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I am wondering how to find the linear isometries $T: \mathbb R^2||.||_p\to \mathbb R^2 ||.||_p$ . can anyone help me to show explicitly if i take $p=1 $ and $2$ .

(Not related to the previous qestion ) And if $1 \le p,q \le \infty$ , is it possible to find the miminum $C(p,q)$ such that $||x||_p\le C ||x||_q$ . Thanks for your guidance in advance.

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$\def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|}$Look at @Stefan's answer for the $p=2$ case. For the $p = 1$ case you can argue as follows: Let $T\colon \mathbb R^2 \to \mathbb R^2$ a linear $\norm\cdot_1$-isometry, let $x_1 := Te_1$ and $x_2 := Te_2$. As $T$ is isometric, we have $\norm{x_i}_1 = 1$ and $\norm{x_1 + x_2}_1 = \norm{T(e_1 + e_2)}_1 = \norm{e_1 + e_2}_1 = 2$. Writing $x_i = (x_{i1}, x_{i2})$, hence \begin{align*} \abs{x_{11}} + \abs{x_{12}} &= 1\\ \abs{x_{21}} + \abs{x_{22}} &= 1\\ \abs{x_{11} + x_{21}} + \abs{x_{12} + x_{22}} &= 2 \end{align*} So, by the triangle inequality, this is only possible if \begin{align*} \abs{x_{11} + x_{21}} &= \abs{x_{11}} + \abs{x_{21}}\\ \abs{x_{12} + x_{22}} &= \abs{x_{12}} + \abs{x_{22}}\\ \end{align*}
So $x_{11}$ and $x_{21}$ have the same sign, the same for $x_{12}$ and $x_{22}$. Now look at $\norm{x_1 - x_2}_1 = \norm{e_1 - e_2}_1 = 2$, so \begin{align*} \abs{x_{11} - x_{21}} + \abs{x_{12} - x_{22}} &= 2 \end{align*} By the same argument as above, $x_{11}$ and $-x_{21}$ need to have the same sign. But now we can coclude that one of them has to be 0, same for $x_{12}$ and $x_{22}$ (to fulfill the requirements, the other is $\pm 1$ then), leaving us with the possibilities:

  • $x_{11} = 0$, then $x_{12} = \pm 1$, hence $x_{22} = 0$, $x_{21} = \pm 1$, giving \[ T = \begin{pmatrix} 0 & \pm 1\\ \pm 1 & 0 \end{pmatrix} \]
  • $x_{11} = \pm 1$, then $x_{12} = 0$, hence $x_{22} = \pm 1$, $x_{21} = 0$, so \[ T = \begin{pmatrix} \pm 1 & 0\\ 0 & \pm 1 \end{pmatrix} \]

So these eight mapping are the only linear isometries of $(\mathbb R^2, \norm\cdot_1)$.

For your last question, note first that Hölder gives for $p \le q$ with $\alpha = \frac qp$ and $\frac 1\alpha + \frac 1\beta = 1$ \begin{align*} \norm x_p &= \norm{\abs x^p}_1^{1/p}\\ &\le \norm{\abs x^p}_\alpha^{1/p}\norm{1}^{1/p}_\beta\\ &= \norm{\abs x^{p\alpha}}^{1/p\alpha}2^{1/p\beta}\\ &= 2^{1/p\beta}\cdot\norm x_q \end{align*} Now $\alpha = \frac qp$, hence $\frac 1{p\beta} = \frac 1p\cdot (1 - \frac pq) = \frac 1p - \frac 1q$, so we have $C(p,q) \le 2^{1/p - 1/q}$ in this case. And looking at $x = (1,1)$ gives us $C(p,q) = 2^{1/p - 1/q}$ for the $p\le q$-case.

For $q \le p$ let $x \in \mathbb R^2$ with $\norm x_q = 1$, then we have $\abs{x_i} \le 1$ and hence $\abs{x_i}^p \le \abs{x_i}^q$ giving \begin{align*} \norm x_p^p &= \abs{x_1}^p + \abs{x_2}^p\\ &\le \abs{x_1}^q + \abs{x_2}^q\\ &= \norm x_q^q\\ &= 1\\ \iff \norm x_p &\le 1 \end{align*} Now let $x \in \mathbb R^2$ arbitrary, then $y := \frac x{\norm x_q}$ has $\norm y_q = 1$, hence $\norm y_p \le 1$, giving \[ 1 \ge \norm{\frac x{\norm x_q}}_p \iff \norm{x}_q \ge \norm x_p \] This gives $C(p,q) \le 1$ for this case with the unit vectors proving equality. Hence we have \[ C(p,q) = \begin{cases} 2^{1/p - 1/q} & p \le q\\ 1 & q \le p \end{cases} \]

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thanks a lot . For $||.||_2$ the only isometries are reflection and rotation and identity . Because these are the only orthogonal transformations which conserve norm. –  Theorem Nov 2 '12 at 12:14
    
No. For $\norm\cdot_2$, as @Stefan wrotes, all orthogonal transformations are isometries, for: If $T\colon \mathbb R^2 \to \mathbb R^2$ is orthogonal, we have $\norm{Tx}_2^2 = \left<Tx,Tx\right> = \left<x,T^tTx\right> = \left<x,x\right> = \norm x_2^2$. –  martini Nov 2 '12 at 12:19
    
Sir , that means there are infinitely many isometries because i can take any values for $cos\theta, sin\theta$ between [0,1] and form a orthogonal matrix. I don't know if i am making sense of what i am talking . Please correct me . Thanks –  Theorem Nov 2 '12 at 12:25
    
Exactly. $(\mathbb R^2, \norm\cdot_2)$ has infinitely many isometries, namely \[ \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin\theta & \cos \theta \end{pmatrix} \text{ and } \begin{pmatrix} \cos \theta & \sin \theta \\ \sin\theta & -\cos \theta \end{pmatrix} \] for any $\theta$. –  martini Nov 2 '12 at 12:27
    
sir Thank you so much sir –  Theorem Nov 2 '12 at 12:31

As you are going from $\mathbb R^2$ to $\mathbb R^2$ and both are finite dimensional, every Isomorphism can be identified with his matrix form. As you want your Isomporphism to be isometric, e.g. $||Tx|| = ||x|| \forall x \in \mathbb R^2$, you want to look at the orthogonal matrices ($A^TA = AA^T = I$).

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Yes, i was thinking along the same line but i am having trouble in determining all the isomorphisms which are isometric .rotation, reflection, translation metrics are some of the orthogonal metrices but how do i characterize all of them with respect to the given norm ?? –  Theorem Nov 2 '12 at 11:09

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