## Category Archives: Analysis

10544

### One example of blowing up corner

Consider ${u(x,y)=\sqrt{x^2+y^4}}$ on ${\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}$. Notify ${\mathbb{R}^2_+}$ has a corner at the origin and ${u}$ is not smooth at the origin. ${u\approx x}$ when ${x\geq y^2}$ and ${u\approx y^2}$ when ${x\leq y^2}$. We want to resolve ${u}$ by blowing up the origin through a map ${\beta}$. After blowing up, ${W}$ looks like the following picture.

Denote ${W=[\mathbb{R}^2_+,(0,0)]}$ and ${\beta:W\rightarrow \mathbb{R}^2_+}$ is the blow down map. On ${W\backslash lb}$(near A), ${\beta}$ takes the form

$\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})$

where ${\xi_1}$ is the boundary defining function for ff and ${\eta_1}$ is boundary defining function for rb. Similarly on ${W\backslash rb}$(near B), ${\beta}$ takes the form

$\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)$

where ${\xi_2}$ is a bdf for lb and ${\eta_2}$ is a bdf for ff. One can verify that ${\beta}$ is a diffeomorphism ${\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}$. Let ${w=\beta^* u}$. Then ${w}$ is a polyhomogeneous conormal function on ${W}$. Its index can be denoted ${(E,F,H)}$ correspond to lb, ff and rb.

$\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}$

Suppose ${\pi_1}$ is the projection to ${x}$ coordinate. Consider ${f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}$, ${f}$ is actually a ${b-}$fibration. Then the push forward map ${f_*}$ maps ${w}$ to a polyhomogeneous function on ${\mathbb{R}^+}$.

$\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy$

In order to make ${u}$ is integrable, let us assume ${u}$ support ${x\leq 1}$ and ${y\leq 1}$. What is the index for ${f_* w}$ on ${\mathbb{R}^+}$?

$\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy$

For the first integral, letting ${y^2/x=t}$

$\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}$

For the second integral, letting ${x/y^2=t}$

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt$

Since the Taylor series

$\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1$

Consequently

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots$

Combining all the above analysis, the index for ${f_*w}$ is ${\{(n,0)\}\cup \{\frac{n}{2},0\}}$

From another point of view, the vanishing order of ${f}$ on each boundary hypersurface of ${W}$ are ${e_f(lb)=1}$, ${e_f(\text{ff})=2}$ and ${e_f(rb)=0}$. ${f}$ maps lb and ff to the boundary of ${\mathbb{R}^+}$. Therefore the index of ${f_*w}$ is contained in

$\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F$

Remark: Daniel Grieser, Basics of ${b-}$Calculus.

### Compensated compactness

Suppose $T$ is a vector field and $\nabla\cdot T = 0$. $E= \nabla \psi$ and $\psi$ is a scalar function. We have following theorem(Coifman-Lions-Meyers-Semmes)

Theorem: If $T\in L^2(\mathbb{R}^n)$ and $T\in L^2(\mathbb{R}^n)$, then $E\cdot T\in \mathcal{H}^1(\mathbb{R}^n)$, which is the hardy space.
Given $f(x)\in L^1(\mathbb{R}^n)$, it has harmonic extension $\mathbb{R}^{n+1}_+=\{(x,t)|x\in\mathbb{R}^n, t>0\}$

$\tilde{f}(x,t)=c_n\int_{\mathbb{R}^n}\frac{ f(x-y)t}{(t^2+|x|^2)^{\frac{n+1}{2}}}dy$

Definition: the non-tangential maximal function

$N(f)=\sup_{(\xi,t)\in \Gamma(x)}|\tilde f(\xi, t)|$

It is easy to prove that $N(f)\leq c_n f^*(x)$ the Hardy-Littlewood maximal function. From this we can Hardy norm as

$||f||_{\mathcal{H}^1}=||f||_{L^1}+||N(f)||_{L^1}$

Hardy space consists of all $f$ having finite hardy norm. There is well know fact that the dual space of $\mathcal{H}^1$ is BMO, which is defined as the following.

Define $f\in L^1_{loc}(\mathbb{R}^n)$, if for any cube $Q$,

$\sup_Q\frac{1}{|Q|}\int_Q|f-f_Q|<\infty,\quad \text{where }f_Q=\frac{1}{|Q|}\int_Qf$

then $f\in BMO$. $L^\infty \subset BMO$ and $\log|x|\in BMO$ but not in $L^\infty$.

