# Compute the norm of bounded operators

Compute the norm of the following operators.

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

$$$$K = \sup_{x\in \ell^2} \frac{\| Ax \|}{\| x \|}$$$$

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. $$$$M^2 = \text{max} \left( \sum_{i=1}^{N} B_{ij}^{\ast} B_{ik} \right), \quad \forall j,k$$$$ where $$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!!

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

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 $$\|A\|=2.$$
• Very helpful! However, by "in finite and infinte dimensions" I meant that $A$ and $B$ are infinite and finite dimensions, respectively. Nov 4, 2022 at 16:30