= Solution
Suppose $J$ is a <Jacobi field> vanishing at both endpoints of a nonconstant <geodesic> segment $[0,t_0]$. Its tangential component $h(t)=\langle J(t),T(t)\rangle$ satisfies
$$
h''(t)=\langle D_t^2J,T\rangle=-\langle R(J,T)T,T\rangle=0,
$$
because $D_tT=0$ and the <Riemann curvature tensor> has the usual skew-adjoint symmetry. The two zero endpoint values imply $h\equiv0$. Both $J$ and $D_tJ$ are therefore perpendicular to $T$.
The <linear map> from these <Jacobi fields> to $T(0)^\perp$ given by $J\mapsto D_tJ(0)$ is injective: $J(0)=0$ together with $D_tJ(0)=0$ forces $J\equiv0$ by uniqueness for the linear <ordinary differential equation> defining a <Jacobi field>. Its target has dimension $n-1$, so
$$
\boxed{\text{multiplicity}\leq n-1.}
$$
This proves the <endpoint-vanishing normal Jacobi fields have dimension at most n-1> bound. For a constant <geodesic> the equation is $J''=0$, and two zero values instead force $J=0$, so there is no <conjugate point>.
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