\mid }}_p$. Suppose now that $v\in S_{p}M$ has finite cut time $c:=c_p(v)<+\mathrm{\infty }$. By [F4] either ${\gamma }_{p,v}(0)$ and ${\gamma }_{p,v}(c)$ are conjugate along ${\gamma }_{p,v}$, which [F3] forbids, or two distinct minimizing unit-speed geodesics join $p$ to ${\gamma }_{p,v}(c)$, which contradicts the uniqueness in [F2] (a minimizing unit-speed geodesic on $[0,c]$ reparametrized affinely to $[0,1]$ is an affinely parametrized geodesic segment from $p$ to the same point, and there is exactly one). Both alternatives are impossible, so $c_p(v)=+\mathrm{\infty }$ for every unit $v$: the cut locus of $p$ is empty, and $r_p({\exp }_{p}w)=\mathrm{\mid }w{\mathrm{\mid }}_p$ for every $w\in T_{p}M$.

The product rule for a squared smooth function. Let $u>0$ be smooth near a point $x$ and set $F:=\frac{1}{2}{u}^2$. Then $\mathrm{grad}F=u\mathrm{grad}u$, so for vector fields $X,Y$, $\mathrm{Hess}F(X,Y)=g({\mathrm{\nabla }}_X(u\mathrm{grad}u),Y)=X(u)g(\mathrm{grad}u,Y)+ug({\mathrm{\nabla }}_X\mathrm{grad}u,Y),$ i.e. $\mathrm{Hess}\left(\frac{1}{2}{u}^2\right)(X,Y)=du(X)du(Y)+u\mathrm{Hess}u(X,Y),$ both sides being smooth and symmetric by [F7].

The second derivative of a smooth function along a geodesic. Let $h$ be smooth and let $\sigma$ be a geodesic with ${\mathrm{\nabla }}_{\dot{\sigma }}\dot{\sigma }=0$. Then $(h\circ \sigma {)}^{\mathrm{\prime }}=dh(\dot{\sigma })=g(\mathrm{grad}h,\dot{\sigma }),$ and differentiating again with metric compatibility and ${\mathrm{\nabla }}_{\dot{\sigma }}\dot{\sigma }=0$ gives $(h\circ \sigma {)}^{\mathrm{\prime }\mathrm{\prime }}=g({\mathrm{\nabla }}_{\dot{\sigma }}\mathrm{grad}h,\dot{\sigma })=\mathrm{Hess}h(\dot{\sigma },\dot{\sigma })\mathrm{.}$ The identity is local in the parameter and holds at every time of $I$.

Smoothness of $f$ and its Hessian at the base point. By step 1.1, $(f\circ {\exp }_p)(w)=\frac{1}{2}\mathrm{\mid }w{\mathrm{\mid }}_p^2$ for every $w\in T_{p}M$, a smooth quadratic form on the vector space $T_{p}M$; since ${\exp }_p$ is a diffeomorphism by [F1], $f$ is smooth on $M$. For $X\in T_{p}M$ the curve $s\mapsto {\exp }_p(sX)$ is an affinely parametrized geodesic, so step 1.1 and step 1.3 give $\mathrm{Hess}f(p)(X,X)=(f\circ {\gamma }_X{)}^{\mathrm{\prime }\mathrm{\prime }}(0)={\frac{d^2}{d{s}^2}\mid }_{s=0}\frac{1}{2}{s}^2\mathrm{\mid }X{\mathrm{\mid }}_p^2=\mathrm{\mid }X{\mathrm{\mid }}_p^2=g_p(X,X)\mathrm{.}$ As $\mathrm{Hess}f(p)$ is symmetric ([F7]), the polarization identity recovers all mixed values, so $\mathrm{Hess}f(p)=g_p$.

The lower bound $\mathrm{Hess}f\ge g$ off the base point, in dimension $n\ge 2$. Fix $q\mathrm{\ne }p$ and let $t_0:=r_p(q)>0$. By step 1.1 the cut locus of $p$ is empty, so $q\mathrm{\notin }\{p\}\cup \mathrm{Cut}(p)$: $r_p$ is smooth near $q$, $\mathrm{\mid }\mathrm{grad}{r}_p\mathrm{\mid }=1$ and $\mathrm{grad}{r}_p$ is the terminal velocity of the minimizing geodesic from $p$ to $q$ ([F5]). By step 1.2 applied to $u=r_p$, $\mathrm{Hess}f=\mathrm{Hess}\left(\frac{1}{2}{r}_p^2\right)=d{r}_p\otimes d{r}_p+r_p\mathrm{Hess}{r}_p\text{near }q\mathrm{.}$ Write an arbitrary $X\in T_{q}M$ as $X=a\mathrm{grad}{r}_p+X^{\perp }$ with $a:=d{r}_p(X)$ and $X^{\perp }\perp \mathrm{grad}{r}_p$. By F6 $\mathrm{Hess}{r}_p(\mathrm{grad}{r}_p,\cdot )=0$, so $\mathrm{Hess}{r}_p(X,X)=\mathrm{Hess}{r}_p(X^{\perp },X^{\perp })\mathrm{.}$ Since $K\le 0$ everywhere, the hypothesis of F6 holds with $k=0$ along the minimizing geodesic from $p$ to $q$, and F6 yields $\mathrm{Hess}{r}_p(X^{\perp },X^{\perp })\ge (1\mathrm{/}{t}_0)\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2$. Therefore $\mathrm{Hess}f(X,X)=a^2+t_0\mathrm{Hess}{r}_p(X^{\perp },X^{\perp })\ge a^2+\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2=g_q(X,X),$ using that $X\mapsto (a,X^{\perp })$ is a $g_q$-orthogonal decomposition.

