The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F)

Question

The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F)

i.e $\Vert Tx\Vert_{\infty}=\Vert x\Vert_{1}$ but after that I can't think , please help.

Answer 1

The statement is true.

Let $\ell^1(2) = (\mathbb{R}^2,\|\cdot\|_1)$ and $\ell^\infty(2)=(\mathbb{R}^2,\|\cdot\|_\infty).$

Define $T:\ell^\infty(2) \to \ell^1(2)$ by $$T \begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{2} \begin{pmatrix} x-y\\ x+y \end{pmatrix}.$$ We claim that $T$ is an onto linear isometry. For any $\begin{pmatrix} x \\ y \end{pmatrix} \in \ell^\infty(2),$ we have $$\left\Vert T \begin{pmatrix} x \\ y \end{pmatrix} \right\Vert_1 = \left\Vert\frac{1}{2} \begin{pmatrix} x-y\\ x+y \end{pmatrix}\right\Vert_1 = \frac{|x-y| + |x+y|}{2} = \max\{|x|,|y| \} = \left\Vert \begin{pmatrix} x \\ y \end{pmatrix} \right\Vert_\infty$$ where the second last identity is standard in any real analysis text. Therefore, $T$ is an isometry.

Clearly $T$ is linear. We claim that $S:\ell^1(2)\to\ell^\infty(2)$ given by $$S\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x+y\\ -x+y \end{pmatrix}$$ is the inverse of $T.$ Indeed, $$(S \circ T)\begin{pmatrix} x \\ y \end{pmatrix} = S \left[ \frac{1}{2} \begin{pmatrix} x-y\\ x+y \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 2x\\ 2y \end{pmatrix} = \begin{pmatrix} x\\ y \end{pmatrix}, $$ $$(T \circ S) \begin{pmatrix} x\\ y \end{pmatrix} = T \left[ \begin{pmatrix} x+y\\ -x+y \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} 2x\\ 2y \end{pmatrix} = \begin{pmatrix} x\\ y \end{pmatrix}.$$ Therefore, $T$ is a bijection. Hence, $\ell^\infty(2) \cong \ell^1(2).$