# Tag Info

4

One of the nice things about Functional Analysis is that you can generally reduce to the scalar case by applying a linear functional to everything, rearranging scalar integrals, and then pulling the functional back outside. Then, knowing that you have enough functionals to separate points allows you to remove the functional from both sides of the resulting ...

2

Write $f_{\epsilon}=cos(\theta_{\epsilon})+i\sin(\theta_{\epsilon})$. Then \begin{align*} \lim_{\epsilon\to0}\theta_{\epsilon}&=0,\\ \Im(a)&=\lim_{\epsilon\to0}\frac{\sin(\theta_{\epsilon})}{\epsilon},\\ ...

2

Let $\mathcal{A}=C[0,1]$, $a\equiv 1$, and let $b$ be $1$ on $[0,1/2]$, taper linearly to $0$ on $[1/2,3/4]$ and be $0$ on $[3/4,1]$. Then $\|a\|=\|a^{-1}\|=1$ and $$\|a-b\| = 1 = \frac{1}{\|a^{-1}\|}.$$ However, $b$ is not invertible.

1

$\dfrac{\lambda}{c} I- A$ isn't invertible if and only if $\lambda I- cA$ isn't invertible. Hence, $\lambda\in Sp(cA)$ if and only if $\dfrac{\lambda}{c}\in Sp(A)$

1

Yes, I would say that you have fully described the relationship between $A$ and $B$.

1

Historically, integral operators are the prototypical compact operators. Compactness came up in the late 1800's when studying differential operators by recasting them as integral operators. While differential operators are very discontinuous, integral operators are especially continuous on bounded regions because they typically map uniformly bounded ...

1

The first part is a standard result (easy to prove also!) available in most books on operator theory for eg: Proposition 4.6 in Banach Algebra Techniques in Operator Theory by R.G. Douglas, 2nd edition states it as follows If $T$ is an operator on the Hilbert space $H$, then $ker T = (ran T^*)^\perp$ and $(ran T)^\perp = ker T^*$ This proposition ...

1

As suggested by a comment, you may first unitarily diagonalise $U_1U_2$ as $A D^2 A^\dagger$. Since $U_1U_2$ is a unitary matrix, the eigenvalue matrix $D^2$ is unitary too. Now define $B = U_1^\dagger AD$, where $D$ is an entrywise square root of $D^2$. Then we are done.

1

I agree with Jack D'Aurizio. If you think you know enough Graph Theory to understand Spectral Graph Theory, grab a Spectral Graph Theory book and start reading it. (My suggestions: Read either Spectral Graph Theory by Fan R. K. Chung, or Graph Spectra by Miegham) If you dont know a graph theory concept mentioned, search it up and learn it. Here's some ...

1

This is an explicit instance of the No Small Subgroups theorem for real or complex Lie groups, namely, that there is a sufficiently small neighborhood of $I$ so that the only subgroup contained in it is $\{I\}$. The specific estimate is perhaps not so crucial in the bigger scheme of things, but is tangible. Indeed, if there were an eigenvalue $\lambda$ ...

1

For 1: note that $g^n \in G \subset B$ for each $n$. It follows that if $\lambda$ is an eigenvalue of $g$, then we have $$|\lambda^n - 1| < 1$$ for every $n \in \Bbb Z$ (since $\lambda$ can't be zero, negative exponents are fine). If $|\lambda| \neq 1$, then either $|\lambda^n|$ or $|\lambda^{-n}|$ diverges to $\infty$ as $n \to \infty$. You'll need ...

1

Here is my understanding. But I know very little about QM, so this explanation may be incorrect. Consider the multiplication operator $A$ defined on $L^2[-a, a]$ by $$A\varphi(x) = x\varphi(x).$$ (In quantum mechanics, $A$ corresponds to the position operator.) Since this operator is bounded with the spectrum $\sigma(A) = [-a, a]$, by the spectral ...

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