Show the equivalence of two norms Let $H$ be a complex vector space and 
$\|u+iv\|_1 := \sqrt{\langle u,u\rangle+\langle v,v\rangle}$, (This norm is induced by a scalar product on $H$)
$\|u+iv\|_2 := \sup_{\alpha\in[0,2\pi)}\|\cos(\alpha) u +\sin(\alpha) v\|$ 
be two norms defined on $H$. Note that $\|\cdot\|$ (without the subscript) is another norm, not explicitly defined.
I want to test if $\|\cdot\|_1,\|\cdot\|_2$ are equal. For equivalence I would have to show that
$\exists c > 0, \forall x\in H\colon \|x\|_1\le c\|x\|_2$ but also 
$\exists c > 0, \forall x\in H\colon \|x\|_2\le c\|x\|_1$ .
All my attempts have failed so far. I tried various estimates to get from one to another (triangle inequalities etc.), but I can not even make one direction seem to work.
 A: There must be something wrong with the question, as $\|\cdot\|_1$ is not well-defined. For example, take $H = \mathbb{C}^2$, and
$$\|(z, w)\| = \sqrt{|z|^2 + 2|w|^2}.$$
We can then compute $\|(1, 0)\|_1$ multiple ways:
\begin{align*}
\|(1, 0) + i(0, 0)\|_1^2 &= \|(1, 0)\|^2 + \|(0, 0)\|^2 = 1 \\
\|(1, i) + i(0, -1)\|_1^2 &= \|(1, i)\|^2 + \|(0, -i)\|^2 = 7.
\end{align*}
A: First we show $||\cdot||_2\leqslant C||\cdot||_1$ with $C=\sqrt{2}$:
$$||\cos(\alpha)u+\sin(\alpha)v||^2=\langle \cos(\alpha)u+\sin(\alpha)v,\cos(\alpha)u+\sin(\alpha)v\rangle\\=\cos^2(\alpha)||u||^2+2\cos(\alpha)\sin(\alpha)\Re(\langle u,v\rangle)+\sin^2(\alpha)||v||^2\\\leqslant\cos^2(\alpha)||u||^2+2|\cos(\alpha)\sin(\alpha)|(||u||\cdot||v||)+\sin^2(\alpha)||v||^2\\\leqslant \cos^2(\alpha)||u||^2+2|\cos(\alpha)\sin(\alpha)|\frac{||u||^2+||v||^2}{2}+\sin^2(\alpha)||v||^2\\=\cos^2(\alpha)||u||^2+|\cos(\alpha)\sin(\alpha)|(||u||^2+||v||^2)+\sin^2(\alpha)||v||^2\\\leqslant ||u||^2+(||u||^2+||v||^2)+||v||^2=2(||u||^2+||v||^2)$$
Therefore 
$$||\cos(\alpha)u+\sin(\alpha)v||\leqslant \sqrt{2}\sqrt{||u||^2+||v||^2}=\sqrt{2}||u+iv||_1$$
Taking supremum on both sides wrt $\alpha$ yields
$$||u+iv||_2=\sup_{\alpha\in[0,2\pi)}||\cos(\alpha)u+\sin(\alpha)v||\leqslant \sqrt{2}||u+iv||_1$$
For the other direction $c||\cdot||_1\leqslant ||\cdot||_2$ with $c=1/\sqrt{2}$:
$$\sup_{\alpha\in[0,2\pi)}||\cos(\alpha)u+\sin(\alpha)v||\geqslant ||\cos(0)u+\sin(0)v||=||u||$$
and $$\sup_{\alpha\in[0,2\pi)}||\cos(\alpha)u+\sin(\alpha)v||\geqslant ||(\cos(\pi/2)u+\sin(\pi/2)v||=||v||$$
Then square both sides of last two inequalities, add them and take square roots. Hence
$$\sup_{\alpha\in[0,2\pi)}||\cos(\alpha)u+\sin(\alpha)v||\geqslant \frac{\sqrt{||u||^2+||v||^2}}{\sqrt{2}}=\frac{||u+iv||_1}{\sqrt{2}}$$
