Tag Info

New answers tagged

2

Since $(V(f))' = f$, it suffices to see that $\lVert V(f)\rVert_{L^2} \leqslant C\lVert f\rVert_{1,2}$. But that is a direct consequence of the continuity of the Volterra operator on $L^2([0,1])$, $$\begin{align} \int_0^1 \lvert V(f)(t)\rvert^2\,dt &=\int_0^1\left\lvert \int_0^t f(s)\,ds\right\rvert^2\,dt\\ &\leqslant \int_0^1 \left( \int_0^t \lvert ...


1

The estimate $$ |Vf(t)| \leq \int_0^1|f| \leq \left(\int_0^1|f|^2\right)^{1/2} $$ gives $\|Vf\|_{L^2}\leq\|f\|_{L^2}$. Since $(Vf)'=f$, this gives $$ \|Vf\|_{1,2}^2 = \|Vf\|_{L^2}^2 + \|(Vf)'\|_{L^2}^2 \leq 2\|f\|_{L^2}^2 \leq 2\|f\|_{1,2}^2. $$ This gives continuity $V:H^1\to H^1$ (and $L^2\to H^1$). If you have trouble bounding the value of a function by ...


1

As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator ...


0

My own version is as follows: if $k$ is the Fourier transform of a certain function from $L^1$, then $K$ belongs to the trace class. Proof. $\,$ Suppose that $k=\mathcal F u$ with $u\in L^1$. Then $$ k(x-y)=\int_{-\infty}^{+\infty} u(t)\exp(it(x-y))\,dt =\int_{-\infty}^{+\infty} u(t)\exp(itx)\exp(-ity)\,dt. $$ For $t\in\mathbb R$, denote by $A_t$ the ...


2

An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of ...


0

This question is positively answered on MO. Even more is true. If $A$ have right divisor of all its elements, then $A$ is unital.


0

The bounded Fourier multiplier operators on $L^\infty$ correspond to convolutions with finite Borel measures. The Fourier transform of $1$ is a Dirac located at the origin. Therefore the unital operators correspond to convolutions with measures $\mu$ satisfying \begin{equation*} 1 = \widehat{\mu}(0) \end{equation*}


2

If $\widehat{\mu}$ is non-negative, then its square root defines a bounded Fourier multiplier operator $S$ on $L^2$. In this case $S$ is self-adjoint, so $T=S^*S$. Conversely, any linear and shift invariant $S$ is a Fourier multiplier operator with an $L^\infty$ symbol. The adjoint $S^*$ would have a symbol that is the complex conjugate of the symbol for ...


2

Assume $T=T^{\star}\in\mathcal{L}(H)$, where $H$ is a complex Hilbert Space. Implication 1: Show $(Tx,x) \ge 0$ for all $x \in H$ implies $\sigma(T)\subseteq [0,\infty)$. To do this, assume that $(Tx,x) \ge 0$ for all $x \in H$, and let $\lambda < 0$. Then $$ 0 \le -\lambda(x,x) \le ((T-\lambda I)x,x) $$ implies $$ |\lambda|\|x\|^{2} \le ...


3

If $A$ is any unital C*-algebra, the hypotheses are satisfied with $C=A$ and $B=\mathbb C1$.


2

If $t\gt0$, then we want $a\ge2|A|$ and $$ \gamma_r=a+ir[-1,1]\cup[a,-r]+ir\cup-r+ir[1,-1]\cup[-r,a]-ir $$ If $t\lt0$, then we want $a\le-2|A|$ and $$ \gamma_r=a+ir[1,-1]\cup[a,r]-ir\cup r+ir[-1,1]\cup[r,a]+ir $$ In each case, $\gamma_r$ is counterclockwise. When $t\gt0$, the finite part of $\gamma_r$ is $[a-i\infty,a+i\infty]$. When $t\lt0$, the finite ...


2

You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional $x^{\star}$ to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed ...


0

For the parabolic PDE this is usually done by an energy estimate: test the PDE with the solution to find that the norm of the solution is bounded in terms of the data. But this is probably not sufficiently abstract.


2

Note that if $\mathcal{H}$ is real then the notion of a unitary operator does not make any sense. However, you can do that with five orthogonal operators instead. See e.g. Theorem 4.3 in A. Böttcher, A. Pietsch, Orthogonal and Skew-Symmetric Operators in Real Hilbert Space, Integral Equations and Operator Theory 74 (2012), 497-511. It should be added ...


