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No. I'm elaborating on my comment: By CS, it would follow that $\left|\ln |x-y|\right| \le A(x)B(y)$, with $A,B\in L^2(0,1)$. So $\left| \ln |x-y|\right| \le C A(x)$ for a.e. $x$ and all $y$ from a se …

No. If $A=P$ is a projection, then $P^2=P$, but in general you won't have $N(P)=0$ or $1$. In fact, you could just take $\Omega=\{ 1,2\}$, so $\mathcal H\cong\mathbb C^2$, and $P$ as the projection on …

Yes, this works. By the closed graph theorem, applied to $\textrm{id}:D(A)\to D(B)$ endowed with the operator norms, your assumption that $D(A)\subseteq D(B)$ implies that $B$ is $A$-bounded. So $$ \| …

The first version does not work, as you suspected. For example, if $Te_n= (1/n)e_{n+1}$ for $n\ge 0$ on $H=\ell^2(\mathbb Z)$, then the $T^n e_0$, $n\ge 0$ are linearly independent. However, we can ea …

This is false. The dual $K(H)^*=B_1(H)$ is given by the trace class operators, and an $S\in B_1(H)$ acts on a $T\in K(H)$ by $(S,T)=\textrm{tr}\, ST$. Consider now $$ S_n = n2^{-n} \sum_{2^n\le j<2^{n …

From the definition of the adjoint, we have that $(v_1,v_2)\in D(T^*)$ precisely if there are $y_1,y_2$ such that $$ \langle v_1, u'_1+au'_2 \rangle = \langle y_1, u_1 \rangle + \langle y_2, u_2 \rang …

The operator (on $L^2(\mathbb R)$) has purely absolutely continuous spectrum $\sigma_{ac}=(-\infty,\infty)$ of multiplicity $2$. I don't think that's very easy to see, and it definitely depends on the …

This is false. Let $A=S+S^*$ be the free Jacobi matrix on $\ell^2$; I write $S$ for the shift $(Sy)_n = y_{n+1}$. Then $\sigma_{ac}(A)=[-2,2]$. Let $B$ be multiplication by a bounded sequence $b_n\ge …

Here's a more explicitly worked out version of my comment above. Consider $P(\sin nx)=\sin 2^n x$. This is an isometry, so in particular bounded in $L^2(0,\pi)$. It can not be approximated in the desi …

We probably don't need another answer, but here it is anyway, especially since Giorgio raised the question about projections. Lemma: There is an ONB $\{ v_j\}$ of $\mathbb C^n$ such that $|\langle e_1 …

No, this fails already when $\Omega=\{ 1, 2\}$, so $H=\mathbb C^2$ and $\widehat{k}_j=e_j$. We can take $$ A= \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} . $$ Then $N(A)=\sqrt{2}$, $N(A^*)=1$.

This is always true. Consider $P(t)=1-U(t)U(t)^*$. Then $T=\{t\in\mathbb R: P(t)=0\}$ is open and closed and $0\in T$. The set is open because if $t_0\in T$, then $\|P(t)\|<1$ for all $t$ sufficiently …

No, this isn't working at all. Since $N,P$ commute (you would have to be more specific what exactly you mean by this since the operators are unbounded, but let's just assume we have the right version) …

Let $B$ be the non-negative square root of the usual self-adjoint realization of $-\Delta$. Then I claim that $A^*A=-\Delta$ precisely if $A=UB$ for a $U$ satisfying $U^*U=1$. In other words, $U$ vari …

Let's first of all recall that the minimal operator of differentiation $$ D(S) = \{ f\in L^2\cap AC : f'\in L^2, f(\pm 2\pi)=0 \} , \quad\quad Sf = -if' $$ is closed and has the maximal operator as it …