# How to interpret the Maximum Modulus Principle for non-constant polynomials?

I was reading Conway's complex variables textbook and had a confusion regarding his remarks about the Maximum Modulus Principle.

Maximum Modulus Theorem. If $$G$$ is a region and $$f: G\to\mathbb C$$ is an analytic function such that there is a point $$a$$ in $$G$$ with $$|f(a)|\ge|f(z)|$$ for all $$z\in G$$, then $$f$$ is constant.

Conway's Remarks:

• There is a property of analytic functions which is not so transparent for polynomials.
• According to the Maximum Modulus Theorem, a non-constant analytic function on a region cannot assume its maximum modulus; this fact is far from obvious even in the case of polynomials.

According to Conway's remark the MMP is true for polynomials but not very obvious. But while trying to find examples of polynomials that satisfy the MMP, I seems to be not true for many polynomials. For example:
Consider $$p(x)=-x^2$$. This is a polynomial of a real variable and has a global maximum at $$x=0$$. But $$p(x)$$ is definitely not constant on its domain $$(\mathbb R)$$.

So what is to be understood about the MMP when applied to polynomials? Why is it far from obvious? Does this mean the MMP is, in general, false for polynomials? Or, something else?

• The real answer is that MMP is a theorem about analytic functions on regions in the complex plane, not segments of the real axis. But note that your counterexample isn't even a counterexample, because you're examining $p(x)$ while MMP is a statement about $|p(x)|$. Jul 29, 2022 at 2:54
• @GregMartin : What's the explicit difference between $p(x)$ and $|p(x)|$ if $p$ is a real polynomial? I thought they're the same things for real polynomials. Jul 29, 2022 at 4:21
• $|{-3}|=3$ in the reals or the complexes. Jul 29, 2022 at 5:41

There is a big difference between real and complex polynomials. Conway is referring to complex polynomials, that is $$p(z) = \sum_{i=0}^n c_i z^i.$$ Complex polynomials are (complex) analytic and hence the Maximum Modulus Theorem (MMT) applies. His remark is merely saying that it is not obvious that $$p(z)$$ satisfies the MMT, just by virtue of being a polynomial (though there is a simpler proof of MMT for polynomials).
By the way, the real polynomial $$p(x) = -x^2$$ is not complex analytic (e.g., check the Cauchy-Riemann equations). It is not a polynomial in $$z=x+iy$$ so there is no contradiction here.
• The analog in the real variable case is the maximum principle for harmonic functions. If $\Delta u=0$, where $\Delta$ is the Laplacian (sum of pure second derivatives), then $u$ satisfies both a maximum and minimum principle (it cannot have interior max or min, unless it is constant). The Cauchy-Riemann equations imply that the real and imaginary parts of a complex analytic function are harmonic, so there is a close connection here.