# Tag Info

5

An idempotent $e$ is always equivalent to $e^*$. On the unitization of $A$, let $z=1+(e-e^*)^*(e-e^*)$. This is positive and invertible. By noticing that $z=1-e-e^*+ee^*+e^*e$, it is clear that $z$ commutes with $e$ and $e^*$, and a fortiori so does $z^{-1}$. Let $x=ee^*\in A$, $y=e^*z^{-1}e\in A$ (recall that $A$ is an ideal in its unitization). Note ...

3

I don't know what the OP means by "the derivative of $[A,B] = 0$", but the equation $$e^{A+B} = e^A e^B [A,B]$$ is certainly not on. This would imply in particular that $[A,B]$ has determinant $1$, which is destroyed by scaling of $A$ or $B$. It's also false in a limit as, say, $B$ approaches something that commutes with $A$.

3

I recently saw it asserted that the second dual of $C[0,1]$ was $\mathcal L^\infty[0,1]$, the space of bounded Borel functions. I was surprised at this, tried to prove it, couldn't quite make the details work. Finally saw a simple proof it was not so. (See the Amusing Note below regarding what $\mathcal L^\infty[0,1]$ actually is.) Started to post something ...

2

If $A$ is unbounded, that means that for any $n$ there exists a $v\in D(A)$ such that $\|Av\|> n\|v\|$. Dividing such a $v$ by $\|v\|$, we may assume $\|v\|=1$, so we can find a sequence $(v_n)$ of unit vectors in $D(A)$ such that $\|Av_n\|>n$ for each $n$. The sequence $(v_n/n)$ then converges to $0$ but $Av_n/n$ does not converge to $A(0)=0$ since ...

1

Its always a bit hard to guess what another person might find intuitive, but here are my two cents on the topic. You can interpretate the elements of $\mathbb{R}^n$ as functions from the set $\{1,...,n\}$ to $\mathbb{R}$, where for $f \in \mathbb{R}^{n}$, $f(i)$ would just be the $i$-th component of the vector. We know from linear algebra that any linear ...

1

Let us consider the family of operators $$T_\ell ((x_n)_n) = x_\ell \cdot e_1,$$ where $e_1 = (1, 0,0,\dots) \in \ell^2$ is the first basis vector. This family satisfies the first property, since for $x = (x_n)_n \in \ell^2$ with $\| x\| = 1$, we have $$\sum_{\ell=1}^\infty \|T_\ell x\|^2 = \sum_\ell |x_\ell|^2 = \|x\|^2 \leq 1.$$ But it is not too ...

1

Because $x$ commutes with $e$, then $e$ commutes with $E$ as well, including with $f=E_x(\mathbb{R}\setminus\{0\})$. Because $fx=xf=x$ and $fe=ef$, then $ef$ is also a projection such that $(ef)x=x(ef)=x$. Therefore $ef=fe=e$ follows from the minimality of $e$. Furthermore $x(f-e)=0$, which puts the range of $f-e$ in $\mathcal{N}(x)=\mathcal{R}(E_x\{0\})$; ...

1

$i(A+1)$ is a symmetric tri-diagonal matrix on each of the subspaces spanned by the even and odd coordinates. These are equivalent to Jacobi matrices (real symmetric with positive off-diagonal elements). It is a classical result that these define essentially self-adjoint operators if the off-diagonal elements grow no more rapidly than $n$. By Theorem 2.7 of ...

1

The subspace $U$ of $V$ has codimension $1$ if and only if $V/U$ has dimension $1$. In particular, there exists $v\in V$ such that $v\notin U$. Consider $\{v_\alpha\}$, a basis of $U$; then it is easy to prove that $\{v\}\cup\{v_\alpha\}$ is a basis of $V$. Can you define the requested functional? Remark. For infinite dimensional spaces this requires the ...

1

Solution 2 (without assuming $D$ to be linear) This is continuation of your solution: you got to the point where $\forall y\in Y\,\exists x\in X: Tx=y$ and therefore $T$ is an open mapping. This means that $T$ maps the open unit ball $B_X(0,1)\subset X$ in an open set in $Y$ which contains \$0_Y\Rightarrow\exists r>0:\,\overline{B_Y(0,r)}\subset ...

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