0

Sketch: The square root is operator monotone. Now, if $Q_n\rightarrow Q$, then, given $\epsilon>0$, there is an $N$ such that for all $n\geq N$, $$ (Q-\epsilon I)_+\leq Q_n\leq Q+\epsilon I. $$ It follows that $$ (Q-\epsilon I)_+^{1/2}\leq Q_n^{1/2}\leq (Q+\epsilon I)^{1/2}. $$ It therefore suffices to show that $$ (Q\pm\epsilon I)_+^{1/2}\rightarrow ...


0

First note that since $T: K \rightarrow X^{*}$ is A-pseudomonotone and hemicontinuous, it also follows that $T: M \rightarrow X^{*}$ is also A-pseudomonotone and hemicontinuous, where $M := K \cap [x,y]$. Note that since $K$ is assumed nonempty, convex and closed it follows that $M$ is also nonempty, closed and convex. We want to show that $$\bigcap_{z \in M ...


1

Both definitions concern restrictions to lines. The intersection of a line with $K$ is a closed line segment. The second definition requires continuity on the whole segment. The first requires one-sided continuity at one point, but it can be any point, and we can pick the side. So, both definitions require continuity on lines. The only (possibly) real ...


1

This is just an intuition for the much simpler case of finite dimension. Let $A$ be any matrix and $v$ an associated eigenvector, i.e. $Av = \lambda v$. Then $(A-I)v = Av-v = (\lambda-1)v$, and thus $\sigma(A-I)= \sigma(A)-1$. And now an attempt for infinite dimension, suppose that $(A-i\lambda I)$ is not invertible for every $\lambda \in (-\infty, 0]$ but ...


2

Yes, there are quite a lot easy examples, e.g. consider the momentum operator $-i\frac{d}{dx}$ on $L^2(-a,a)$ where $a$ is some finite number and the Sobolev Space $H^1$ as domain. In this case, $e^{ikx}$ is an eigenfunction for every $k$ and therefore the spectrum is whole $\mathbb{C}$. However, if you choose $H_0^1$ as domain, this operator will be ...


1

Yes, I have seen cases where an operator on $L^1(R^3,d^3x)$ has a much larger spectrum than on $L^2(R^3,d^3x)$. If my memory serves me well this was the case for a Fokker-Planck operator, the $L^2$ spectrum was real but the $L^1$ spectrum contained contributions outside the real axis.


0

The answer is no, the sequence need not converge in the completely bounded norm to $x$ even if $x$ and $x_n$ are completely bounded. Let $W=V=B(\mathcal H)$, where $\mathcal H$ is a Hilbert space considered with some fixed orthonormal basis $\{ e_1, e_2, \dots \}$. Let $P_n$ be the orthogonal projection onto the space spanned by the first $n$ elements of ...


2

Suppose $A$ is symmetric. Equivalently, assume $(Ax,x)$ is real for all $x\in\mathcal{D}(A)$. Claim: If $A-\lambda I$ is surjective for some $\lambda\notin\mathbb{R}$, then $A$ is densely-defined. Proof: Suppose $x \perp \mathcal{D}(A)$. If $(A-\lambda I)$ is surjective, then $(A-\lambda I)y=x$ for some $y\in\mathcal{D}(A)$, which gives $$ 0 = ...


0

I think this problem is Theorem 2.4(iii) of the book written by, L. N. Trefethen, M. Embree, spectra and pseudospectra, 2005.


0

Hint: Since $T$ is finite rank, you can pick an orthonormal basis $g_1,\ldots,g_n$ of the range of $T$, $$ Tv=\sum_{j=1}^n \langle Tv,g_j\rangle g_j. $$ Show that $\mbox{ker } T\supset \mbox{span }\{T^*g_1,\ldots,T^*g_n\}^\perp$. Pick and orthonormal basis of $\mbox{span }\{g_1,\ldots,g_n,T^*g_1,\ldots,T^*g_n\}$, say $e_1,\ldots,e_m$. Show that $m\leq 2n$. ...


1

For $\lambda \in \mathbb{R}$ and $k > 0$, $$ \frac{k^{2}}{2}\left[\frac{1}{\lambda+ik}+\frac{1}{\lambda-ik}\right] =\lambda \frac{k^{2}}{\lambda^{2}+k^{2}} $$ So the above function converges to $\lambda$ as $k\rightarrow\infty$. You have $$ e^{iT_{k}}x = ...


