The rotation matrix is non-expansive

Question

Define for each $\alpha\in \mathbb{R},$ $$A(\alpha)=\begin{bmatrix}\cos\alpha& -\sin\alpha\\ \sin\alpha &\cos\alpha\end{bmatrix}.$$ Then $A$ is non-expansive if $$\|A(\alpha) -A(\beta)\|\leq \|\alpha -\beta\|.$$

So, I'm choosing $\|A\|_{\infty}=\max\limits_{1\leq i\leq 2}\sum^{2}_{j=1}|a_{ij}|.$ Thus, \begin{align} \|A(\alpha) -A(\beta)\|_{\infty}&=\max\{|\cos\alpha-\cos\beta|+|\sin\alpha-\sin\beta|, |\sin\alpha-\sin\beta|+|\cos\alpha-\cos\beta| \}\\&= \max\{\left|-2\sin\left(\dfrac{\alpha+\beta}{2} \right) \sin\left(\dfrac{\alpha-\beta}{2} \right) \right|+\left|2\sin\left(\dfrac{\alpha-\beta}{2} \right) \cos\left(\dfrac{\alpha+\beta}{2} \right) \right|, \\&\quad\quad\quad\quad \left|2\sin\left(\dfrac{\alpha-\beta}{2} \right) \sin\left(\dfrac{\alpha+\beta}{2} \right) \right|+\left|-2\sin\left(\dfrac{\alpha+\beta}{2} \right) \cos\left(\dfrac{\alpha-\beta}{2} \right) \right| \} \\&\leq \max\{2\left|\dfrac{\alpha-\beta}{2} \right| +2\left|\dfrac{\alpha-\beta}{2} \right|, 2\left|\dfrac{\alpha-\beta}{2} \right| +2\left|\dfrac{\alpha-\beta}{2} \right| \} \\&= 2\|\alpha -\beta\|.\end{align} How I'm I getting $2$ as my constant instead of $1$? Any help in fixing this? Thanks for your help!

Answer 1

The inequality $(|\cos\, \alpha -\cos\, \beta|+|\sin\, \alpha -\sin\, \beta|) \leq \alpha -\beta $ is false. If thus is true for all $\alpha, \beta$ we can divide by $\alpha -\beta $ and let $\beta \to \alpha$ to see that $|sin\, \alpha| +|cos\, \alpha|\leq 1$. But this is false when $\alpha =\frac {\pi} 4$.

Answer 2

An "abstract" way to prove the same result is to observe that your operator norm $\|A\|_{\infty}$ is nothing but the norm of $A$ as an operator from $\ell_1^n$ to $\ell_{\infty}^n$, where $n=2$. Now you can decompose the operator as follows: $$\ell_1^n\overset{\text{id}}{\longrightarrow}\ell_2^n\overset{A}{\longrightarrow}\ell_2^n\overset{\text{id}}{\longrightarrow}\ell_{\infty}^n$$ It is easy to check that the identity map from $\ell_1^n$ to $\ell_2^n$ has norm $1$, as well as the identity map from $\ell_2^n$ to $\ell_{\infty}^n$. Since $A$ is an isometry of $\ell_2^n$, we deduce that $$\|A\|_{\infty}\leq \|\text{id}\|_{1\to 2}\|A\|_{2\to 2}\|\text{id}\|_{2\to\infty}\leq 1$$.

The advantage of this method is that it proves the same result for any dimension, whenever $A$ is an orthonormal matrix in $\mathbb{R}^n$.

Edit: As pointed out by gerw, this does not answer the original question.