Prove that $f(x,y)= \|x\|^2+\|y\|^2+2|\langle x,y\rangle |$

Question

Let $x,y\in \Bbb C^n$.Consider

$f(x,y)=\sup _{\theta,\phi} \{||e^{i\theta}x+e^{i\phi }y||^2:\theta ,\phi \in \Bbb R\}$

Prove that

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

$||e^{i\theta}x+e^{i\phi}||^2=\langle e^{i\theta}x+e^{i\phi}y,e^{i\theta}x+e^{i\phi}y\rangle =\langle e^{i\theta }x,e^{i\theta }x\rangle +2 \text{Re}\langle e^{i\theta }x,e^{i\phi }y\rangle+\langle e^{i\phi }y,e^{i\phi }y\rangle \le\langle e^{i\theta }x,e^{i\theta }x\rangle +2 |\langle e^{i\theta }x,e^{i\phi }y\rangle|+\langle e^{i\phi }y,e^{i\phi }y\rangle =||x||^2+2|\langle x,y\rangle|+||y||^2$

On taking supremum over $\phi,\theta $ we have $f(x,y)\le ||x||^2+||y||^2+2|\langle x,y\rangle |$

But I am not getting how the equality comes into play.Please help.

Answer 1

You already showed $f(x,y) \leq \|x\|^2 + 2 | \langle x,y \rangle |+ \|y\|^2$, and you also showed $$\|xe^{i\theta}+ye^{i\phi}\|^2=\|x\|^2 + 2 \Re \langle xe^{i\theta} ,ye^{i\phi} \rangle +\|y\|^2$$

Now, $f(x,y) \geq \|xe^{i\theta} + ye^{i\phi}\| $ for all $\theta,\phi$. Can you choose them wisely so that the R.H.S will be equal $\|x\|^2 + 2 | \langle x,y \rangle |+ \|y\|^2$ (This will yield the desired equality).

Hint: You are left to pick $\theta,\phi$ such that $\Re \langle xe^{i\theta} ,ye^{i\phi} \rangle = |\langle x,y \rangle |$. Notice that $\langle x e^{i\theta} , ye^{i\phi} \rangle = \langle x , y \rangle \cdot e^{i (\theta - \phi)}$, and one can certainly make sure the right side is real and positive for some choices of $\theta,\phi$.