1

For example, let $X=L^{2}[0,2\pi]$ and let $A=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(A)$ consisting of all functions $f$ which are continuously differentiable on $[0,2\pi]$ and for which $f'$ is absolutely continuous on $(0,2\pi)$ with $f'' \in L^{2}[0,\pi]$. This is as large a domain as you can reasonable expect for $A$. This operator is not ...


1

I found the answer to my specific problem: the keywords are Taylor series and Implicit Function Theorem. Imagine the x the input, y the output, f a non-linear function: \begin{equation} y= f(x-b \cdot y)\\ \end{equation} (this is as above, just renamed). Then we can do the following: \begin{equation} 0= f(x-b \cdot y)-y\\ F(x,y(x))=0 \end{equation} Means we ...


0

It seems you are thinking of the Fredholm alternative, which ensures that the index of $I+K$ is zero whenever $K$ is compact.


1

You don't need linearity to write down Neumann series. Neumann series is just a fancy name for the geometric series: as long as the common ratio is less than 1 (in some norm), then it does converge (in that same norm). Namely, if $\ ||AF||_*<1$ (for some subadditive norm $\|\cdot\|_*$), then $$ \left\|(I-AF)^{-1} - \sum_{k=0}^n (AF)^k\right\|_* \to 0 ...


0

Here are my results without the rigorous details. Let me know if you come up with different results. Also, I'm nbt sure about the solution set for $\lambda=0$. Consider $\lambda\neq 0$, We note that $K$ is Hilbert-Schmidt so $-\lambda(I+(-\lambda^{-1}K))$ is Fredholm of index $0$. Thus the existence of a solution is equivalent to its uniqueness. So we ...


1

If $A$ is selfadjoint with $\sigma(A)=\{\lambda\}$, then we may assume that $\lambda = 0$ by replacing $A$ with $A-\lambda I$ if necessary. However, the norm and spectral radius for a selfadjoint operator are the same because $\|A\|^{2}=\|AA^{\star}\|=\|A^{2}\|$. Therefore, $\|A\|=0$. The counter-example is any niplotent operator $N$ of order $n > 1$.


1

As a counterexample consider $$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$over $\mathbb C^2$. It's spectrum is $\{0\}$.


1

Since $T$ is self adjoint, the residual spectrum is empty, so $\lambda\notin \sigma_r(T)$. This means either $\lambda \in \sigma_p(T)$ (point spectrum) or $\lambda \in \sigma_c(T)$ (continuous spectrum). Note: Since $T$ is self adjoint, $<Tu_n,v>=<u_n,Tv>$ and $\lambda$ will be real, so $<\lambda u_n,v>=< u_n,\overline \lambda ...


2

Giuseppe Negro already mentioned it, but I wanted to go into a bit more detail with his first point, the Sturm-Liouville methods. A strategy which is used very often in practice is: Calculate the Greens function or respectively the fundamental solution for $L-\lambda \mathbf{1}$ (you can actually allow $G$ to be a distribution), this induces a resolvent ...


3

There is a whole theory dedicated to this, so the short answer is: there are a lot. I can think of three: Solving analytically the resolvent differential equation (i.e. the equation $Lu - \lambda u = v$). This tends to work when the geometrical domain is one-dimensional (Sturm-Liouville's theory) or when it is very symmetrical (separation of variables, ...


1

The subspace $S_{n}$ spanned by $\{ e^{2\pi i kx}\}_{k=-n}^{n}$ is invariant under $A$, and $A$ is bounded on $S_{n}$ because its a matrix. However, $A$ is not bounded on the union of these nested subspaces. That puts a limit on any Zorn's lemma argument. I suppose you could find a maximal subspace on which $A$ is bounded by a given fixed constant $M$.


3

The simplest case I know is the translation $T(t)$ semigroup on $L^{2}[0,\infty)$ defined by $(T(t)f)(x)=f(x+t)$. There's no way to invert once you've lost part of the function.


3

Consider the Heat semigroup on $L^2(\mathbb{R}^n)$: for $t>0$, $$U(t)g=K(t)\ast g$$ where $$K(t)=(4\pi t)^{-n/2}e^{-|x|/4t}$$ is the heat kernel. $U(t)$ is a strongly continuous semigroupo (of contractions). Let's show that it cannot be extended to a group ($U(t)$, $t\in\mathbb{R}$) on $L^2$. If such a group eventually exists, it should satisfy ...


