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Given$$x,y\in\mathbb{C}^n,\quad f(x,y)=\sup_{\theta,\phi}\{\|e^{i\theta}x-e^{i\phi}y\|^2,\theta,\phi\in\mathbb{R}\}$$

Then which is/are the following are true?

$1.\ f(x,y)\le \|x\|^2+\|y\|^2-2Re|\langle x,y\rangle|$

$2.\ f(x,y)\le \|x\|^2+\|y\|^2+2Re|\langle x,y\rangle|$

$3.\ f(x,y)= \|x\|^2+\|y\|^2+2Re\langle x,y\rangle$

$4.\ f(x,y)\ge \|x\|^2+\|y\|^2-2Re\langle x,y\rangle$

I have no idea how to start with or how to solve, could any one help me?

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Myshkin
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    I changed multiple instances of $<x,y>$ to $\langle x,y\rangle$. That is standard usage. I also changed multiply instances of $||x||$ to $|x|$. That is subtler but is quite visible when $\TeX$ or $\LaTeX$ is used in the normal way, as opposed to the way it is used on web pages. And that is also standard. – Michael Hardy May 27 '13 at 04:49

1 Answers1

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First, an important equality!

$$\begin{align*}\|a-b\|^2 &= \langle a - b, a - b\rangle \\ &= \langle a,a \rangle - \langle a,b \rangle - \langle b,a \rangle + \langle b ,b \rangle \\ &= \|a\|^2 - (\langle a,b\rangle + \langle b,a\rangle) + \|b\|^2 \\ &= \|a\|^2 - (\langle a,b\rangle + \overline{\langle a,b\rangle}) + \|b\|^2 \\ &= \|a\|^2 - 2\mbox{Re}(\langle a,b\rangle) + \|b\|^2\end{align*}$$

Now, let's attack the norm inside the sup.

$$\begin{align*}\|e^{i\theta}x - e^{i\phi}y\|^2 &= \|x\|^2 - 2\mbox{Re}(\langle e^{i\theta}x, e^{i\phi}y\rangle) + \|y\|^2 \\ &= \|x\|^2 - 2\mbox{Re}(e^{i(\theta-\phi)}\langle x, y\rangle) + \|y\|^2 \end{align*}$$

Finally, use this nice fact about complex numbers: given any complex number $\lambda$, there is some $e^{ip}$ so that $\mbox{Re}(e^{ip}\lambda) = |\lambda|$. Geometrically, we are rotating the vector $\lambda$ so that it lies on the real axis. (We could also rotate to be $-|\lambda|$...)

Put these facts together, noting that $\mbox{Re}|\langle x,y\rangle| = |\langle x,y\rangle|$, since the norm is already real!

Zach L.
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