### Abstract

The following notion of bounded index for complex entire functions was presented by Lepson. function f ( z ) is of bounded index if there exists an integer N independent of z , such that max{l:0≤ l≤ N}|f(l)(z)|l!≥ |f(n)(z)|n!for alln. $$ \max\limits_{\{l: 0\leq l\leq N\}} \left \{ \frac{|{f^{(l)}(z)}|}{l!}\right \} \geq \frac{|{f^{(n)}(z)}|}{n!}\quad\text{for all}\,\, n. $$ The main goal of this paper is extend this notion to holomorphic bivariate function. To that end, we obtain the following definition. A holomorphic bivariate function is of bounded index, if there exist two integers M and N such that M and N are the least integers such that max{(k,l):0,0≤ k,l≤ M,N}|f(k,l)(z,w)|k!l!≥ |f(m,n)(z,w)|m!n!for allmandn. $$ \max\limits_{\{(k,l): 0,0\leq k, l\leq M, N\}} \left \{ \frac{|{f^{(k,l)}(z,w)}|}{k!\,l!}\right \} \geq \frac{|{f^{(m,n)}(z,w)}|}{m!\,n!}\quad\text{for all}\,\, m \, \text{and}\,\, n. $$ Using this notion we present necessary and sufficient conditions that ensure that a holomorphic bivariate function is of bounded index.