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Let $T$ be a linear operator on $\mathbb{R}$ defined by the $2\times2$ matrix $(a_{jk})_{j,k=1}^{2},\: a_{jk}\in\mathbb{R}$. Let $l_{\infty}^{2}$ denote $\mathbb{R}^{2}$ equipped with the norm $||x||_{\infty}=\max(|x_{1}|,|x_{2}|)$ for $x=(x_{1},x_{2})\in\mathbb{R}^{2}$. Let $T:l_{\infty}^{2}\rightarrow l_{\infty}^{2}$. Find the operator norm $||T||$.

I guess it should be $||T||=\max\{|a_{11}|+|a_{12}|,|a_{21}|+|a_{22}|\}$, but I am not sure how to show that. Thank you.

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up vote 2 down vote accepted

So let $x \in \ell^2_\infty$, $||Tx|| = \max_i|(Tx)_i|$. Then we have $$ \begin{align*} ||Tx|| &= \max_i|(Tx)_i|\\ &= \max_i|a_{i1}x_1 + a_{i2}x_2|\\ &\leq \max_i(|a_{i1}| + |a_{i2}|)||x||. \end{align*} $$ So $||T|| \le \max\{|a_{11}| + |a_{12}|, |a_{21}| + |a_{22}|\}$. For the other bound, choose $i$ such that $\max\{|a_{11}| + |a_{12}|, |a_{21}| + |a_{22}|\} = |a_{i1}| + |a_{i2}|$. Let $x_j = \frac{|a_{ij}|}{a_{ij}}$, then $||x|| = 1$ and $$ \begin{align*} ||Tx|| &\ge \max_i|(Tx)_i|\\ &= \max_i|a_{i1}x_1 + a_{i2}x_2|\\ &= \max_i(|a_{i1}| + |a_{i2}|). \end{align*} $$ So $||T|| = \max\{|a_{11}| + |a_{12}|, |a_{21}| + |a_{22}|\}$,

AB,

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Your align* environments do not render properly for me. I tried to fix them, but for some reason I didn't manage to achieve that –  t.b. Mar 14 '12 at 9:22
    
I have tried also to fix. Perhaps its an issue of a MathJax-Update or sth like that? –  martini Mar 14 '12 at 9:24
    
Thank you! It is very helpful. –  user16859 Mar 16 '12 at 3:29
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