Let us see how do we use the main theorem. Suppose on $\mathbb{R}^2$, $u$ is the solution of the following elliptic equation

$\displaystyle\frac{\partial}{\partial x_i}\left(a_{ij}(x)\frac{\partial u}{\partial x_j}\right)=\frac{\partial f}{\partial x_1}\frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2}\frac{\partial f}{\partial x_2}$

where $||\nabla f||_{L^2}<\infty$, $||\nabla g||_{L^2}<\infty$ and $(a_{ij})$ is uniform elliptic. YanYan Li and Sagun Chanillo proved that the green function of this elliptic operator belongs to BMO. The right hand side of this equation can be rewritten as $T\cdot E$, where

$T=\left(\frac{\partial f}{\partial x_2}, -\frac{\partial f}{\partial x_1}\right),\quad E=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_a}\right)$

therefore the right hand side belong to $\mathcal{H}^1$. Since

$u(x)=\int G_x(y)T\cdot E(y)dy$

therefore from the theorem we stated at the beginning, we get

$||u||_\infty\leq C||\nabla f||_{L^2}||\nabla g||_{L^2}$

### BV function and its property involves translation

Theorem 1 Suppose ${u\in L^1(\mathbb{R})}$, then ${u\in \text{BV}}$ if and only if ${\exists\, C}$ such that

$\displaystyle ||\tau_hu-u||_{L^1(\mathbb{R})}\leq C|h|,\quad \forall\, h$

Moreover, one can take ${C=|u|_{BV}}$. Here ${\tau_hu(\cdot)=u(\cdot+h)}$ is the tanslation operator.

Proof: Firstly suppose ${u\in \text{BV}}$. Let us prove

$\displaystyle \left|\int_{\mathbb{R}}(\tau_hu(x)-u(x))\phi(x)dx\right|\leq |u|_{BV}|\phi|_{L^\infty}|h|,\quad \forall \phi\in C^\infty_c(\mathbb{R}) \ \ \ \ \ (1)$

To show that

$\displaystyle LHS=\left|\int_{\mathbb{R}}u(x)(\phi(x-h)-\phi(x))dx\right|$

$\displaystyle =\left|\int_{\mathbb{R}}u(x)\psi(x)hdx\right|$

$\displaystyle \leq |u|_{BV}|\psi|_{L^\infty}|h|$

where

$\displaystyle \psi(x)=\int^x_{-\infty}\frac{\phi(s-h)-\phi(s)}{h}ds\in C_c^\infty(\mathbb{R})$

it is easy to verify ${|\psi|_\infty=|\phi|_\infty}$ therefore (1) is proved. Next one can choose such ${\phi_n\rightarrow sign(\tau_hu-u)\in L^1}$ with ${|\phi_n|\leq 1}$(it is easy to show by mollification). By dominating theorem, one get

$\displaystyle \int_{\mathbb{R}}|\tau_hu(x)-u(x)|dx\leq |u|_{BV}|h|.$

The other direction need more analysis. $\Box$

### Automorphism of annulus

Suppose $\mathbb{R}_\mu=\{z\in \mathbb{C}:1 is the annulus for $\mu>1$. Then

$\bf{Thm:}\mathbb{R}_\mu$ is not biholomorphic to $\mathbb{R}_{\mu'}$ whenever $\mu\neq \mu'$.

The proof goes like following: Suppose there exists a biholomorphic map between them, then this map can be extended to the boundary by the so called Kellog theorem, then one can reflect the annulus the inner part and outside part. Near the origin, one can use the removable singularity theorem to prove the extended map is holomophic. Thus we can get a biholomorphic map between $\mathbb{C}$ and $\mathbb{C}$ which must have the form $az+b$. First $b=0$, because the map keeps the origin. Second $a$ must have the unit length to keep the inner boundary to be $|z|=1$. Thus the map can only be rotation around origin.

This actually tells us the automorphism of an annulus can only be the rotation. Let us verify this fact by another perspective.

As we all know, $\mathbb{R}_\mu$ is biholomorphic to $\mathbb{H}/\Gamma_\lambda$, where $\mathbb{H}$ is the upper half plane and $\Gamma_\lambda(z)=\lambda z$. Here $\ln \lambda\ln \mu=2\pi^2$. $\Gamma_\lambda$ acts on the $\mathbb{H}$ freely discontinuously.

