Let $f$ be an entire function such that $|f|$ follows the triangle inequality for every pair of complex numbers, i.e. $$ |f(z_1+z_2)|\le |f(z_1)| + |f(z_2)| $$ Show that $f$ must be a polynomial of degree less than $2$.
It's 1 month that I failed an exam with this question and it's haunting me since then. Please relieve me from the angst of trying to solve this exercise every day I can.
Let $$ f(z)=\sum_{k=0}^\infty a_kz^k $$ be an entire function with the desired property. W.l.o.g. let $f \not=0$, that is at least one $a_k$ is not zero. By setting $z_1=z_2=z$ we have $|f(2z)| \le 2|f(z)|$ $(z \in \mathbb{C})$. Hence $$ g(z):= \frac{f(2z)}{f(z)}, \quad |g(z)|\le 2. $$ Note that points $z\in \mathbb{C}$ with $f(z)=0$ are removable singularities of $g$ (since $g$ is bounded near every singularity, see Riemann's theorem on removable singularities), so $g$ is a bounded entire function. By Liouville's Theorem $g$ is constant. Hence $$ \exists c \in \mathbb{C} \forall z \in \mathbb{C}: ~ f(2z)=cf(z) $$ Comparing the coefficients from the power series of $f$ yields $$ 2^ka_k=ca_k \quad (k \in \mathbb{N}_0). $$ Thus $a_k \not=0$ implies $2^k=c$, so only one $a_k$ can be $\not=0$. Thus, $f(z)=a_kz^k$ for some $k$. The cases $k=0,1$ are possible, and $k>1$ is not possible, since for $z\not=0$ $$ |f(2z)|=|a_k(2z)^k|\le 2|a_kz^k|=2|f(z)| \iff 2^k \le 2. $$ Thus, the entire functions with $|f(z_1+z_2)| \le |f(z_1)|+|f(z_2)|$ are exactly the functions $f(z)=a$ and $f(z)=az$, $a \in \mathbb{C}$.
This is an application of Cauchy's inequality, that is: If $M(r)$ is the maximum modulus of $f$ on the circle with radius $r$ centered at some point $z_0$ (say $0$), then $|f^{(n)}(z_0)|\leq \frac{M(r)n!}{r^n}$.
Now from the triangle inequality it follows that $|f(2e^{i\varphi})|=|f(e^{i\varphi}+e^{i\varphi})|\leq|f(e^{i\varphi})|+|f(e^{i\varphi})|=2|f(e^{i\varphi})|$ and by induction $|f(ke^{i\varphi})|\leq k|f(e^{i\varphi})|$. This is for all $\varphi\in\mathbb R$. It follows that $M(k)\leq kM(1)$. But then
$$|f^{(n)}(0)|\leq \frac{M(k)n!}{k^n}\leq \frac{k M(1)n!}{k^{n}}=k^{-(n-1)}M(1)n!$$
This is true for all $k\in\mathbb N$. As long as $(n-1)$ is positive - that is, $n\geq2$ - this goes to $0$ as $k\to\infty$, and thus $|f^{(n)}(0)|\leq0$, making $f^{(n)}(0)=0$ for all $n\geq2$. The only analytic functions satisfying this are the polynomials of degree less than $2$.
The intuition here is that the growth rate of an entire function puts bounds on its coefficients: If it grows at least linearly, quadratically, cubically along some path to infinity, then the coefficient of the linear, quadratic, cubic part can be nonzero. But the triangle inequality enforces at most linear growth, so the only nonzero coefficients are those of the constant and linear parts of the power series.
