The following comes from the discussion of Laplace transformation in ODE.
Let $f(t)$ be piecewise continuous on $[0, \infty)$ and of exponential order. Prove that there exist constants $K$ and $\alpha$ such that $|f(t)| \leq K e^{\alpha t}$ for all $t \geq 0$.
I find this result amazing. A function having an exponential order only requires $|f(t)| \leq K e^{\alpha t}$ for all $t \geq M$. The above result means that if such $M$ exists, the whole function should be also of exponential order. Then $f(x)=\frac 1x$ is also of exponential order? How do we prove the above statement?
Thanks in advance.
Piecewise continuous on $[0, M]$ implies bounded on $[0,M]$. Note that $f(x) = \frac{1}{x}$ is not piecewise continuous on $[0,\infty)$, since it is not defined at 0. But moreover, it cannot be extended to a piecewise continuous function.