Any automorphism of $\mathbb{H}/\Gamma_\lambda$ can be lift to an automphism of $\mathbb{H}$ which commutes with the action $\Gamma_\lambda$. In the matrix language, these automphisms are the matrix in $PSL(2,\mathbb{R})$ commutes with $\left(\begin{matrix}\sqrt{\lambda}&0\\0&\sqrt{\lambda^{-1}}\end{matrix}\right)$. Some elementary calculation shows that these matrix have the form $\left(\begin{matrix}a&0\\0&a^{-1}\end{matrix}\right)$ where $a\in \mathbb{R}$. Any one of these  is corresponding to a rotation of the annulus.

### Dual spaces

Suppose ${(A,||\cdot||_A)}$ and ${(B,||\cdot||_B)}$ are Banach space, ${A^*}$ and ${B^*}$ are their dual spaces. If ${A\subset B}$ with ${||\cdot||_B\leq C||\cdot||_A}$, then

$\displaystyle i:A\mapsto B$

$\displaystyle \quad x\rightarrow x$

is an embedding. Let us consider the relation of two dual spaces. For any ${f\in B^*}$

$\displaystyle |\langle f,x\rangle|=|f(x)|\leq ||f||_{B^*}||x||_B\leq C||f||_{B^*}||x||_A\quad \forall\, x\in A$

Then ${f|_{A}}$ will be a bounded linear functional on ${A}$

$\displaystyle i^*:B^*\mapsto A^*$

$\displaystyle \qquad f\rightarrow f|_A$

is a bounded linear operator.

In a very special case that ${A}$ is a closed subset of ${B}$ under the norm ${||\cdot||_B}$, one can prove ${i^*}$ is surjective. In fact ${\forall\,g\in A^*}$ can be extended to ${\bar{g}}$ on ${B}$ by Hahn-Banach thm such that ${i^*\bar{g}=g}$. Then

$\displaystyle A^*=B^*/\ker i^*.$

Let us take ${A=H^1_0(\Omega)}$ and $\displaystyle B=H^1(\Omega)$ for example. Define the inner product to be

$\displaystyle \langle u,v\rangle=\int_\Omega uv+D_iuD_iv$

When ${\Omega}$ is a bounded subset of ${\mathbb{R}^n}$, ${H^1_0(\Omega)}$ is a proper closed subset of ${H^1(\Omega)}$. From the above explanation,

$\displaystyle H^{-1}(\Omega)=(H^1_0(\Omega))^*=(H^1(\Omega))^*/\ker i^*$

Define the continuous linear functional ${f(u)= \int_{\partial \Omega}u}$ for ${u\in H^1(\Omega)}$. One can see that ${i^*f=0}$.

$\textbf{Remark:}$ Should attribute this to Lun Zhang. I am always confused that $B^*\subset A^*$.

### Sobolev functions on puncture ball

Let us first see the prove of sobolev embedding of ${W^{1,1}\rightarrow L^2}$ on the plane.

Lemma: Suppose ${f\in W^{1,1}(\mathbb{R}^2)}$ with compact support. Then

$\displaystyle ||f||_{L^2}\leq ||\nabla f||_{L^1}$

Proof: Let us suppose ${f\in C^\infty_c(\mathbb{R}^2)}$, the general case can be proved by approximation. Since ${f}$ has compact support, then

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^x\frac{\partial f}{\partial x}(t,y)dt\right|\leq \int_{-\infty}^{\infty}|\nabla f|(t,y)dt=F(y)$

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^y\frac{\partial f}{\partial y}(x,s)ds\right|\leq \int_{-\infty}^{\infty}|\nabla f|(x,s)ds=G(x)$

Then

$\displaystyle \int_{\mathbb{R}^2}f^2(x,y)dxdy\leq \int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(x)dxdy$

$\displaystyle =\int_{-\infty}^\infty F(y)dy\int_{-\infty}^\infty G(x)dx\\ =\left(\int_{-\infty}^\infty\int_{-\infty}^\infty|\nabla f|dxdy\right)^2$

$\Box$

Suppose we have a function ${u\in W^{1,1}_{loc}(D\backslash\{0\})}$, where ${D}$ is the unit disc in ${\mathbb{R}^2}$, ${u}$ can blow up wildly near the origin. However if we know ${\nabla u\in L^1(D)}$, then actually ${u\in L^2(D)}$ and ${u\in W^{1,1}(D)}$.

