# 微积分网课代修|函数代写Function theory代考|MAEN5060 Genesis and Development

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Genesis and Development

The classical Schwarz lemma is part of the grist of every complex analysis class. A version of it says this:

Lemma 2.1.1. Let $f: D \rightarrow D$ be holomorphic. Assume that $f(0)=0$. Then
(a) $|f(z)| \leq|z|$ for all $z \in D$;
(b) $\left|f^{\prime}(0)\right| \leq 1$

At least as important as these two statements are the cognate uniqueness statements:
(c) If $|f(z)|=|z|$ for some $z \neq 0$, then $f$ is a rotation: $f(z)=\lambda z$ for some unimodular complex constant $\lambda$;
(d) If $\left|f^{\prime}(0)\right|=1$, then $f$ is a rotation: $f(z)=\lambda z$ for some unimodular complex constant $\lambda$.
There are a number of ways to prove this result. The classical argument is to consider $g(z)=f(z) / z$. On a circle $|z|=1-\epsilon$, we see that $|g(z)| \leq 1 /(1-\epsilon)$. Thus $|f(z)| \leq|z| /(1-\epsilon)$. Since this inequality holds for all $\epsilon>0$, part (a) follows. The Cauchy estimates show that $\left|f^{\prime}(0)\right| \leq 1$.

For the uniqueness, if $|f(z)|=|z|$ for some $z \neq 0$, then $|g(z)|=1$. The maximum modulus principle then forces $g$ to be a unimodular constant, and hence $f$ is a rotation. If instead $\left|f^{\prime}(0)\right|=1$, then $|g(0)|=1$ and again the maximum modulus principle yields that $f$ is a rotation.

## 微积分网课代修|偏微分方程代写Partial Differential Equation代考|Other Versions of Schwarz’s Lemma

Here we present some fascinating but less well known versions of the Schwarz lemma concept.

Proposition 2.2.1. Let $f$ be holomorphic on $D(0, r)$, and assume that $|f(z)| \leq$ $M$ for all $z$. Then
$$\left|\frac{f(z)-f(w)}{z-w}\right| \leq \frac{2 M r}{\left|r^{2}-\bar{z} w\right|} .$$
Proof. Define
$$g(z)=\frac{f(r z)}{M} .$$
Then $g: D \rightarrow D$ and we may apply Schwarz-Pick to $g$. The result is
$$\left|\frac{g(z)-g(w)}{1-\overline{g(z)} g(w)}\right| \leq\left|\frac{z-w}{1-\bar{z} w}\right|,$$
which translates to
$$\left|\frac{f(r z) / M-f(r w) / M}{1-\overline{f(r z) / M} \cdot f(r w) / M}\right| \leq\left|\frac{z-w}{w-\bar{z} w}\right| .$$

## 微积分网课代修|偏微分方程代写Partial Differential Equation代 考|Genesis and Development

(一) $|f(z)| \leq|z|$ 对所有人 $z \in D$;
(二) $\left|f^{\prime}(0)\right| \leq 1$

(c) 如果 $|f(z)|=|z|$ 对于一些 $z \neq 0$ ，然后 $f$ 是一个旋转: $f(z)=\lambda z$ 对于一些单 模复数常数 $\lambda$;
(d) 如果 $\left|f^{\prime}(0)\right|=1$ ，然后 $f$ 是一个旋转: $f(z)=\lambda z$ 对于一些单模复数常数 $\lambda$. 有很多方法可以证明这个结果。经典的论点是考虑 $g(z)=f(z) / z$. 在一个圆圈上 $|z|=1-\epsilon$, 我们看到 $|g(z)| \leq 1 /(1-\epsilon)$. 因此 $|f(z)| \leq|z| /(1-\epsilon)$. 由于这种不 等式适用于所有人 $\epsilon>0$ ，(a) 部分如下。柯西估计表明 $\left|f^{\prime}(0)\right| \leq 1$.

## 微积分网课代修|偏微分方程代写Partial Differential Equation代 考 Other Versions of Schwarz’s Lemma

$$\left|\frac{f(z)-f(w)}{z-w}\right| \leq \frac{2 M r}{\left|r^{2}-\bar{z} w\right|} .$$

$$g(z)=\frac{f(r z)}{M} .$$

$$\left|\frac{g(z)-g(w)}{1-\overline{g(z)} g(w)}\right| \leq\left|\frac{z-w}{1-\bar{z} w}\right|,$$

$$\left|\frac{f(r z) / M-f(r w) / M}{1-\overline{f(r z) / M} \cdot f(r w) / M}\right| \leq\left|\frac{z-w}{w-\bar{z} w}\right| .$$