Here is the question: Suppose that $X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph. Is it true that $T$ maps closed sets to closed sets? Is it true if $T$ is linear? So I think I am over thinking this one, but I am having some […]

The strong operator topology on a Banach space $X$ is usually defined via semi-norms: For any $x \in X$, $|\cdot|_x: B(X) \to \mathbb R, A \mapsto \|A(x)\|$ is a semi-norm, the strong topology is the weakest/coarsest topology which makes these maps continuous. Alternatively it is generated by the sub-base $\left\{B_\epsilon(A;x)=\{B\in X \mid |B-A|_x<\epsilon\}\phantom{\sum}\right\}$. If we define […]

I have just been introduced to the concept of invertibility for bounded linear operators. Specifically, we defined a bounded operator $A$ to be invertible if there exists a bounded $A^{-1}$ which is its right and left inverse, i.e. $AA^{-1}=\mathrm{id}_{\mathrm{Im}A},A^{-1}A=\mathrm{id}_{\mathrm{Dom}A}$. So I was wondering: is the requirement of boundedness (or equivalently of continuity) of the inverse […]

Let $A,B$ be positive self-adjoint bounded operators and $\lambda >0$ then I want to show that if $$A-B \ge 0 $$ that is $\langle x,(A-B)x \rangle \ge 0$ we have that the resolvents (whose existence is clear) satisfy $$(A+\lambda I)^{-1}-(B+\lambda I)^{-1} \le 0,$$ i.e. exactly the opposite relation. Although this is intuitively clear, I got […]

Let $\Omega\subset\mathbb{R}^n $ be bounded smooth domain. Given a sequence $u_m$ in Sobolev space $H=\left \{v\in H^2(\Omega ):\frac{\partial v}{\partial n}=0 \text{ on } \partial \Omega \right \}$ such that $u_m$ is uniformly bounded i.e. $\|u_m\|_{H^2}\leq M$ and given the function $f(u)=u^3-u$. If I know that $u_m\rightharpoonup u(u\in H)$ in $L^2$ sense i.e. $\int_{\Omega}u_m v\to \int_{\Omega}u […]

Note: The question has been cross-posted (and answered) on MathOverflow here. Let $X$ be a Banach space. Is it true that if dual norm of $X^*$ is Frechet differentiable then $X$ is reflexive?

Determine the extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$. My attempt: I know the definition but I don’t know how to find these extreme points.Please help me to solve this problem.Thanks in advance. Extreme point:An element $f$ of the convex subset $K$ of $X$ is said to be an extreme point of […]

Problem Given a Hilbert space $\mathcal{H}$. Consider a spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$ Denote its probability measures by: $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$ Introduce the pure-point space: $$\mathcal{H}_0(E):=\{\varphi:\exists\#\Lambda_0\leq\aleph_0:\nu_\varphi(\Lambda_0)=\nu_\varphi(\Omega)\}$$ Construct its normal operator: $$\varphi\in\mathcal{D}(N):\quad\langle N\varphi,\chi\rangle=\int_\mathbb{C}\lambda\mathrm{d}\langle E(\lambda)\varphi,\chi\rangle\quad(\chi\in\mathcal{H})$$ Regard its eigenspace: $$\mathcal{E}_\lambda=\{\varphi:N\varphi=\lambda\varphi\}:\quad\mathcal{E}(N):=\cup_{\lambda}\mathcal{E}_\lambda$$ Then one has: $$\mathcal{H}_0(E)=\overline{\langle\mathcal{E}(N)\rangle}$$ How to prove this? Reference This thread is related to: Spectral Spaces (I)

I have a question please : Let $f:[0,2\pi]\times \mathbb{R} \rightarrow \mathbb{R}$ a differential function satisfying : $\displaystyle k^2\leq \liminf_{|x|\rightarrow \infty} \frac{f(t,x)}{x}\leq \limsup_{|x|\rightarrow \infty}\frac{f(t,x)}{x} \leq (k+1)^2$ Let $(x_n)\subset H^1([0,2\pi],\mathbb{R})=\lbrace x\in L^2([0,2\pi],\mathbb{R}),x’\in L^2([0,2\pi],\mathbb{R}),x(0)=x(2\pi)\rbrace$ sucht that $\|x_n\|\rightarrow \infty$ when $n \rightarrow \infty$ Why : the sequence $\left(\displaystyle\frac{f(t,x_n)-k^2 x_n}{\|x_n\|}\right)$ is bounded ? Is this answer given by :@TZakrevskiy true […]

Let $[0,1] \subset \mathbb R$ be a the compact interval in the real numbers $\mathbb R$. We know that $C([0,1] \to \mathbb R)$ (the continuous function on $[0,1]$ with values in $\mathbb R$) are dense in $L^{2} ([0,1] \to \mathbb R)$ (the usual Lebesgue space). Now consider the space of continuous functions on $[1,2]$ taking […]

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