# Tag Info

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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