Hot answers tagged operator-theory
8
We have to consider the problem over the completion $\overline X$ of $X$. But that does not hurt our reasoning: if $A$ is not invertible, neither will its extension to $\overline X$ be.
Assuming that $\|A_n^{-1}\|\leq M$ for all $n$, we have
$$
\|A_n^{-1}-A_m^{-1}\|=\|A_n^{-1}(A_m-A_n)A_m^{-1}\|\leq\,M^2\,\|A_m-A_n\|.
$$
As $\{A_n\}$ is Cauchy, we deduce ...
4
Here's a general outline of what the hint in the book is intending: Let $I,J$ be the two integrals. You want to show $A(I) = J$. Let $I_n$ and $J_n$ be the two integrals reduced to the interval $[0,n]$. Then $I_n \rightarrow I$ and $J_n \rightarrow J$. Now, see if you can approximate $I_n$ and $J_n$ by further sequences $I_{n,k}$ and $J_{n,k}$ using Riemann ...
4
Recall that the compact operators form a two-sided ideal in $B(X)$ for any normed vector space $X$.
Writing $T=L^{-1}LT$ shows that every bounded operator is therefore compact. In particular your $(L+B)^{-1}$.
Remark: actually, the existence of a compact invertible operator in $B(X)$ is equivalent to $X$ being finite-dimensional. Indeed, this is equivalent ...
4
We want $L^*$ such that
$$\langle f \vert Lg \rangle = \langle L^*f \vert g \rangle$$
We have
$$S=\int_0^{\infty} e^{-x}f(x)\left(x+\dfrac{d}{dx}\right)g(x) dx = \int_0^{\infty}x e^{-x}f(x) g(x) dx + \int_0^{\infty} e^{-x} f(x) dg(x)$$
We have
$$\int_0^{\infty} e^{-x} f(x) dg(x) = \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right) - \int_0^{\infty} g(x) ...
3
I'm not in position to explain any $K$-theory. But it is not hard to see that $M_n(K(H))\simeq K(\bigoplus_1^nH)$ canonically. So you can see matrices of compact operators as compact operators on a bigger Hilbert space, and you can calculate its trace if it is trace-class.
More concretely, if you have a matrix of operators $T=(T_{kj})_{k,j}\in M_n(B(H))$, ...
3
This proof is (sort of) from Advanced Linear Algebra by Steven Roman.
Let $V$ be a complex inner product space and let $A$ be a linear operator on $V$ with the given property. For all $v\in V$ and $r\in\mathbb{C}$ we have
\begin{align}
0 &= \langle A(rv+Av),rv+Av \rangle \\
&= |r|^2\langle Av,v\rangle + \langle A^2v,Av\rangle + r\langle Av, ...
3
Suppose that $\{F_n\}$ is a sequence of finite-rank operators such that $F_n\to I$ strongly. Note that by the uniform boundedness principle the sequence is bounded, i.e. there exists $k>0$ with $\|F_n\|<k$ for all $n$ (thanks julien for reminding me of this). I will assume that all $F_n$ are selfadjoint (I need for my estimates, but didn't think if ...
3
From Gelfand's formula, which holds for every element of a unital complex Banach algebra more generally,
$$
\lim_{n\rightarrow +\infty}\|A^n\|^\frac{1}{n}=0\quad\iff\quad \rho(A)=0.
$$
This characterizes the set of quasinilpotent operators, which extends the set of nilpotent operators.
There is nothing you can say about $\rho(A+A^*)$ without further ...
2
You can prove this using the following observation: Let $T$ be a bounded linear operator on a Hilbert space. Then $T$ is a partial isometry if and only if the spectrum of $T^{\ast} T$ consists only of $\{0,1\}$. Furthermore, let $A$ be a bounded, self-adjoint operator. Then $\sigma(A) \subset \{0,1\}$ if and only if $A = A^2$; this follows from the spectral ...