0

Suppose $A\in L(X),A'\in L(X')$ such that $A'$ is the dual, i.e. $\forall x,x'\;(A'x')(x)=x'(Ax)$. Then of course $(Ax)(x')=x'(Ax)=(A'x')(x)=x(A'x')$, and $A$ satisfies the property that defines the dual operator. Note that it can only work when $X$ is reflexive, since we assume that any element in $X''$ can be written as $x\in X$.


0

The subspace $M$ spanned by $\{ 1, x\}$ is invariant under this operator. And, if $u \perp M$, then $Tu=0$. So this completely reduces to a $2\times 2$ matrix problem. You could use an orthonormal basis for $M$, but there's no real need that I can tell. Using the ordered basis $b=\{ 1,x\}$, $$ Tu = x(u,1)+1(u,x),\;\;\; (f,g) = ...


1

Consider $\lambda=0$. $$\int^1_0 (x+y)u(y)dy = 0 $$ or $$x\int^1_0 u(y)dy+\int^1_0 yu(y)dy = 0 $$ So $u(x)$ is an eigenfunction if both $<y,u>=0$ and $<1,u>=0$ (by linear independence). It seems me that this implies $0$ is an eigenvalue of infinite multiplicity because any polynomial of degree greater than 2 can be made orthogonal to ...


3

Lemma: Let $B : \mathcal{D}(B)\subseteq X\rightarrow X$ be a symmetric linear operator on a Hilbert space $X$ for which $(Bx,x) \ge 0$. If $I+B$ is surjective, then $B$ is closed, densely-defined and selfadjoint. Proof: Let $B$ as stated. Suppose that $B+I$ is surjective. To show that $B$ must be densely-defined, suppose that $y \perp \mathcal{D}(B)$ ...


2

In example 2, page 52, Weidmann does not consider multiplication operators, but the functional $T \colon L^2(\mathbb R) \to \mathbb C$ $$ Tf = \int_{\mathbb R} \psi(x)f(x) \, dx $$ which of course is defined on the whole of $L^2(\mathbb R)$ for $\psi \in L^2(\mathbb R)$. It's just the scalar product with $\bar\psi$. The multiplication operator is defined on ...


1

Assume that $(A-\lambda I)^{-1}$ is defined on all of $X$ and is bounded. Then $(A-\lambda I)^{-1}$ is closed. The graph of $A-\lambda I$ and the graph of $(A-\lambda I)^{-1}$ are transposes of each other in $X\times X$, which guarantees that $A$ is closed. So you can deduce that $A$ is closed under those circumstances. You can also deduce that the domain ...


3

If you use unbounded operators, then $$ H = \int_{0}^{\infty}\lambda dE(\lambda). $$ The spectrum theorem for unbounded selfadjoint operators has the excellent provision that $$ \mathcal{D}(H) = \left\{ x \in X : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty\right\}. $$ You have $H=A^{2}$, where ...


1

The domain of $A^{\star}$ is identical to the domain of $A$ in this case. However, when you compose the two, then you get $\mathcal{D}(A^{\star}A)$ as $$ \mathcal{D}(A^{\star}A)= \{ f \in L^{2} : f\in \mathcal{D}(A) \mbox{ and } Af \in \mathcal{D}(A^{\star}) \}. $$ In particular, $f$ is twice absolutely continuous with $f(0)=f(1)$ and $f'(0)=f'(1)$ . ...


1

If $H:X \to X$ is a bounded operator, there is indeed always such a decomposition available. We may construct one such decomposition as follows: By the spectral theorem, we have $H = UTU^*$ Where the operator $T$ is given by $$ [T(\phi)](x) = f(x)\phi(x) $$ For some $f:\Bbb R \to \sigma(H) \subset [0,\infty)$. We can simply define $\sqrt T$ by $$ [\sqrt ...


2

See the MathOverflow question http://mathoverflow.net/questions/5303/basis-of-linfinity to learn why it is consistent with ZF (without the axiom of choice) that there is no such operator. Discontinuous everywhere defined operators can be defined using Hamel bases. In particular, as Daniel Fischer points out, one can use a Hamel basis for $X$ to define ...


3

This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along ...


4

The classical operator $$ L = -\frac{d^{2}}{dx^{2}}+V,\;\;\; a \le x \le b, $$ is different if $V$ is very singular. If $V \in L^{1}[a,b]$, then things are nice because there are 2 linearly-independet classical solutions of $Lf = \lambda f$ for every $\lambda$. That is, such solutions are continuous on $[a,b]$, their first derivatives are ...



Top 50 recent answers are included