# linear operator from $\ell^1$ to $\ell^1$

Let $0 < \lambda <1$ and define $T: \ell^1 \to \ell^1$ with $T(x_1,x_2,x_3, \cdots ) =(\lambda x_2 ,\lambda x_3 ,\lambda x_4 , \cdots)$.

(a) Prove that $T$ is bounded and find $||T||$.

(b) For every $n \in \mathbb{N}$ find $||T^n||$ where $T^n=T \circ T \circ T \cdots \circ T$ where the composition of $T$ is $n-$times.

(c) Prove that the series $\sum_{n=0}^{\infty} T^n$ converges where ($T^0=Id$).

(d) Find the norm of the operator $S$ where $S=\sum_{n=0}^{\infty} T^n$.

I need some help for (c) and (d).

Here it is what I did in (a) and (b).

It is easy to see that $T$ is linear.

(a) For every $x=(x_n)_n \in \ell^1$ it is $||T(x)||_{1}^{2} =\sum_{n=2}^{\infty} \lambda |x_n| \leq \lambda \sum_{n=1}^{\infty} \lambda |x_n|= \lambda ||x||_1^{2} < + \infty$

so $T(x) \in \ell^1$ and $||T|| \leq \lambda$. Also it is $||T|| \geq \frac{||T(e_2)||_1}{||e_2||_1} = \lambda$. Thus we have that $||T||= \lambda$.

(b) Using induction we see that $T^n(x)= (\lambda^n x_{n+1} , \lambda^n x_{n+2} , \cdots )$

So as in (a) it is$||T^n||= \lambda^n$.

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Why are you squaring your norms? You shouldn't do this here. $\Vert T^n\Vert = \lambda^n$, not $\lambda^{n/2}$. – David Mitra Feb 21 '12 at 14:58
@DavidMitra: Yes you are right. I edit it. – passenger Feb 21 '12 at 15:07

Set $S=\sum T^n$. Given $x\in l_1$, formally $$Sx = ( \sum_{n=1}^\infty \lambda^{n-1} x_n, \sum_{n=2}^\infty \lambda^{n-2} x_n, \sum_{n=3}^\infty \lambda^{n-3} x_n,\ldots )$$

Since $\sum|x_i|$ is finite, $Sx$ is well defined. Moreover, $Sx$ is an element of $\ell_1$:

$$\Vert Sx\Vert_1 =\sum_{j=1}^\infty | \sum_{n=j}^\infty \lambda^{n-j} x_n| \le \sum_{n=1}^\infty \sum_{j=n}^\infty \lambda^{n -1}| x_j| \le \sum_{n=1}^\infty \lambda^{n-1}\Vert x\Vert_1={1\over 1-\lambda}\Vert x\Vert_1$$

The above also shows $\Vert S\Vert\le{1\over 1-\lambda}$. To show that the norm of $S$ is $1\over 1-\lambda$, consider the image of the vectors $$w_n=(\underbrace{0,0,\cdots,0\vphantom{\textstyle{1\over n}}}_{n\text{-terms}},\underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}},\cdots, {\textstyle {1\over n}}}_{n\text{-terms}},0,0,\cdots)$$

We have \eqalign { T^0w_n &=\lambda^{0} (\underbrace{ 0,0,0,\ldots,0}_{n\text{-terms}}, \underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots ) \cr T^1w_n &=\lambda^{1} ( \underbrace{ 0,0, \ldots,0}_{(n-1)\text{-terms}}, \underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots ) \cr T^2w_n &=\lambda^2 ( \underbrace{ 0,\ldots,0}_{(n-2)\text{-terms}},\underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots ) \cr &\vdots\cr T^nw_n &=\lambda^{n} ( \underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, {\textstyle {1\over n}}, \ldots, {\textstyle {1\over n}}}_{n\text{-terms}}, 0,0\ldots ) ;\cr }

from which it follows that $$\Vert Sw_n\Vert \ge\Bigl\Vert \sum_{j=0}^n T^jw_n\Bigr\Vert= 1+\lambda+\lambda^2+\cdots+\lambda^n={1-\lambda^{n+1}\over 1-\lambda}.$$

Since $\Vert w_n\Vert=1$ for all $n$, and since $\lim\limits_{n\rightarrow\infty} {1-\lambda^{n+1}\over 1-\lambda}={1\over 1-\lambda}$, we have $\Vert S\Vert\ge {1\over 1-\lambda}$.

We have already shown that $\Vert S\Vert\le {1\over 1-\lambda}$; thus $\Vert S\Vert= {1\over 1-\lambda}$.

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Can you help we to find $||Sw|$ where $w=(\underbrace{0,0,\cdots,0\vphantom{\textstyle{1\over n}}}_{n\text{-terms}},\underbrace{ {\textstyle {1\over n}}, {\textstyle {1\over n}},\cdots, {\textstyle {1\over n}}}_{n\text{-terms}},0,0,\cdots)$? – passenger Feb 21 '12 at 15:05
Thank you very much for your time! – passenger Feb 21 '12 at 15:45