Post History
#8: Post edited
by
Snoopy
·
2024-08-17T14:48:25Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where
$$\psi(z) = \varphi(z + z_0) - z_0,$$ with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where
- $$\psi(z) = \varphi(z + z_0) - z_0,$$ with $\tilde{\Omega} = \Omega - z_0 := \\{z - z_0 : z \in \Omega\\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#7: Post edited
by
Snoopy
·
2024-08-17T14:47:56Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
If $z_0e 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0
- e 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where
- $$\psi(z) = \varphi(z + z_0) - z_0,$$ with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#6: Post edited
by
Snoopy
·
2024-08-16T14:41:06Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
Suppose the above expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#5: Post edited
by
Snoopy
·
2024-08-15T21:34:10Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Suppose the above expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Suppose the above expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#4: Post edited
by
Snoopy
·
2024-08-15T21:32:30Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Suppose the above expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#3: Post edited
by
Snoopy
·
2024-08-15T21:30:23Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
|k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!} ight| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n.- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!} ight| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n\ .
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#2: Post edited
by
Snoopy
·
2024-08-15T21:29:20Z (3 months ago)
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n.
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
- First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
- Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
- If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
- It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
- Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
- Given the conditions, we can express $\varphi(z)$ near $0$ as:
- $$
- \varphi(z) = z + a_n z^n + O(z^{n+1}),
- $$
- where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
- By induction, as suggested in the hint, we find:
- $$
- \varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
- $$
- Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
- $$
- |k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n.
- $$
- Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
#1: Initial revision
by
Snoopy
·
2024-08-15T21:28:41Z (3 months ago)
First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $\psi(z) = \varphi(z + z_0) - z_0$, with $\tilde{\Omega} = \Omega - z_0 := \{z - z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z - z_0) + z_0 = z$, which is the desired result.
It's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
Given the conditions, we can express $\varphi(z)$ near $0$ as:
$$
\varphi(z) = z + a_n z^n + O(z^{n+1}),
$$
where $a_n$ is the coefficient of the first nonzero higher-order term. We aim to show that $a_n = 0$, implying there can be no higher-order terms in the power series.
By induction, as suggested in the hint, we find:
$$
\varphi_k(z) = z + k a_n z^n + O(z^{n+1}).
$$
Using Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
$$
|k a_n| = \left|\frac{\varphi_k^{(n)}(0)}{n!}\right| \le \frac{\|\varphi_k\|_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} |\varphi(z)| / R^n.
$$
Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.