Finding an example of a linear operator

Question

I am asked to give an example of a linear operator $T$ on a normed space $X$ such that $$\Vert T\Vert = 1$$ and $$\Vert T(x)\Vert < \Vert x\Vert, \;\forall x \in X\setminus\{0\}.$$

I consider $X = (\ell^1, \Vert \cdot \Vert _1)$, and define $T: X\to X$ by $$\forall x = (x_n)_{n=1}^\infty,\;T(x) = \left( \left( 1-\frac1{2^n} \right)x_n\right)_{n=1}^\infty.$$ I can prove that $T$ is in fact a linear operator from $X$ to $X$, but I can't prove that $\Vert T\Vert = 1$ and $\Vert T(x)\Vert < \Vert x\Vert, \;\forall x \in X\setminus\{0\}.$

I would like some tips if my example is correct, and if so how to prove the statement. Thanks a lot!!!

Answer 1

Let us denote $a=(a_n)_{n=1}^\infty$, we notice that $$Tx=(a,x)=\left(\underbrace{\left(1-\frac1{2^n}\right)}_{a_n}x_n\right)_{n=1}^\infty\in\ell_1^*.$$ Since $\ell_1^*=\ell_\infty$, we can use $\|a\|_\infty=\|T\|_1$, which is easy to find $$\|a\|_\infty=\sup_{n\in \mathbb{N}} \left(1-\frac1{2^n}\right)=1.$$

Answer 2

$$||T(x)|| = \sum_{n = 1}^\infty \left( 1-\frac1{2^n} \right)|x_n| < \sum_{i = 1}^\infty 1 |x_n| = ||x|| \, \forall x \in X \setminus \{0\}$$

Including the case $x = 0$, we have $||T(x)|| \le x \, \forall x \in X$, from which we deduce $||T||\le1$.

\begin{align} & ||T|| \\ \ge& \sup_{n \in \Bbb N} ||T(e_n)|| \quad \text{, where } e_n = (\underbrace{0,\dots,0}_{n - 1 \text{ zeros}},1,0,0,\dots) \\ =& \sup_{n \in \Bbb N} 1-\frac1{2^n} = 1 \end{align}

Hence $||T|| = 1$.

Answer 3

For $x \ne 0$ we have

$||T(x)||= \sum_{n=1}^{\infty}(1-\frac{1}{2^n})|x_n|=\sum_{n=1}^{\infty}|x_n|-\sum_{n=1}^{\infty}\frac{1}{2^n}|x_n|=||x||-\sum_{n=1}^{\infty}\frac{1}{2^n}|x_n|<||x||$.

Furthermore we have $||T|| \le 1$

Let $e^{k}$ denote the unit vectors in $l^1$

Then $||T(e^{k})||=1-\frac{1}{2^k} \to 1$ for $k \to \infty$.

hence $||T|| =\sup\{||T(x)||: x \in l^1, ||x||=1\} \ge 1$