Proof: Because the bad thing happened only at origin, we can suppose ${u}$ has spt inside ${\frac{1}{4}D}$ or ${D_{1/4}}$. Put a substantially large square box ${B_\epsilon}$ with length ${\frac 12}$ inside the left half of the disc ${D^-}$whose distance to the origin is ${\epsilon}$ see the picture.

puncture disc

Then on the three sides, ${l_1}$, ${l_2}$, ${l_3}$, ${u=0}$. Using the proof of the above, one can prove

$\displaystyle ||u||_{L^2(B_\epsilon)}\leq ||\nabla u||_{L^1(B_\epsilon)}\leq ||\nabla u||_{L^1(D)}$

Letting ${\epsilon\rightarrow 0}$, we get ${u\in L^2(D^-)}$. The same proof works for the right part ${D^+}$. Finally ${u\in L^2(D)}$. Choose a cut off function ${\zeta_\epsilon=\zeta(x/\epsilon)}$. Then

$\displaystyle u\zeta_\epsilon\rightarrow u\text{ in }L^1$

$\displaystyle \nabla(u\zeta_\epsilon)\rightarrow \nabla u\text{ in }L^1$

So ${u\in W^{1,1}(D)}$. $\Box$

Remark: This is called the removable singularity. There is a more general theorem related to this. Assume ${n\geq 2}$, ${K\subset\subset\Omega}$ such that ${\mathcal{H}^{n-2}(K)=0}$. Suppose ${u\in W^{1,1}_{loc}(\Omega\backslash K)}$ and ${\int_{\Omega\backslash K}|\nabla u|dx<\infty}$. Then ${u\in W^{1,1}(\Omega)}$.

I learned this from Prof. Brezis’s class. Also see his book: Sobolev maps with values into the circle.

### Hardy inequality in dimesion 2

Suppose ${u}$ is a smooth function defined in ${B^c=\{|x|>1\}}$ in the plane ${\mathbb{R}^2}$, assume ${u=0}$ on ${\partial B^c}$ and also has compact support, then

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx\leq 4\int_{|x|>1}|\nabla u|^2dx$

There is a way presented by my advisor to explain why the strange function ${|x|^2\ln^2|x|}$ pops up here. First we transform LHS to polar coordinates

$\displaystyle \int_{|x|> 1}\frac{u(x)^2}{|x|^2\ln^2|x|}dx=\int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{r^2\ln^2 r}rdrd\theta$

which leads us to start with a very general ${f(r)}$

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta$

Suppose ${F'(r)=r/f(r)}$, then

$\displaystyle \int_{0}^{2\pi}\int_{1}^\infty\frac{u(r,\theta)}{f(r)}rdrd\theta=\int_{0}^{2\pi}\int_{1}^\infty u(r,\theta)dF(r)d\theta$

$\displaystyle =\int_{0}^{2\pi}\left(u(r,\theta)F(r)|_1^\infty-\int_{1}^\infty F(r)\partial_r u(r,\theta)dr\right)d\theta$

$\displaystyle =-\int_{0}^{2\pi}\int_{1}^\infty F(r)\partial_r u(r,\theta)drd\theta=-\int_{|x|>1} \frac{F(r)\partial_r u(x)}{r}dx$

$\displaystyle \leq 2\left(\int_{|x|>1}\frac{u(x)^2}{f(r)}dx\right)^{1/2}\left(\int_{|x|>1}\frac{f(r)F^2(r)}{r^2}|\nabla u|^2dx\right)^{1/2}$

So we only need to find ${f(r)}$ and ${F(r)}$ such that

$\displaystyle \frac{f(r)F^2(r)}{r^2}\leq C$

Actually one can solve the ODE

$\displaystyle \frac{f(r)F^2(r)}{r^2}=1, \quad F'(r)=\frac{r}{f(r)}$

to get ${F(r)=\frac{-1}{\ln r}}$, ${f(r)=r^2\ln^2 r}$. Plugging in this function back to the above proof gives you the desired inequality.

### Polynomial on each variable

Problem: Suppose ${f\in \mathcal{O}(\mathbb{C}^2)}$. If ${f(z_1,z_2)}$ is a polynomial of ${z_1}$ for every fixed ${z_2}$ and a polynomial of ${z_2}$ for every fixed ${z_1}$. Then ${f}$ must be a polynomial.

