If $\mathscr{L}$ is a normal operator, mathematically, we have the formula
$$\mathscr{L}^{-1}=i \int_0^\infty \exp(-i t \mathscr{L})dt.\tag{1}$$
My question:
1. On the basis of the above equation, how can I see the following inequation?
$$\| \mathscr{L}^{-1} \| \le \int_0^\infty \| \exp (-i t \mathscr{L}) \| dt;$$
2. In practice, why this inequality is typically within a factor of 2 of equality?
I understand that the RHS of Eq.(1) is equivalent to rotate the complex number $\int_0^\infty \exp(-i t \mathscr{L})dt$ by $90^\circ$ anticlockwise. And a key point of my question is related to the the norm of an operator? Thank you for any suggestion.
It's just the triangle inequality. An integral is a limit of sums. And for a sum you have $$ \left\| \sum_{j=1}^n \exp(-it_j\mathscr L)\,\Delta_j\right\|\leq\sum_{j=1}^n\|\exp(-it_j\mathscr L)\|\,\Delta_j $$