Suppose $\gamma:[a,b]\to M$ is geodesic, $T:=\dot\gamma$ is its velocity vector.
A vector field $V$ along $\gamma$ is Jacobi field if it satisfies the following Jacobi equation $$\nabla_{T}\nabla_{T}V+R(V,T)T=0.$$
How can I prove a piecewise smooth Jacobi field must be smooth everywhere?
A hint says it's similar in spirit to show a piecewise smooth curve $c$ is geodesic if first variation of energy is $0$ for every its proper variation, which means for arbitrary variation $V$, $$\sum_{i=1}^k\langle V(t_i), \dot c(t_i^+)-c(t_i^-)\rangle=0$$ at all discontinuous points $t_i$. But I don't know such an equivalent characterizing for Jacobi field.
I searched this result in books, some define Jacobi fields to be smooth vector fields, and some simply says $\nabla_T V$ is continuous without proof.
My aim is to use this to characterise Jacobi field as null space of index form for piecewise smooth vector fields vanishing at end points.
Thanks for your time and patience.
This cannot be true. Consider the simplest example where $M = \mathbb R^2$ with the flat metric and $\gamma( t) = (t, 0)$. Then
\begin{equation} V(t) = \begin{cases} 0 & \text{ if } t \le 0, \\ t & \text{ if }t >0\end{cases} \end{equation}
is a piecewise smooth Jacobi fields which is not smooth.
Unless by piecewise smooth, you mean that both $V$ and $\nabla _TV$ agrees at the endpoints. Then as you described in the comment, $\nabla_T\nabla_ T V$ also agrees at the endpoints. Then one can argue inductively, by differentiating the Jacobi equation, that $V$ is smooth.