Proof: Since ${f}$ is holomorphic on the ${\mathbb{C}^2}$, we have the power series expansion

$\displaystyle f(z_1,z_2)=\sum_{i,j\geq 0}a_{ij}z_1^iz_2^j=\sum_{i\geq 0}a_i(z_1)z_2^i$

where ${a_i(z_1)}$ is a holomorphic function on ${\mathbb{C}}$ for any ${i\geq 0}$. Suppose there exists a sequence ${i_1,i_2,\cdots}$ with ${\lim_{k\rightarrow \infty}i_k=\infty}$ satisfying ${a_{i_k}(z_1)\neq 0}$. Since the set of roots of ${a_{i_k}}$ are countable in ${\mathbb{C}}$, then all the union of these sets are also countable. Therefore, ${\exists\, w\in\mathbb{C}}$ such that ${a_{i_k}(w)\neq 0}$ for all ${k\geq 0}$. Then ${f(w,z_2)}$ can never be a polynomial. So

$\displaystyle f(z_1,z_2)=\sum_{i=0}^Ma_i(z_1)z_2^i=\sum_{i\geq 0}b_i(z_2)z_1^i$

where every ${b_i}$ is polynomial of ${z_2}$. Repeating the above proof we get this is also a finite sum, which means ${f}$ is a polynomial. $\Box$

Remark: Exercise from Shabat, 1.10. Idea comes from Hanlong Fang.

### Some uniqueness sets of holomorphic function

Suppose ${f\in \mathcal{O}(\mathbb{C}^2)}$, if ${f}$ is zero on the following set
(a) A real hyperplane in ${\mathbb{C}^2}$
(b) The real two dimensional plane ${\{z_1=\bar{z}_2\}}$
(c) the arc ${\{z_1=\bar{z}_2,y_1=x_1\sin(1/x_1)\}}$
Show that ${f}$ must be zero on ${\mathbb{C}^2}$.

Proof: (a) Suppose the hyperplane has normal vector ${(1,0)\in \mathbb{C}^2}$ and passing through origin, then ${(iy,z_2)}$ will be in this hyperplane for any ${y\in \mathbb{R}}$.

Fix ${z_2}$, ${f(\cdot,z_2)}$ is a holomorphic function in ${\mathbb{C}}$. Because zero point of holomorphic function is isolated unless a constant, ${f(iy,z_2)=0\quad\forall\,y\in \mathbb{R}}$ will imply ${f(z_1,z_2)\equiv 0}$.

(b) Let ${w_1=\frac{z_1+z_2}{2}}$, ${w_2=\frac{z_1-z_2}{2i}}$, then

$\displaystyle \{z_1=\bar{z}_2\}=\{\text{Im}\,w_1=\text{Im}\,w_2=0\}$

${g(w_1,w_2)=f(z_1,z_2)}$ is also a holomorphic function in ${w_1,w_2}$. As we did in part (a), fix ${w_1=x\in \mathbb{R}}$, ${g(x,y)=0}$ imply ${g(x,w_2)=0}$ for any ${w_2\in\mathbb{C}}$. Since ${x}$ is arbitrary, then ${g(w_1,w_2)\equiv 0}$.

(c) Use the transformation of part (b), ${\{z_1=\bar{z}_2, y_1=x_1\sin \frac{1}{x_1}\}}$ is equivalent

$\displaystyle \gamma=\{\text{Im}\,w_1=\text{Im}\,w_2=0,w_2=w_1\sin\frac{1}{w_1}\}$

Consider ${g(w_1,w_2)}$, then ${g(\cdot,0)=0}$ has a sequence of roots ${w_1}$ converging to 0, then ${g(w_1,0)=0}$ for any ${w_1\in \mathbb{C}}$. From the Taylor expansion, there exists ${g_1\in \mathcal{O}(\mathbb{C}^2)}$ such that

$\displaystyle g(w_1,w_2)=w_2g_1(w_1,w_2)$

Since ${g(w_1,w_2)}$ is zero on ${\gamma}$, then ${g_1(w_1,w_2)}$ is zero on ${\gamma}$ when ${w_2\neq 0}$. By continuity, ${g_1(w_1,w_2)}$ is zero on ${\gamma}$.

Then we can repeat the above procedure, ${\exists\, g_2}$ such that

$\displaystyle g_1(w_1,w_2)=w_2g_2(w_1,w_2),\quad g=w_2^2g_2$

Then we must have ${g\equiv 0}$, otherwise we can continue this process infinitely. $\Box$

Remark: The idea of part (b) is due to Lun Zhang.

### Complex lines in C^2

Problem: (a) Suppose ${l_1}$ and ${l_2}$ are two complex lines in ${\mathbb{C}^2}$ which are not orthogonal. Prove that the projection of circles ${\Gamma\subset l_1}$ to ${l_2}$ are also circles.

