Prove that function $A\colon \mathbb{R} \to \mathbb{R^2}$ defined by $A(x)=(x,\sin(x))$ is isometry of $\left(\mathbb{R},|\cdot |\right)$ to $ \left(\mathbb{R^2}, |\cdot|^{(2)}_{\infty}\right)$ .
This is an exercise that I encountered, but I am not sure which norm is denounced by $|\cdot|^{(2)}_{\infty}$
If this would be an isometry, then it would need to follow $\|A(x)-A(y)\|=\|x-y\|$.
I think what the norm denotes there is $||[x ,y]||_{\infty} = \max\{a,b\}$. Assuming this, here is a proof. The distance in domain space is $|x-y|$. For the range, $$ A(x) = [x , \sin(x)] \ \text{and} \ A(y) = [y,\sin(y)]. $$ Hence, $A(x)-A(y) = [x-y, \sin(x) - \sin(y)]$. Now,
Claim : For any $x,y\in\mathbb{R}$, $|\sin(x)-\sin(y)|\leq |x-y|$.
Proof : $|\sin(x)-\sin(y)|=2\left|\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)\right|\leq2\left|\sin\left(\frac{x-y}{2}\right)\right| \leq |x-y|$ as $|\sin(\alpha)|\leq |\alpha|$.
With the claim in hands, we are done. $\blacksquare$.