Liouville's theorem (complex analysis)

In mathematics, and in particular in complex analysis, theorem of Liouville asserts that if a function is holomorphic throughout the complex plane and is sheltered, then it is constant. Note that this claim is false in real numbers (take, for example, the function ${displaystyle cos(x)}$which is sheltered but not constant.

Statement of the theorem

Sea ${displaystyle f:mathbb {C} to mathbb {C} }$an entire and accompanied function, i.e., exists ${displaystyle Mgs0}$ such as

${displaystyle Șf(z)}{quad forall zin mathbb {C}}}$;

Then it turns out that ${displaystyle f}$ It's constant.

A more general version of this theorem states that if ${displaystyle f:mathbb {C} to mathbb {C} }$ It's a whole function and if ${displaystyle forall zin mathbb {C} }$ You have to ${displaystyle ⋅f(z)$, with ${displaystyle C,;D 2005}$ for some ${displaystyle nin mathbb {N} _{0}}$, then ${displaystyle f}$ Must be a grade polynomial to the most ${displaystyle n}$. As a direct consequence of the above, if ${displaystyle ⋅f(z)$, with ${displaystyle p(z)in mathbb {C} [z]}$, a grade polynomial ${displaystyle n}$, then ${displaystyle f}$ It's a grade polynomial to the most ${displaystyle n}$.

Demo

Cauchy's integral formula says that

${displaystyle f'(z)={frac {1}{2pi ,i}}}{oint _{ overseer}{frac {f(zeta)}{(z-zeta)^{2}dzeta. !$

So that

${displaystyle Δf'(z)ёleq {frac {1}{2pi }{frac {M}{r^{2}}}}},2pi ,r={frac {M}{r}}}}}. !$

How we can choose ${displaystyle r}$ As big as we want, we concluded that ${displaystyle f'(z)=0}$ for everything ${displaystyle z}$ in ${displaystyle mathbb {C} }$. Finally, as ${displaystyle f}$ is defined on a simply related set, then f must be constant.

Liouville's theorem and fundamental theorem of algebra

The theorem of Liouville gives a simple demonstration of the fundamental theorem of algebra, that is, that every polynomial does not constant to coefficients in ${displaystyle mathbb {C} }$ has a root in ${displaystyle mathbb {C} }$. The demonstration is as follows:

Sea ${displaystyle P(z)}$ a non-constant polynomial, and suppose it has no roots. Then, as all polynomials are whole functions, you have to ${displaystyle Q(z)={frac {1}{P(z)}}}}}$ it turns out to be also an entire function.

We rewrite (or factor) to ${displaystyle P(z)=a_{n}z^{n}+a_{n-1z^{n-1}+ldots +a_{0}=a_{n}z^{n}{n}{a_{n-1}{a_n1}{a_{n}z}$. Using a ${displaystyle R}$ big enough, so ${displaystyle Δz국R}$ each of the terms of ${displaystyle m(z)}$ will be less than ${displaystyle {frac {1}{2n}}}}$.

So. ${displaystyle leftё1+m(z)right presupposegeq leftnutrition1-information(z))$where it comes from ${displaystyle ⋅Q(z)$.

Later. ${displaystyle Q(z)}$ it is sheltered for ${displaystyle Δz국R}$, but as it is continuous, it is also attached to the disk ${displaystyle Șz R!$. Applying the theorem of Liouville, the function ${displaystyle Q(z)}$ is constant, so ${displaystyle P}$ It will, too. That contradicts our initial hypothesis, and it concludes that then ${displaystyle P}$ must have a root.

Note that it follows easily then ${displaystyle P}$ has as many roots as its degree (counting multiplicity), because it is enough to divide each time ${displaystyle P}$ for ${displaystyle (z-z_{0}}}}$Where ${displaystyle z_{0}}$ It's the root just found.

Consequences

Spectrum of an operator

One of the interesting consequences of Liouville's theorem is that the spectrum of an operator is necessarily a non-empty set. To see it, let's see that the fact that it was empty contradicts Liouville's theorem. If said spectrum were empty then the norm of the resolving function:

${displaystyle rho:mathbb {C} longrightarrow mathbb {C} qquad lambda mapsto rho _{lambda }=pos(lambda I-B)^{-1}associated}$

Where ${displaystyle B;}$ it is a coupled operator of a Banach space, it would be defined in the entire complex plane and would be holomorphic and sheltered. And that implies that the function would be constant by the Liouville theorem. And since:

${displaystyle lim _{lambda to infty }rho _{lambda }=0}$

Because it is constant, it would have to be 0 everywhere and that would contradict the fact that the solver is a bounded linear operator.