Find the norm $\|T\|$, $T^2$ and compare $\|T^2\|$ with $\| T \|^2$

functional-analysislp-spacesnormed-spaces

Consider the normed vector space $\ell^2$ of complex numbers with norm

$\|x\| =\left(\sum_{n=1}^{+\infty}|x_n|^2\right)^{1/2}$.

For every $x=(x_1, x_2, \dots) \in \ell^2$ and let $T(x) = (0,2x_1, x_2, 2x_3, x_4, \dots)$

Find the norm $\|T\|$.

It is easy to see that $T$ is linear operator and we show that $T$ is bounded:

\begin{align*}
||T(x_n)||_2 = ||(a_nx_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2}.
\end{align*}

where $a=(1, 2, 1, 2, \dots)$

Hence norm should be

$||T||=\sup_{n \in \mathbb{N}}|a_n|$ ?

Also, how do I find $T^2(x)$ and what is the difference between $\|T^2\|$ and $\| T \|^2$.

Best Answer

Hint. Let $x=(x_1,x_2,\ldots)$ and $a=(a_1,a_2,a_3,\ldots)=(2,1,2,\ldots)$.

We have, since $|a_n|\leq 2$ for all $n$, $$||Tx||_2^2=\sum_{n\geq 1} |a_n x_n|^2\leq 4\sum_{n\geq 1}|x_n|^2,$$ so $$||Tx||_2\leq 2||x||_2.$$ Now let $x=(1,0,0,\ldots)$, then $Tx=(0,2,0,0,\ldots)$, hence we have equality. We conclude that $||T||=2$.

Can you calculate what the operator $T^2$ and its norm is?

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[Math] What are the range and the norm of this bounded linear operator

It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence $\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.

One can easily verify that $$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$

$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).

The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image $\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.

You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).

[Math] Show that $T$ in $\ell^2(\mathbb{N})$ is a bounded linear operator and compute the operator norm $\|T\|$

This question is Old but I reckon its still worth answering since I can't find any other related questions. We had this question in a Hilbert space course I am currently doing at Uni. It is in a slightly different form but it should be more or less the same for this question.

Proposition.

let $(a_n)_{n \in \mathbb{N}}\subset \mathbb{C}$ be a fixed bounded sequence. Then the operator \begin{align*} T:&\ell_2(\mathbb{N},\mathbb{C}) \rightarrow \ell_2(\mathbb{N},\mathbb{C})\\ &(x_n) \mapsto(a_nx_n). \end{align*} Is a bounded linear operator and $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$.

Proof:

Throughout this proof we will endow $\ell_2$ with the usual 2-norm.

We start with Linearity.

let $\lambda \in \mathbb{C}$ and $(x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}}\in \ell_2(\mathbb{N},\mathbb{C})$ then we compute; \begin{align*} T(\{\lambda x_n+y_n\}_{n \in \mathbb{N}})&= \{a_n(\lambda x_n+y_n)\}_{n \in \mathbb{N}} & \text{[Definition of $T$]}\\ &=\{a_n\lambda x_n+a_ny_n\}_{n \in \mathbb{N}}\\ &=\{\lambda a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}} & \text{[Point wise addition]}\\ &=\lambda \{a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}}\\ & = \lambda T(\{x_n\}_{n \in \mathbb{N}})+T(\{y_n\}_{n \in \mathbb{N}}). \end{align*} Thus we can see that $T$ is linear.

Next we show that $T$ is bounded.

Let $(x_n)_{n \in \mathbb{N}},\in \ell_2(\mathbb{N},\mathbb{C})$ be a arbitrary sequence. Then we compute; \begin{align*} ||T(x_n)||_2 = ||(a_nx_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2}. \end{align*} Now The boundedness of $(a_n)$ becomes crucial. Since $(a_n)$ is bounded we can find $\sup_{n \in \mathbb{N}}|a_n|$ (if we knew more about $(a_n)$ we could find this exactly). Now let $M = \sup_{n \in \mathbb{N}}|a_n|$, then we have \begin{align*} ||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq. \sqrt{\sum_{i=1}^{\infty}|Mx_n|^2} = M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2. \end{align*} Since $(x_n)$ was arbitrary, this is true for all sequences in $\ell_2$. Consequently $T$ is bounded.

Now finally we find $||T||_{\text{op}}$: Recall that $||T||_op$ is the minimum of the bounds of $T$. To show that $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$ we need to show the following;

If $M \geq 0$ is a finite constant such that \begin{align*} ||T(x_n)||_2 \leq M||(x_n)||_2\ \ \ \ \ \ \ \ (1) \end{align*} for all $(x_n)\in \ell_2$ then $\sup_{n \in \mathbb{N}}|a_n| \leq M$. So we fix $M \geq 0$ that satisfies $(1)$. Now for all $(x_n) \in \ell_2$ we have \begin{align*} ||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2. \end{align*} Now this holds in particular when $||(x_n)||_2 =1$. For every $n \in \mathbb(N)$ we define the sequence \begin{align*} x_n &= (x_{n,k})_{k \in \mathbb{N}}\\ x_{n,k} &:= \delta_{k,n}. \end{align*} where $\delta_{k,n}$ is the Kronecker delta. Then we have \begin{align*} ||x_n||_2 = \sqrt{\sum_{i=1}^{\infty}|x_{n,i}|^2}=\sqrt{|x_{n,n}|^2} = 1 \end{align*} thus $x_n \in \ell_2$. Then we have \begin{align*} Tx_n &= (a_k x_{n,k})_{k \in \mathbb{N}}\\ Tx_{n,k} &:= a_k \delta_{k,n}. \end{align*} Now we also have \begin{align*} ||Tx_n||_2 = \sqrt{\sum_{i=1}^{\infty}|a_kx_{n,k}|^2}=\sqrt{|a_n|^2} =|a_n|. \end{align*} then by the way we chose $M$ we have \begin{align*} ||Tx_n||_2 \leq M||x_n||_2 \implies |a_n| \leq M. \end{align*} However remember that there was nothing special about $n$ so that $|a_n| \leq M$ for all $n \in \mathbb{N}$ and in particular this means $\sup_{n \in \mathbb{N}}|a_n| \leq M$, which proves the result.

Best Answer

Related Solutions

[Math] What are the range and the norm of this bounded linear operator

[Math] Show that $T$ in $\ell^2(\mathbb{N})$ is a bounded linear operator and compute the operator norm $\|T\|$

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