I read in Kolmogorov-Fomin’s (p. 430 here) the statement, sadly left without a proof, that if function $f:(a,b)\to\mathbb{C}$, measurable almost everywhere on $(a,b)$, where $-\infty\leq a<b\leq\infty$, is not equal to 0 [i.e. $\forall x\in(a,b)\quad f(x)\ne 0$, as Daniel, whom I deeply thank, explains in his answer] and satisfy the condition $|f(x)|\leq Ce^{-\delta|x|}$ with $\delta>0$, then […]

Here is a proof which I do not fully understand. Theorem : Let $H$ be a Hilbert space. A continuous linear map $T : H \rightarrow H$ is self-adjoint (hermitian) if and only if $$\big\langle T(x), x\big\rangle \in \mathbb{R},~~~~~(\forall x \in H)$$ Proof : ($\Rightarrow$) $$ \overline{\big\langle T(x), x\big\rangle} = \big\langle x, T(x)\big\rangle = \big\langle […]

Let $H$ be a $\mathbb C$-Hilbert space, $U\subseteq H$ be a closed subspace of $H$, $\operatorname P_U$ denote the orthogonal projection from $H$ onto $U$ and $x\in H$. I want to show that $$\tilde x=\underset{u\in U}{\operatorname{arg min}}\left\|u-x\right\|_H\Leftrightarrow\tilde x=\operatorname P_Ux\tag1\;.$$ “$\Leftarrow$”: $\tilde x=\operatorname P_Ux$ $\Rightarrow$ $$\langle\tilde x-u,u\rangle_H=0\;\;\;\text{for all }u\in U\tag2$$ and hence (since $t\tilde x\in U$) […]

Let $(X,<.>)$ is an inner product space prove that $x$ and $y$ are orthogonal if and only if $||x+αy|| \ge ||x||$ for any scalar $α$ . The first direction if $x$ and $y$ are orthogonal then$||x+αy||^2=||x||^2+|α|^2||y||^2 \ge ||x||^2 $ so $||x+αy|| \ge ||x||$ . But the second direction if $||x+αy|| \ge ||x||$ for any scalar […]

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 […]

Let $X_{n} \in \mathbb{R}^{\infty}$ be a tight sequence of processes in metric space $(\mathbb{R}^{\infty}, l_{2})$ and for each $x\in\mathbb{Z}_{+}$ we have that $X_{n,x}\stackrel{d}{\to} Y_{x}$. Does it follow the weak convergence of the process $X_{n}$ to $Y$?

Is it true that $$|| a R+b P||\leq\max \{|a|,|b|\},$$where $a$ and $b$ are complex numbers and $P,R$ are (orthogonal) projection operators on finite-dimensional closed subspaces of an infinite-dimensional hilbert space ? In finite dimensions this would be true, since a projection there has always norm one.

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)

If $T$ is an operator on $l_2$, and $\lambda>r(T)$ (where $r(T)$ denoted the spectral radius of $T$), the resolvent $(\lambda I-T)^{-1}$ can be expanded as $$ (\lambda I-T)^{-1}=\frac{1}{\lambda}I+\frac{1}{\lambda^2}T+\frac{1}{\lambda^3}T^2+\dots $$ If we fix some $x\in l_2$, and denote $y:=(\lambda I-T)^{-1}x$ the above expansion implies that $y$ is in the closed span of $(T^nx)_{n=0}^{\infty}$. Is $y$ in […]

let’s see if someone can help me with this proof. Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. And let $L^2\left(\Omega\right)$ be the space of equivalence classes of square integrable functions in $\Omega$ given by the equivalence relation $u\sim v \iff u(x)=v(x)\, \text{a.e.}$ being a.e. almost everywhere, in other words, two functions belong to the same equivalence […]

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