I think the argument is, first of all, I assume \{t_{n}\} take values in \mathbb{R}, then, due to the existence of limit, \inf is indeed \min and so it should be > - \infty. Somehow I think it is also an if-and-only-if statement.
The existence of the limit does not imply infimum is minimum.
It's a general fact that if a sequence of points has a limit, the sequence is bounded. The proof can be sketched as follows: Only finitely many points can be a distance greater than 1 away from the limit (by the definition of a limit). So a lower bound of the set is either one of the values farther away than 1 from the limit, or one less than the limit is a lower bound
#4
wayneckm
66
0
Argh, you are right.
I made a mistake in assuming that the bound can be attained within the finite set in the infimum but indeed it is not necessary true. Thanks.
I'm reviewing Meirovitch's "Methods of Analytical Dynamics," and I don't understand the commutation of the derivative from r to dr:
$$
\mathbf{F} \cdot d\mathbf{r} = m \ddot{\mathbf{r}} \cdot d\mathbf{r} = m\mathbf{\dot{r}} \cdot d\mathbf{\dot{r}}
$$