2
Look at some of the hits from a google search for "exponentiation of operators". Among other things, you'll see that exponentials of various differentiation operators, e.g. $e^{\frac{d}{dx}}$, arise in quantum mechanics. More general than exponentiation of operators are operators of operators defined by various power series expansions of an operator.
Also, ...
2
Yes. If $\varphi$ is a holomorphic map of unit disk into itself, the composition operator $f\mapsto f\circ \varphi$ is bounded on $H^p$. In fact,
$$\|f\circ \varphi\|_{H^p}\le \left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{1/p}\|f \|_{H^p} \tag1$$
Original source:
John V. Ryff, Subordinate $H^p$ functions, Duke Math. J. Vol. 33, no. 2 (1966), ...
2
1) There exist bijection between bounded bilinear operators and bounded opeartors. The proof of this fact requires Banach-Steinhaus theorem. If bilinear form is symmetric, then the respecitive opeartor is (obviously) symmetric too
2) Spectrum of any bounded operator is compact, and as the consequence closed
3) No, consider bilinear form
$$
...
2
In general, the integral triangle inequality holds for functions $f$ from a measure space $\Omega$ to a normed linear space $X$. It says that
$$\left\|\int_\Omega f \right\|\le \int_\Omega \|f\| \tag1$$
One can then follow up with the pointwise inequality $\|f\|\le \sup_\Omega \|f\|$ and conclude that
$$\left\|\int_\Omega f \right\|\le \sup_\Omega \|f\| ...
1
You need to specify the domain of $L$ more precisely, as the spectrum depends on it very much, e.g. the resolvent set is empty unless $L$ is a closed operator. I assume henceforth that $\text{dom}L = \{f \in L^{2}[0,\infty): f' \in L^2[0,\infty)\}$. In that case, if I haven't made any mistake, $\text{dom}L^{\ast}=\{f \in L^{2}[0,\infty): f(0)=0, f' \in ...
1
Since $B$ is linear, we need only show that $B$ is weakly continuous at zero.
Suppose $V$ is a weak neighborhood of $0$. Then $V$ contains a set of the form $W = \{ x \,|\, |\phi_k(x)| < \epsilon, \ k \in I\}$, where $I$ is a finite set of indices and $\phi_k \in \mathbb{H}^*$. Then we need to find a weak neighborhood $U$ of $0$ such that $BU \subset ...
1
Here is a proof from scratch, in particular, without the Lax-Milgram.
Lemma. Let $\widetilde H$ be a real vector space with positive definite inner product (completeness not assumed). Suppose that $K$ is a subspace of $\widetilde H$ which is complete. Then any element $v\in \widetilde H$ can be written as $v_1+v_2$ where $v_1\in K$ and $v_2\in K^\perp$.
...
1
The subsequences you took must not have any index in commom: For example, let $K_1(x)=K_2(x)=x$ in $\mathbb{C}$ (or $\mathbb{R}$, if you prefer) and take $x_n=(-1)^n$. Then $K_1$ and $K_2$ are compact operators and $\left\{x_n\right\}_n$ is a bounded sequence
If $n_k=2k$ and $n_l=2l+1$, then $\left\{K_1(x_{n_k})\right\}$ and $\left\{K_2(x_{n_l})\right\}$ ...
1
Recall principle of uniform boundedness:
If $\{T_n\}$ are linear bounded operators from $X$ to $X$ where $X$ is a Banach space, for which $\sup_n\lVert T_nx\rVert<\infty$ for all $x$, then
$\sup_n\lVert T_n\rVert<\infty$.
Here, the sequence $(T_ny,n\geqslant 1)$ is bounded for all $y$ (as it converges weakly to $0$), which gives the result.
1
Counterexample. Let $\lambda\in c_0$ with non zero entries. Consider diagonal operators
$$
T:\ell_1\to\ell_1:x\mapsto\lambda x\\
S:\ell_\infty\to\ell_\infty:x\mapsto\lambda x
$$
This is straightforward to check that $S=T^*$.