The lower bound in dimension $n=1$. If $\dim M=1$ then at every $q\mathrm{\ne }p$ the tangent space is spanned by the unit vector $\mathrm{grad}{r}_p$, so $X^{\perp }=0$ and the bound $\mathrm{Hess}{r}_p(X^{\perp },X^{\perp })\ge (1\mathrm{/}{t}_0)\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2$ holds with both sides equal to $0$; the gradient is a unit geodesic field by [F5], so ${\mathrm{\nabla }}_{\mathrm{grad}{r}_p}\mathrm{grad}{r}_p=0$ and $\mathrm{Hess}{r}_p(\mathrm{grad}{r}_p,\mathrm{grad}{r}_p)=\frac{1}{2}\mathrm{grad}{r}_p(\mathrm{\mid }\mathrm{grad}{r}_p{\mathrm{\mid }}^2)=0$ by [F7]. Hence, by step 1.2, $\mathrm{Hess}f=d{r}_p\otimes d{r}_p=g$ off $p$, and the computation of step 2.2 applies verbatim without invoking the dimension-$n\ge 2$ comparison bound.

The bound $\mathrm{Hess}f\ge g$ on all of $M$. Off $p$ the pointwise inequality was proved in step 2.2 for $n\ge 2$ and in step 2.3 for $n=1$, and at $p$ the equality $\mathrm{Hess}f(p)=g_p$ of step 2.1 gives the bound as well. In dimension zero $M$ is a point and the tensor inequality is vacuous. In positive dimension the inequality was checked on a decomposition spanning each tangent space, so $\mathrm{Hess}f\ge g$ holds on $M$.

The second-derivative bound along a geodesic. For the nonconstant affinely parametrized geodesic $\gamma$, put $\phi (t):=f(\gamma (t))=\frac{1}{2}{d}_g(p,\gamma (t){)}^2$. By step 1.3, ${\phi }^{\mathrm{\prime }\mathrm{\prime }}(t)=\mathrm{Hess}f(\dot{\gamma }(t),\dot{\gamma }(t))\ge g(\dot{\gamma }(t),\dot{\gamma }(t))=c>0,$ where $c=\mathrm{\mid }\dot{\gamma }{\mathrm{\mid }}^2$ is a positive constant by [F8]. In particular $\phi$ is twice differentiable on the open interval $I$ with ${\phi }^{\mathrm{\prime }\mathrm{\prime }}\ge c$ everywhere.

Strict convexity. Define $\psi (t):=\phi (t)-\frac{c}{2}{t}^2$ on $I$. Then ${\psi }^{\mathrm{\prime }\mathrm{\prime }}={\phi }^{\mathrm{\prime }\mathrm{\prime }}-c\ge 0$, so $\psi$ is convex by [F9]; and $t\mapsto \frac{c}{2}{t}^2$ is strictly convex by [F9], since $c>0$. For distinct $s,t\in I$ and $\lambda \in (0,1)$, put $m:=(1-\lambda )s+\lambda t$ and $q(z):=\frac{c}{2}{z}^2$; convexity of $\psi$ and strict convexity of $q$ give, one of the two summed inequalities being strict, $\phi (m)=\psi (m)+q(m)<[(1-\lambda )\psi (s)+\lambda \psi (t)]+[(1-\lambda )q(s)+\lambda q(t)]=(1-\lambda )\phi (s)+\lambda \phi (t),$ the strict inequality because $s\mathrm{\ne }t$ in [F9]. Multiplying by $2$, the function $t\mapsto d_g(p,\gamma (t){)}^2$ is strictly convex on $I$, as claimed. A constant geodesic has ${\phi }^{\mathrm{\prime }\mathrm{\prime }}=0$ and $\phi$ constant, so no strict convexity can be asserted there; this is the only case excluded, and the affine parametrization was used only to identify $c=\mathrm{\mid }\dot{\gamma }{\mathrm{\mid }}^2$ as a positive constant in step 4.1. The proof selects no object of its own: the geodesics and the minimizing segments it uses are those supplied one pair at a time by the cited consequences of completeness, and the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ of [A1] is consumed exactly through them.