In my stochastic analysis class we've defined the following local convergence metric in the space of continuous square-integrable martingales $$m(M,N) = \sum^\infty_{n=1} 2^{-n} \Big(1\wedge \big(E(M_n-N_n)^2\big)^{1/2}\Big)$$
My issue is mainly with the triangle inequality. I can see how we could prove a relaxed version of the triangle inequality with $$m(M,N) \leq \sqrt 2 m(M,P) + \sqrt 2 m(P,N)$$ using the fact that $(a+b)^2 \leq 2a^2 + 2b^2$, but I don't see how to obtain the usual triangle inequality. Am I missing something, or am I indeed correct?
Thank you.
By applying the triangle inequality for the $L^2$ norm we see \begin{align*} & m(M,N) = \sum^\infty_{n=1} 2^{-n} \Big(1\wedge \big(E(M_n-N_n)^2\big)^{1/2}\Big) \\ & \leq \sum^\infty_{n=1} 2^{-n} \Big(1\wedge \big(\big(E(M_n-P_n)^2\big)^{1/2} + \big(E(N_n-P_n)^2\big)^{1/2} \big)\Big) \\ & \leq \sum^\infty_{n=1} 2^{-n} \Big(1\wedge \big(E(M_n-P_n)^2\big)^{1/2} \big) + \sum^\infty_{n=1} 2^{-n} \Big(1\wedge \big(E(N_n-P_n)^2\big)^{1/2} \big) \\ & = m(M,P) + m(N,P) \end{align*} giving the result.