Which of following inequality holds?

Question

$f(x,y) = _{\theta,\phi}^{{sup}}$ $ ||e^{i\theta}x - e^{i\phi}y||_{2}^2$ ,where $x,y \in \mathbb{C^2}$ and $\theta,\phi \in \mathbb{R}$

1. $f(x,y) \leq ||x||^2 + ||y||^2 - 2Re (\langle x,y \rangle )$
2. $f(x,y) \leq ||x||^2 + ||y||^2 + 2Re ( \langle x,y\rangle )$
3. $f(x,y) \leq ||x||^2 + ||y||^2 - 2 \mid \langle x,y\rangle \mid$
4. $f(x,y) \geq ||x||^2 + ||y||^2 - 2Re ( \langle x,y\rangle )$

My claim :

Take n=2. let $x= (x_1 ,x_2), y = (y_1,y_2)$

then $||e^{i\theta}x - e^{i\phi}y||_{2} = \mid \mid (e^{i\theta}x_1-e^{i\phi} y_1,e^{i\theta}x_1-e^{i\phi} y_2) \mid \mid _{2}.$

Is that any two norms in $\mathbb{C^2}$ are also equivalent ?

Now we have to find the partial derivative w.r.t $\theta$ and $ \phi$ of both the co-ordinates and equate to zero and find the critical points to find the maximum.

Is that a correct way to proceed ?

Answer 1

Inner products

In $\mathbb C:\;\,\qquad \langle x, y\rangle=\bar yx$

In $\mathbb C^2:\qquad\langle x, y\rangle=\bar y^\top x$

Norms

In $\mathbb C:\;\,\qquad ||x||_1^2=x\bar x$

In $\mathbb C^2:\qquad||x||_2^2=||x_1||_1^2+||x_2||_1^2$

Calculation

$$ ||e^{\theta}x-e^{i\phi}y||_2^2=(e^{i\theta}x_1-e^{i\phi}y_1)\overline{(e^{i\theta}x_1-e^{i\phi}y_1)}+(e^{i\theta}x_2-e^{i\phi}y_2)\overline{(e^{i\theta}x_2-e^{i\phi}y_2)}=\\ (e^{i\theta}x_1-e^{i\phi}y_1){(e^{-i\theta}\bar x_1-e^{-i\phi}\bar y_1)}+(e^{i\theta}x_2-e^{i\phi}y_2)(e^{-i\theta}\bar x_2-e^{-i\phi}\bar y_2)=\\ ||x||^2_2+||y||_2^2-e^{i(\theta-\phi)}\langle x,y\rangle-e^{-i(\theta-\phi)}\overline{\langle x,y\rangle}=\\ ||x||^2_2+||y||_2^2-2\Re\left(e^{i(\theta-\phi)}\langle x,y\rangle\right) $$

Conclusion

$$ f(x,y)=\sup_{\theta,\phi}||e^{i\theta}x-e^{i\phi}y||_2^2\ge\sup_{\phi}||e^{i\phi}x-e^{i\phi}y||_2^2=||x||^2_2+||y||_2^2-2\Re\left(\langle x,y\rangle\right) $$