Is $f$ Identically Zero on the Punctured Unit Disc?

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In summary, a function $f$ is identically zero on the punctured unit disc if it equals zero for all points in the disc except for the origin. The punctured unit disc is related to the unit disc by excluding the origin, and studying functions that are identically zero on this disc can provide insights into complex-valued functions and have various applications. It is possible for a function to be identically zero on the punctured unit disc but not the unit disc, and to prove this, one can use methods such as the Cauchy-Riemann equations or the maximum modulus principle.
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Euge
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Here's this week's problem!

_________

Let $f$ be a holomorphic function on the punctured unit disc $\Bbb D \setminus \{0\}$ such that

\(\displaystyle |f(z)| \le \log \frac{1}{|z|} \quad \text{for all} \quad z \in \Bbb D \setminus \{0\}.\)

Prove that $f \equiv 0$.

_________Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
No one answered this week's problem. Here is my solution.
Since $f$ is holomorphic on $\Bbb D\setminus\{0\}$ with $|zf(z)| \le -|z|\log|z| \to 0$ as $z\to 0$, it has a removable singularity at $z = 0$. Let $a_n$ denote the $n$th coefficient of the power series expansion of $f$ near $z = 0$. By the Cauchy estimates, for all $r\in (0,1)$,

\(\displaystyle |a_n| \le r^{-n} \max_{|z| = r} |f(z)| \le r^{-n}\log \frac{1}{r} \quad (n = 0, 1, 2, 3,\ldots)\)

Letting $r \to 1^-$, the result follows.
 

FAQ: Is $f$ Identically Zero on the Punctured Unit Disc?

What does it mean for $f$ to be identically zero on the punctured unit disc?

For a complex-valued function $f(z)$ defined on the punctured unit disc $\mathbb{D}^{\ast}$, being identically zero means that $f(z)$ equals zero for all points $z$ in the punctured unit disc, except possibly at the point $z = 0$. In other words, $f(z) = 0$ for all $z \neq 0$ in $\mathbb{D}^{\ast}$.

How is the punctured unit disc related to the unit disc?

The punctured unit disc $\mathbb{D}^{\ast}$ is obtained by removing the point $z = 0$ from the unit disc $\mathbb{D}$. This means that $\mathbb{D}^{\ast}$ contains all points within and on the boundary of the unit disc, except for the origin.

What is the importance of studying functions that are identically zero on the punctured unit disc?

Functions that are identically zero on the punctured unit disc have interesting properties and can provide insights into the behavior of complex-valued functions. They can also be used to construct other complex functions and have applications in various fields of mathematics, such as complex analysis and differential equations.

Can a function be identically zero on the punctured unit disc but not on the unit disc?

Yes, it is possible for a function to be identically zero on the punctured unit disc but not on the unit disc. This is because the punctured unit disc excludes the point $z = 0$, which may have a different value for the function compared to other points on the unit disc.

How can one prove that a function is identically zero on the punctured unit disc?

To prove that a function $f(z)$ is identically zero on the punctured unit disc $\mathbb{D}^{\ast}$, one can use techniques such as the Cauchy-Riemann equations or the maximum modulus principle. These methods involve showing that $f(z) = 0$ for all points $z \neq 0$ in $\mathbb{D}^{\ast}$, which can be done by manipulating the given function and using known properties of complex numbers and functions.

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