Suppose $x,y\in\mathbb{R}^2$ and let $\left\|\cdot\right\|$ denote the Euclidean norm. I am trying to find a bound for $\|x\otimes x - y\otimes y\|$ as a multiple of $\|x-y\|$.
We have a special easy case when $y = 0$. In that case $$ \|x\otimes x\|^2 = \sum_{i=1}^{2}\sum_{j=1}^{2}(x_ix_j)^2 = \left(\sum_{i=1}^{2}x_{i}^{2}\right)^2 = \|x\|^4. $$ Can anyone help me with the case $y\neq0$?
Note that $$ \|x\otimes x - y \otimes y\|^2 = (x\otimes x - y \otimes y)^T(x\otimes x - y \otimes y) = \\ (x\otimes x)^T(x\otimes x) - 2 (x\otimes x)^T(y \otimes y) + (y \otimes y)^T(y \otimes y) = \\ x^Tx \otimes x^Tx - 2x^Ty \otimes x^Ty + y^Ty \otimes y^Ty =\\ \|x\|^4 - 2\langle x,y \rangle^2 + \|y\|^4 $$ Compare this to $$ \|x - y\|^2 = \|x\|^2 - 2 \langle x,y \rangle + \|y\|^2 $$
Suppose that $\|y\| = \alpha \|x\|$ for some $\alpha > 0$. The ratio of these two quantities looks like $$ \frac{\|x\|^4 - 2\langle x,y \rangle^2 + \|y\|^4}{\|x\|^2 - 2 \langle x,y \rangle + \|y\|^2} = \\ \frac{(1 + \alpha^4)\|x\|^4 - 2\beta^2} {(1 + \alpha^2)\|x\|^2 - 2 \beta}, \qquad |\beta| < \alpha \|x\|^2 =\\ \alpha \|x\|^2 \frac{(1 + \frac 1{\alpha^4}) - 2\gamma^2} {(1 + \frac 1{\alpha^2}) - 2 \gamma}, \qquad |\gamma| < 1 $$ It is clear that, with a fixed $\alpha = \frac{\|y\|}{\|x\|}$ and a fixed $\gamma = \frac{\langle x,y \rangle}{\|x\|\,\|y\|}$, this quantity is unbounded as $\|x\| \to \infty$.