Exactly! Because the function inside that square determines the values outside the square. But the function is entire so it is continous, but continous functions on compact sets are bounded. So this function is bounded everywhere. Does that make sense?
that does kind of make sense but ive just kind of started learning this lol. how would you show it as a proof mathematically though?
Let be the square . The function is continous everywhere because entire functions (differenciable) are continous. Thus, is bounded on because it is a closed and bounded set. But then is bounded by that same number because the value of is determined by within the rectangle. So for example, . Thus, any value outside the rectangle is known from the value inside the rectangle. Since the function is bounded on this rectangle is is bounded outside it too because it has the same values as inside the rectangle.