Following on from the previous blog post, it is worth giving a cleaned-up version of the proof. While putting this together, I also found a way to obtain the optimal constant of $2$, although I am not entirely happy with the trick needed to achieve this.

## Restating the problem (with the sharp constant)

As in the previous post, we define a function $\phi: \mathbb C \to \mathbb C$ by $\phi(z) = |z|^{-1/2}z$. Alternatively: $\phi(re^{i\theta}) = r^{1/2}e^{i\theta}$ for all $r\geq 0$ and all $\theta\in\mathbb R$. Our goal is now to prove the following inequality

Proposition 1. $|\phi(w)-\phi(z)|^2 \leq 2 |w-z|$ for all $w,z\in \mathbb C$.

## Reducing the problem to one with fewer parameters

We can simplify/reduce the problem slightly using symmetry arguments. To maintain a more formal style than the previous post, let’s spell this out more explicitly.

Lemma 2. Let $t\geq 1$. Then $|t-e^{i\theta}|^2 \leq 2 |t^2-e^{i\theta}|$ for all $\theta\in [0,\pi]$.

Proof of Proposition 1, using Lemma 2. The desired inequality is trivial if either $w$ or $z$ is zero.
Let $w,z\in \mathbb C\setminus\{0\}$, and assume without loss of generality that $|w|\leq |z|$.

Since $\phi$ is angle-preserving (and orientation-preserving) we may apply a rotation or reflection of the complex plane which maps $z$ to the real-line and $w$ to a point in the closed upper-half plane. We then let
$t= |z|^{1/2}|w|^{-1/2} \geq 1$ and choose $\theta\in [0,\pi]$ such that $w=|w|e^{i\theta}$.

Then

$(1)\qquad\qquad \displaystyle |\phi(w)-\phi(z)|^2 = | |w|^{1/2} e^{i\theta} - |z|^{1/2} |^2 = |w| | e^{i\theta} - t |^2$

which, by Lemma 2, is bounded above by

$(2)\qquad\qquad \displaystyle 2|w| |t^2-e^{i\theta}| = 2 |w| \left\vert |z| |w|^{-1} - e^{i\theta} \right\vert = 2|z -w|$

as required. Q.E.D.

## The proof of Lemma 2

The proof of Lemma 2 will be divided into two cases.

### Case 1. $0\leq\theta \leq \pi/2$.

In this case $|t- e^{i\theta}| \leq |t+e^{i\theta}|$; this is evident from the geometry of this configuration, but also follows from squaring both sides and using $\cos\theta\geq 0$. Therefore

$(3)\qquad\qquad \displaystyle | t-e^{i\theta}|^2 \leq |t-e^{i\theta}| |t+e^{i\theta}| = |t^2-e^{2i\theta} |.$

Also, observe that

$(4)\qquad\qquad \displaystyle |e^{i\theta} -e^{2i\theta} | = |1-e^{i\theta} | \leq |t^2-e^{i\theta} |\;.$

The last inequality is once again evident from a geometrical argument. To give a more algebraic/Cartesian argument: the two complex numbers $t^2-e^{i\theta}$ and $1-e^{i\theta}$ have the same imaginary part; their real parts are $t^2-\cos\theta$ and $1-\cos\theta$ respectively; and $t^2-\cos\theta\geq 1-\cos\theta \geq 0$ since $t\geq 1$.

Combining (3) and (4) we obtain

$(5)\qquad\qquad \displaystyle |t-e^{i\theta}|^2 \leq |t^2-e^{2i\theta} \leq |t^2-e^{i\theta} | + |e^{i\theta}- e^{2i\theta} | \leq 2|t^2-e^{i\theta}|$

as required. Q.E.D.

### Case 2. $\pi/2 \leq\theta \leq \pi$.

Let $d= -\cos\theta\in [0,1]$, so that

$(6)\qquad\qquad\displaystyle |t-e^{i\theta}|^2 = t^2+1 +2td \quad\hbox{ and } \quad |t^2-e^{i\theta}| = (t^4+1+2t^2d)^{1/2}\;.$

Note that as a special case of the Cauchy–Schwarz inequality,

$(7)\qquad\quad\displaystyle x_1+x_2+x_3+x_4 \leq 2 \sqrt{x_1^2+x_2^2+x_3^2+x_4^2} \quad\hbox{ for all }x_1,x_2,x_3,x_4 \geq 0.$

Applying (7) with $x_1=t^2, x_2 = 1, x_3 =x_4 = td^{1/2}$, we obtain

$(8)\qquad\qquad\displaystyle t^2+1+2td^{1/2} \leq 2 ( t^4+1+2t^2d)^{1/2}\;.$

Since $0\leq d\leq 1$ and $t\geq 0$, $2td \leq 2td^{1/2}$. Combining this with (6) and (8), we conclude that

$(9)\qquad\qquad\displaystyle |t-e^{i\theta}|^2 \leq t^2+1+2d^{1/2} \leq 2 |t^2-e^{i\theta}|$

as required. Q.E.D.

## Remarks

1. In comparison with the approaches taken in the previous blog post, note that in Case 2 we have improved the constant from $2\sqrt{2}$ to $2$, but with a slightly more ad hoc argument that is less geometric (although we are able to sidestep any messy calculus with Cauchy–Schwarz).
2. It is also worth noting that for much of the proof, the formulation in terms of complex numbers is merely for convenience, and that most of the arguments merely take place in $\mathbb R^2$ with its usual norm — or, to be more abstract and less co-ordinate dependent, in 2-dimensional Euclidean space. The exception is one part of Case 1, namely the inequality (3), where we made use of the algebraic structure in $\mathbb C$ and the multiplicative property of $| \cdot |$. Can we find an alternative approach, which only uses the Euclidean geometry of $\mathbb R^2$?
3. On a related note: the definition of the map $\phi$ can be extended to any normed space $V$, as a radial square-root map: we define $\phi(\mathbf{0})=\mathbf{0}$ and $\phi(v) = \Vert v \Vert^{-1/2} v$ for any non-zero vetor $v\in V$. If $V$ is Euclidean (i.e. its norm satisfies the parallelogram law) then, since 2-dimensional subspaces of any Euclidean real vector space are Euclidean, the previous remarks and Proposition 1 show that $\Vert \phi(v_1)-\phi(v_2)\Vert^2 \leq 2 \Vert v_1-v_2\Vert$ for all $v_1,v_2\in V$. What happens for more general norms?