| dbp:proof
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- If is an entire function, it can be represented by its Taylor series about 0:
:
where
:
and is the circle about 0 of radius . Suppose is bounded: i.e. there exists a constant such that for all . We can estimate directly
:
where in the second inequality we have used the fact that on the circle . But the choice of in the above is an arbitrary positive number. Therefore, letting tend to infinity gives for all . Thus and this proves the theorem. (en)
- Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since is bounded, the averages of it over the two balls are arbitrarily close, and so assumes the same value at any two points. (en)
- Suppose for the sake of contradiction that there is a nonconstant polynomial with no complex root. Note that as . Take a sufficiently large ball ; for some constant there exists a sufficiently large such that for all .
Because has no roots, the function is entire and holomorphic inside , and thus it is also continuous on its closure . By the extreme value theorem, a continuous function on a closed and bounded set obtains its extreme values, implying that for some constant and .
Thus, the function is bounded in , and by Liouville's theorem, is constant, which contradicts our assumption that is nonconstant. (en)
- Suppose for all in the complex plane, we can apply the Cauchy estimate to a disk center at any of any radius to obtain:
.
Let tend to , we obtain . Since This is true for all , is a constant. (en)
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