Exponential type

Remarks

Example: $\abs{\sin z} \le e^{\abs{z}}$ for all $z$, and $M(r) = \sinh r$ for $\sin z$, so $\sin z$ is of exponential type with type $1$. Non-example: $f(z) = e^{z^2}$ has $\abs{f(x)} = e^{x^2}$ on the real line, and $e^{x^2} \le C e^{\tau \abs{x}}$ fails for large $x$ for every $C$ and $\tau$. In terms of the order, $f$ is of exponential type exactly when its order is less than $1$, or equal to $1$ with finite type. For $f(z) = \sum_n a_n z^n$, the type is $\sigma = \frac{1}{e} \limsup_{n \to \infty} n \abs{a_n}^{1/n}$; for $e^z$ this gives $1$. The notion is also used for functions holomorphic in a sector or a half-plane, with the bound required only there.