Let $v^i$ be a vector on a Riemannian 3-manifold with metric $g_{ij}$ embedded inside a 3+1 space-time such that for some constant $N_M$ it satisfies the inequality $g_{ij}v^iv^j \leq N_M ^2$. Let $K$ be a symmetric rank-2 tensor on the 3-manifold. Then apparently the following holds:
$$\vert K_{ij} v^i v^j \vert \leq \vert K \vert _g N_M ^2.$$
This looks like some sort of a Cauchy-Schwarz inequality but given that $K$ is a tensor as described, I don't understand what the notation on the RHS means. For a rank-2 symmetric tensor $K$ what does $\vert K \vert _g$ mean?
If one knows that for some function $N$, $g_{ij}v^iv^j \leq N^2,$ where the function $N$ is itself bounded between constants
$$N_m \leq N \leq N_M,$$
then using inequalities like the above one can apparently show the following bound:
$$\int _{t_1} ^t \frac{1}{N} \Big(-v^i \partial _i N - \frac{dN}{dt} + K_{ij}v^iv^j\Big) dt \leq -2\log N_m + \frac{1}{N_m} \int _{t_1}^t (\vert \nabla N \vert _g N_M + \vert K \vert _g N_M ^2 )dt,$$
for some fixed $t_1$ and $t$.
I can't understand that first "log" term in the above.
Also once the above bound is shown does it follow that the integral can be unbounded above or below depending solely on the property of the function $N$? If yes then what would be needed of $N$ to make the integral unbounded above or below?