Norm, adjoint operator and compactness os some operators

Question

Let $A_i:\ell^2\rightarrow \ell^2$ be two operators given as follows:

$A_1x=(0,x_1,0,\frac{x_2}{2},0,\frac{x_3}{3},...)$ and $A_2x=(x_1,x_1,x_2,x_2,x_3,x_3,...)$

Compute the norm and the adjoint operator and decide whether these operators are compact or not.

1)

Norm: We have obviously $||A_1x||_{\ell^2}\leq ||x||_{\ell^2}$ for all $x\in \ell^2$, hence $||A_1||<1$. But for $x=(0,0,0,..)$ we have equality, hence $||A_1||=1$.

Adjoint operator: $A_1^*$ must fulfill $<A_1x,y>=<x,A_1^*y>$ for all $x,y\in \ell^2$. So we must have $\sum_{n=1}^{\infty}|(A_1x)_ny_n|=\sum_{n=1}^{\infty}|x_n(A_1^*y)_n|$.

But I don't know how to continue from here..

Compactness: The sequence of operators $A_nx=(0,x_1,0,\frac{x_2}{2},0,...,\frac{x_n}{n},0,0,..)$ is a sequence of finite rank operators, which converges to $A_1$ because:

$||A-A_n||^2=\sup_{||x||_{\ell^2}\leq1}\sum_{m=n+1}^{\infty}(|\frac{x_m}{m}|)^2\leq \sup_{||x||_{\ell^2}\leq1}(|\frac{1}{n+1}|)^2\sum_{m=n+1}^{\infty}|x_m|^2\leq(\frac{1}{n+1})^2\rightarrow 0$

2)

Norm: Each addend of $||A_2x||$ appears exactly twice in $||x||$ and both values are of course finite, hence $||A_2||=\frac{1}{2}$

Adjoint: No idea as in 1)

Compactness: The method from 1) did not work here for me

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

Answer 1

For every $x \in \ell^2$ we have \begin{eqnarray} \|A_1x\|^2&=&\sum_{n=1}^\infty\frac{x_n^2}{n^2} \le \sum_{n=1}^\infty x_n^2=\|x\|^2 \quad \mbox{ since } \quad n \ge 1,\\ \|A_2x\|^2&=&x_1^2+x_1^2+x_2^2+x_2^2+x_3^2+x_3^2+\ldots=2\sum_{n=1}^\infty x_n^2=2\|x\|^2, \end{eqnarray} and so
\begin{eqnarray} \|A_1\|&=&\sup_{\|x\|=1}\|A_1x\|\le \sup_{\|x\|=1}\|x\|=1,\\ \|A_2\|&=&\sup_{\|x\|=1}\|A_2x\|=\sup_{\|x\|=1}\sqrt{2}\|x\|=\sqrt{2}. \end{eqnarray} Because the vector $e_1=(1,0,\ldots)$ belongs to the unit ball of $\ell^2$ and $\|A_1e_1\|=1$, we deduce that $\|A_1\|=1$.

Unlike $A_1$, the operator $A_2$ is not compact. In fact, the unit ball $B_1=\{x\in \ell^2:\quad \|x\|=1\}$ is bounded, and its image through $A_2$ is the ball $B_{\sqrt{2}}=\{x\in\ell^2:\quad \|x\|=\sqrt{2}\}$ which, obviously is not relatively compact. As a matter of fact, the sequence $\{\sqrt{2}e_n\}_{n\ge 1}$, where $e_1=(1,0,\ldots),e_2=(0,1,0,\ldots),\ldots$, belongs to $B_{\sqrt{2}}$ but it has no convergent subsequence.

The idea is to find an operator $A_i^*$ such that $$ \langle A_ix,y\rangle=\langle x,A_i^*y\rangle \quad \forall x,y\in \ell^2. $$ For every $x,y\in \ell^2$ we have \begin{eqnarray} \langle A_1x,y\rangle&=&x_1y_2+\frac12x_2y_4+\frac13x_3y_6=\sum_{n=1}^\infty\frac{x_ny_{2n}}{n}=\langle(y_2,\frac{y_{4}}{2},\ldots),x\rangle,\\ \langle A_2x,y\rangle&=&x_1(y_1+y_2)+x_2(y_3+y_4)+x_3(y_5+y_6)+\ldots=\sum_{n=1}^\infty x_n(y_{2n-1}+y_{2n})\\ &=&\langle(y_1+y_2,y_3+y_4,\ldots),x\rangle, \end{eqnarray} hence $$ A_1^*x=2(\frac{x_2}{2},\frac{x_4}{4},\frac{x_6}{6},\ldots),\quad A_2^*x=(x_1+x_2,x_3+x_4,x_5+x_6,\ldots) \quad \forall x \in \ell^2. $$

Answer 2

$A_1$:

Norm: $||A_1||=1$, because it is reached with $x=(1,0,0,\ldots)$ (note that for $x$ the null vector is not relevant. maybe it was a typo..) Adjoint: \begin{align} \langle A_1 x, y \rangle &= \langle (0,x_1,0,x_2/2,0,x_3/3,0,\ldots), (y_1,y_2,y_3,\ldots) \rangle \\ &= x_1 \bar{y_2} + \frac{x_2}{2} \bar{y_4} + \frac{x_3}{3} \bar{y_6} + \ldots \\ &= \langle (x_1,x_2,x_3,\ldots),(y_2,y_4/2,y_6/3,\ldots) \rangle, \end{align} hence $A_1^* x = (x_2,x_4/2,x_6/3,\ldots)$. Compactness: your argument works.

$A_2$:

Norm: $||A_2||=\sqrt{2}$, it's commented by T.A.E. Adjoint: a similar calculation as for $A_1$, it gives $A_2^* x = (x_1+x_2,x_3+x_4,\ldots)$. Compactness: No. Recall that $A$ is compact iff for every bounded sequence $(x_n)_n$, $(Ax_n)_n$ has a convergent subsequence. But the sequence $x_n=e_n$, $n\geq 1$ is bounded but $(Ax_n)_n$ does not have any convergent subsequence (two one's moving to the right). (Note aside: this operator is somehow similar to the identity operator, which is never compact in an infinite dimensional Hilbert space).