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3

The `basis' you described in 3 is given by the Dirac measures, and it is not in $L^2$, meaning that $\langle x|x\rangle$ is not defined. In other words, states concentrated at a single point is not allowed in the theory from a strict mathematical point of view. So 4 is not true. If you want to include states like $|x\rangle$ there is a formalism called ...


3

$L^2$ spaces should not be sensitive to the topology or shape of whatever underlying space you're working with. Indeed, given a "manifold" (a generalization of circles and surfaces), one way of defining an $L^2$ space on it is to pick a chart $D^n \to M$, where $D$ is the unit disc, that is injective except at a set of measure zero, and then pull back ...


3

Yes. For a positive linear operator $A$, $\|A\| = \sup \{(Ax, x): \; \|x\| = 1\}$. If $A \ge B$, $(Ax, x) \ge (Bx, x)$, so we must have $\|A\| \ge \|B\|$.


2

1) You know that $$f(x)=\sum_{i=1}^\infty \lambda_ie_i$$ where $\lambda_i\in\mathbb R$. Now, since $\{e_i\}$ is an orthonormal basis, you'll get, $$\left< f(x),e_i\right>=\sum_{i=1}^\infty \left<\lambda_j e_j,e_i\right>=\sum_{j=1}^\infty \lambda_j\underbrace{\left<e_j,e_i\right>}_{=\delta_{ij}}=\lambda_i.$$ Notice that $\left<f(x),e_i\...


2

Start with $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^\perp. $$ As you noted, $\mathcal{N}(T^{1/2})\subseteq\mathcal{N}(T)$. Because $T^{1/2}$ is selfadjoint, $$ \|T^{1/2}x\|^2=(T^{1/2}x,T^{1/2}x)=(Tx,x). $$ Therefore $\mathcal{N}(T)\subseteq\mathcal{N}(T^{1/2})$. So, $$ \overline{\mathcal{R}(T)}=\mathcal{N}(T)^{\perp}=\...


2

Of course as Hilbert spaces $L^2(S^1)$ and $L^2\bigl([0,1]\bigr)$ are isomorphic, and you could also say that $L^2\bigl([0,1]\bigr)$ is the prime example of a Hilbert space arising from Lebesgue theory. But note that $L^2(S^1)$ is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between $L^2\bigl([0,1]\...


1

You seem to be under the impression that $||T||$ is the absolute value of the largest eigenvalue. This is not so: $$\left[\begin{matrix}1 &1\\0&1\end{matrix}\right].$$But it's clear from the definition that $|\lambda|\le||T||$ for any eigenvalue $\lambda$, and given that you've given the proof. I don't see how what you wrote is "informal"...


1

The "basis" you imply in 4. and 5. is the set of all shifted delta distributions $\{\delta_x (f) = f(x)\,\forall\,f\in \mathscr{S}(\mathbb{R}^N)\}$ where $\mathscr{S}(\mathbb{R}^N)\subset L^2(\mathbb{R}^N)$ is the Schwartz Space of bounded smooth functions $f:\mathbb{R}^N\to\mathbb{R}^N$ such that any derivative $D^\alpha (p\,f)$ of any multiple $f p$ of $f$ ...


1

Without orthogonality this is false: an example is given by Robert Israel. Orthogonality implies that $\|u+v\|^2 = \|u\|^2+\|v\|^2$ for $u\in M$, $v\in N$. Thus, if a sequence $(u_n+v_n)$ converges, the inequalities such as $$\|u_n-u_m\|\le \|(u_n+v_n)-(u_m+v_m)\|$$ imply convergence of both $u_n$ and $v_n$. So, $u_n\to u\in M$ and $v_n\to v\in N$, which ...



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