(b) Suppose ${l_1}$ is a 2 dimensional real plane in ${\mathbb{C}^2}$ and ${l_2}$ is a complex line. If for any circle ${\Gamma\subset l_1}$, its projection to ${l_2}$ is a circle on ${l_2}$, then ${l_1}$ must be a complex or anticomplex line. Proof: (a) Under some translation and rotation which preserves the projection and circles, we can assume

$\displaystyle l_1=\{(z,wz)|z\in \mathbb{C}\}$

$\displaystyle l_2=\{(\xi,0)|\xi\in \mathbb{C}\}$

where ${w}$ is a fixed constant in ${\mathbb{C}}$. Here we have used the property that ${l_2}$ is not orthogonal to ${l_1}$. Suppose ${\Gamma}$ is circle in ${l_1}$ with center ${(z_0, wz_0)}$ with radius ${r}$, then points on ${\Gamma}$ satisfy

$\displaystyle r=|(z-z_0,w(z-z_0))|=|z-z_0|\sqrt{1+|w|^2}$

which means ${|z-z_0|=const}$. This exactly means the projection on ${l_2}$ is a circle.

(b) Any 2 dimensional real space through origin can be respresented by

$\displaystyle l_1=\{\alpha\zeta+\beta\bar{\zeta}|\zeta\in \mathbb{C}\}$

where ${\alpha=(\alpha_1,\alpha_2)\in\mathbb{C}^2,\beta=(\beta_1,\beta_2)\in\mathbb{C}^2}$. WLOG, assume ${l_2}$ is the same as (a).

If ${|\alpha_1|\neq |\beta_1|}$, then let ${z=\alpha_1\zeta+\beta_1\bar{\zeta}}$,

$\displaystyle l_1=\left\{\left(z,\frac{(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}{|\alpha_1|^2-|\beta_1|^2}\right)\right\}$

By projecting to ${l_2}$, we get ${|z|=const}$ must imply ${(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}$ has constant norm also. One can verify that this is only true when ${\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2=0}$ or ${\alpha_1\beta_2-\alpha_2\beta_1=0}$, which means ${l_1}$ must be an anti-complex line or complex line.

If ${|\alpha_2|\neq |\beta_2|}$, we can porject circles on ${l_1}$ to ${l_3=\{(0,\zeta)\}}$, which are also circles by assumption. So we can apply all the above proof.

Next suppose ${|\alpha_1|=|\beta_1|\neq 0}$ and ${|\alpha_2|= |\beta_2|\neq 0}$, first note that

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}\equiv const$

In fact, for any ${\zeta}$, there exists ${c_1>0}$ such that

$\displaystyle |\alpha_1\zeta+\beta_1\bar{\zeta}|+|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1$

then ${\zeta/c_1}$ will be in a circle in ${l_1}$ whose center is origin and radius 1. So ${|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1c_0}$ for some constant ${c_0}$. Then

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{c_1c_0}{c_1-c_1-c_0}=const$

Choosing ${\zeta}$ in particular, this constant must be ${|\alpha_1|/|\alpha_2|=|\beta_1|/|\beta_2|}$. From

$\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{|\alpha_1|}{|\alpha_2|}$

Squaring it and write it as equality, we get

$\displaystyle [|\alpha_2|^2\alpha_1\bar{\beta}_1-|\alpha_1|^2\alpha_2\bar{\beta}_2]\zeta^2=[|\alpha_1|^2\bar{\alpha_2}\beta_2-|\alpha_2|^2\bar{\alpha_1}\beta_1]\bar{\zeta}^2$

Then

$\displaystyle |\alpha_2|^2\alpha_1\bar{\beta}_1=|\alpha_1|^2\alpha_2\bar{\beta}_2$

$\displaystyle |\alpha_1|^2\bar{\alpha_2}\beta_2=|\alpha_2|^2\bar{\alpha_1}\beta_1$

This two imply ${\alpha_1/\beta_1=\alpha_2/\beta_2=\eta^{-1}}$. Hence

$\displaystyle l_1=\{(\alpha_1,\alpha_2)(\zeta+\eta\bar{\zeta})|\zeta\in\mathbb{C}\}$

However, ${|\zeta|=|\eta\bar{\eta}|}$ implies that

$\displaystyle \text{Arg}\,(\zeta+\eta\bar{\zeta})=\frac{1}{2}\text{Arg}\,\eta$

then ${l_1}$ is in fact only one dimesional in this case. Contradiction. $\Box$

Remark: Shabat, Introduction to complex analysis Part II. Functions of several variables.