Let $T:l^\infty\rightarrow l^\infty$ be defined by $(x_1,x_2, \dots)\mapsto (x_2,x_3,\dots)$.

I have seen a claim without justification that $\|T\|=1$, but I am not convinced. I know that

$\|T\| = \sup_{{\|x\|=1}}\|Tx\|$ . If $\|x\|=1$ then surely $\|Tx\|\le 1$. So

$$\|T\| = \sup\limits_{\|x\|=1}\|Tx\|\le 1.$$

I don't see how it can be claimed that $\|T\| = 1$

Is it true or I am missing something?


$||T||=\sup_{||x||=1} ||Tx||$.

Since $||x||=1\implies \sup \{|x_1|,|x_2|,\ldots |x_n|,\ldots\}=1$.

Note that $A\subset B\implies \sup A\le \sup B$.

So $||Tx||=\sup \{|x_2|,|x_3|,\ldots |x_n|,\ldots\}\le \sup \{|x_1|,|x_2|,\ldots |x_n|,\ldots\}=1\implies ||T||\le1$

Now ,Take $x_0=(1,1,1,\ldots,1)$ , then $Tx_0=(1,1,1,\ldots,1)\implies ||T||\ge 1 $

Combining $||T||=1$.

  • $\begingroup$ add absolute values all over $\endgroup$ Apr 7 '17 at 4:36
  • $\begingroup$ Thank you so much@VictorZurkowski; $\endgroup$
    – Learnmore
    Apr 7 '17 at 4:39
  • $\begingroup$ nice answer! Welcome to the site :) $\endgroup$
    – Ant
    Apr 7 '17 at 7:12

Clearly $\|Tx\| \le \|x\|$, hence $\|T\| \le 1$. Since $T(0,1,0,...) = (1,0,...)$ we have $\|T\| = 1$.


To show that $\|T\|\geq c$ it suffices to find one $x$ such that $\|Tx\|\geq c\|x\|$, by the definition of the operator norm.

In particular, in your example, to show that $\|T\|\geq 1$ it suffices to find one $x$ such that $\|Tx\|=1$. It is easy to find such examples.


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