Question about the order of entire function $f+g$

Question

My question is based on this old question : Order of entire function $f + g$. I try to better understand the inequality we get for the order of $f+g$. It is obvious that there are two cases. The one where $\rho=\max \lbrace \lambda _{1}, \lambda _{2}\rbrace$ and the one where $\rho <\max \lbrace \lambda _{1}, \lambda _{2}\rbrace$. I have already managed to prove the first case, but I am looking for examples for the second. Can anyone help me find an example that shows the second case (for $f\neq 0$)?

Answer 1

If $f, g$ are entire functions with $\lambda_f \ne \lambda_g$ then the “order inequality” $$ \tag{$*$} \lambda_{f+g} \le \max(\lambda_f, \lambda_g) $$ becomes an equality: $$ \lambda_{f+g} = \max(\lambda_f, \lambda_g) \, . $$

Proof: Assume that $\lambda_g < \lambda_f$. Then $\lambda_{f+g} \le \lambda_f$ and $$ \lambda_f = \lambda_{f+g-g} \le \max(\lambda_{f+g}, \lambda_g) \le \max(\lambda_f, \lambda_g) = \lambda_f \\ \implies\max(\lambda_{f+g}, \lambda_g) = \lambda_f \\ \implies \lambda_{f+g} = \lambda_f \, . $$

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This can be used to construct examples where $(*)$ is a strict inequality. Choose entire functions $f, h$ with $\lambda_h < \lambda_f$ and set $g = h-f$. Then $\lambda_f = \lambda_g$ and $$ \lambda_{f+g} = \lambda_h < \lambda_f = \max(\lambda_f, \lambda_g) \, . $$

A simple example is $f(z) =e^z$ and $h(z) = z$, which have order $1$ and $0$, respectively.