 Solution

Let $\tau >0$ be the first conjugate time and choose a nonzero Jacobi field $J$ with $J(0)=J(\tau )=0$. It is normal by part 1(b), and $D_{t}J(\tau )\ne 0$, since zero value and derivative there would force $J=0$. Fix any $T>\tau$ and extend $J$ by zero to a continuous piecewise smooth vector field $V$ on $[0,T]$.

For endpoint-vanishing fields, the Riemannian index form isIntegration by parts on $[0,\tau ]$ and the Jacobi field equation give $I(V,V)=0$ andChoose a smooth normal field $W$, zero at $0,T$, with $W(\tau )=-D_{t}J(\tau )$. Thenfor sufficiently small positive $\epsilon$. The corner in $V$ causes no difficulty: smooth endpoint-vanishing fields approximate $V+\epsilon W$ in the Sobolev space $H^1$, and the Riemannian index form is continuous in this norm on the compact segment. A smooth approximant therefore still has negative index form.

Realize this smooth field as a fixed-endpoint curve variation using the Riemannian exponential map. The allowed second variation of geodesic energy gives $E^{\prime }(0)=0$ and $E^{\prime \prime }(0)<0$, so some nearby curve has energy less than $T/2$, the energy of the unit-speed reference geodesic. The Cauchy-Schwarz inequality bounds its length byIt is shorter than the reference segment. Thus a geodesic cannot minimize beyond its first conjugate time. This argument deliberately uses $T>\tau$; it does not rule out minimization up to the conjugate point itself.