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Compute the norm of the following operators.

a) Let $\hat{A}: \ell^2 \to \ell^2 $ be the operator given as follows \begin{equation} \hat{A} (x_1, x_2, ..., x_n, ...)^t \mapsto (x_1+x_2, x_2+x_3, ... , x_n+x_{n+1}, ...)^t \end{equation} For some $x = (x_1, x_2, ...)^t \in \ell^2$.

My attempt:

I want to find a relation between the norms $\| Ax \|$ and $\| x \|$, s.t. $\| Ax \| \leq K\| x \|$. Then conclude that

\begin{equation} K = \sup_{x\in \ell^2} \frac{\| Ax \|}{\| x \|} \end{equation}

So for some $x \in \ell^2$

\begin{align} \| Ax \|^2 &= \sum_{i=1}^{\infty} |x_i+x_{i+1}|^2 \\ &= \sum_{i=1}^{\infty} \left( |x_i|^2 + |x_{i+1}|^2 + 2\ \Re [x_i^{\ast} x_{i+1}] \right) \\ \end{align}

where $^{*}$ is the conjugate operation.

I couldn't manage to find a bound for that real part.

b) Let $\hat{B}:\mathcal{H} \to \mathcal{H} $ be an operator defined in a finite Hilbert space ($\text{dim}\ \mathcal{H} =N$), s.t. \begin{equation} M^2 = \text{max} \left( \sum_{i=1}^{N} B_{ij}^{\ast} B_{ik} \right), \quad \forall j,k \end{equation} where $B_{ij}$ are the entries of the matrix in a orthonormal basis.

My attempt:

I tried to do a similar procedure. So for some $x \in \mathcal{H}$ \begin{align} \| Bx \|^2 &= (Bx \ | \ Bx) \\ &= \sum_{j,k=1}^{N} \left( \sum_{i=1}^{N} B_{ij}^{\ast} B_{ik} \right) x_j^{\ast} x_k \\ &\leq M^2 \sum_{j,k=1}^{N} x_j^{\ast} x_k \end{align}

Since

\begin{align} 0 &\leq \Bigg| \sum_{j=1}^{N} x_j - \sum_{k=1}^{N} x_k\ \Bigg|^2 \\ &\leq \Bigg|\sum_{j=1}^{N} x_j\ \Bigg|^2 + \Bigg|\sum_{k=1}^{N} x_k\ \Bigg|^2 - 2\ \Re \left[ \sum_{j=1}^{N} x_j^{\ast} \sum_{k=1}^{N} x_k \right] \end{align}

we have

\begin{align} 2 \sum_{j=1}^{N} x_j^{\ast} \sum_{k=1}^{N} x_k \leq \Bigg|\sum_{j=1}^{N} x_j\ \Bigg|^2 + \Bigg|\sum_{k=1}^{N} x_k\ \Bigg|^2 \end{align}

But I don't know how to put all together.

I hope someone can go through it and help me for the points which are left. Thanks!!

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  • $\begingroup$ How do you define the operator in the finite dimensional case. Perhaps this way $$Ax=(x_1+x_2,x_2+x_3,\ldots, x_n+x_1)$$ $\endgroup$ Nov 4, 2022 at 0:49
  • $\begingroup$ I just want to compute the infinite dimensional case. $\endgroup$
    – Spectree
    Nov 4, 2022 at 16:33

1 Answer 1

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For the infinite dimensional case the operator is if the form $A=I+S,$ where $S$ is the shift operator $$Sx=(x_2,x_3,\ldots )$$ We have $$\|A\|\le 1+\|S\|=2$$ On the other hand for $$x^{(n)}=(1,1,\ldots,1,0,0,\ldots)$$ with $n$ entries equal $1$ we get $$Ax^{(n)}=2x^{(n-1)}+e_n,$$ where $e_n$ denotes the sequence with $0$ entries except for the $n$th one equal $1.$ Thus $$\|Ax^{(n)}\|\ge 2\sqrt{n-1}\quad \|x^{(n)}\|=\sqrt{n}$$ Hence $\|A\|\ge 2,$ i.e.$\|A\|=2.$

For the finite dimensional case we have $A=I+S$ where $$Sx=(x_2,x_3,\ldots, x_n,x_1)$$ Then $\|A\|\le 2.$ Moreover $$A(1,1,\ldots,1)=2(1,1,\ldots,1)$$ hence $\|A\|=2.$

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  • $\begingroup$ Very helpful! However, by "in finite and infinte dimensions" I meant that $A$ and $B$ are infinite and finite dimensions, respectively. $\endgroup$
    – Spectree
    Nov 4, 2022 at 16:30

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