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The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...


Suppose you have a non-trivial solution of $$ T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t) $$ Then $\lambda \ne 0$ because the above would give $f=0$ after differentiating a couple of times. For $\lambda \ne 0$, any solution of the above must satisfy $$ \lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0. $$ Any ...


This depends on the domain and range of the the operator $A$. For a linear operator $A : X \to Y$, where $X$ and $Y$ are normed spaces, one can show that if $A$ is bounded and of finite rank, i.e., $\dim A(X) < \infty$, then the operator $A$ is compact. So, if you consider normed spaces, then your reasoning is fine.


You have already done all the necessary work: $$ u=\begin{bmatrix}\begin{matrix}u_1&&&\\&\ddots&&\\&&u_j&\\ &&&I_d\end{matrix} \end{bmatrix}. $$


Let's just try to find an eigenfunction for a given $|\lambda|< 1$: we want an $f$ that solves $$ f\left( \frac{x+1}{2}\right) = 2\lambda f(x)- f(x/2) . \quad\quad\quad\quad (1) $$ Start out with an arbitrary (integrable) function on $(0,1/2)$, then use (1) for $0<x<1/2$ to define $f(t)$ for $1/2<t<3/4$, then reenter (1), with ...


Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that $B(\ell_2)$ as well as the Calkin algebra cannot be decomposed like that no matter which tensor ...

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