Obviously $\operatorname{Ker}(T)^\perp=\{0\}^\perp=\ell_1^*=\ell_\infty$. Since $\lambda\in c_0$ then $\operatorname{Im}(S)\subset ...
1
It relies on several things. First rewriting the derivative as limit:
$$
\int^b_a \lim_{\epsilon\to 0} \frac{\frac{\partial }{\partial x}u(x,t+\epsilon) - \frac{\partial }{\partial x}u(x,t) }{\epsilon} dx\tag{1}
$$
The question is: whether above is the same as
$$
\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\int^b_a \frac{\partial }{\partial ...
1
Your argument does not prove the result at this point. For a bounded linear operator, $T$ is self-adjoint ($T^*=T$) iff $(Tx,y)=(y,Tx)$ for every $x,y\in H$. So you need to do more.
Using the textbook lemma is a good idea. It yields a straightforward argument. The proof of that lemma is no very difficult if you are familiar with polarization. It boils down ...
1
Dirac won a Nobel Prize for finding the square root of a pesky differential operator while working on the relativistic Schrodinger equation. http://www.youtube.com/watch?v=zM-Lc16nyho. <--- That video gives a clear overview of the hacks Dirac used.
1
(A2) is correct. Note that it assumes that $A$ be injective. And that $X$ be a normed vector space suffices.
For (Q1), a natural family of counterexamples is given by non surjective isometries on a Banach space, since their ranges are closed proper subspaces. For instance, the unilateral shift operator $S:\ell^2(\mathbb{N})\longrightarrow ...
1
Try $U_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$, $U_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. Both are unitary, hence normal. However $U_1+U_2 = \begin{bmatrix} 0 & 2 \\ 0 & 0\end{bmatrix}$, and $(U_1+U_2) e_1 = 0$, but $(U_1+U_2)^* e_1 = 2 e_2$, hence $U_1+U_2$ is not normal.
1
Yes.
Operator multiplication is continuous in operator norm. To see this (if you haven't already), suppose $S_n \to S$ and $T_n \to T$. In particular, $\sup_n \|T_n\| < \infty$. Then
$$\|S_n T_n - ST\| \le \|S_n T_n - S_n T\| + \|S_n T - ST\| \le \|S_n\| \|T_n - T\| + \|S_n - S\| \|T\|$$
and both terms go to 0.
Now to show your claim, use induction ...
1
A more general question has been asked and answered on here before. The bounded operators on a Hilbert space form a Banach Algebra. A beautiful argument of H. Wielandt shows that the equation $AB-BA = I$ is not possible in any normed algebra. I won't repeat it here, but it follows by induction that if $AB -BA = I,$ we have $AB^{n}-B^{n}A = nB^{n-1}$ for all ...
1
Yes, that’s exactly what it is. And you derive it in the same way, as a consequence of the product rule for the difference operator:
$$\begin{align*}
\Delta\big(f(x)g(x)\big)&=f(x+1)g(x+1)-f(x)g(x)\\
&=f(x+1)g(x+1)-f(x)g(x+1)+f(x)g(x+1)-f(x)g(x)\\
&=f(x)\Delta g(x)+Eg(x)\Delta f(x)\;.
\end{align*}$$
1
I'm assuming that $E,F$ are Hilbert spaces with some scalar products $\langle \cdot , \cdot \rangle_{E}$ and $\langle \cdot , \cdot \rangle_{F}$.
Let $\begin{pmatrix} e \\ f \end{pmatrix} \in G^\perp$, i.e. $\left\langle \begin{pmatrix} e \\ f \end{pmatrix}, \begin{pmatrix} x \\ Ax \end{pmatrix}\right\rangle_{E\times F}=0$
for all $x\in D(A)$. Using the ...
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