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37

In short terms, kets are vectors on your Hilbert space, while bras are linear functionals of the kets to the complex plane $$\left|\psi\right>\in \mathcal{H}$$ \begin{split} \left<\phi\right|:\mathcal{H} &\to \mathbb{C}\\ \left|\psi\right> &\mapsto \left<\phi\middle|\psi\right> \end{split} Due to the Riesz-Frechet theorem, a ...


28

I would like to extend Alex' answer, as well as answer your question in the comments: "Is the "bra" basically something like the dot product?" If you have a vector space $V$ over a field $F$, which is, for now, finite dimensional, you can create another vector space, $V^*$, called the dual space of $V$, which consists of linear functionals defined on $V$. ...


28

First, the $bra$c$ket$ notation is simply a convenience invented to greatly simplify, and abstractify the mathematical manipulations being done in quantum mechanics. It is easiest to begin explaining the abstract vector we call the "ket". The ket-vector $|\psi\rangle $ is an abstract vector, it has a certain "size" or "dimension", but without specifying ...


11

Here's a pretty direct hint. If there were only rationals, the function could be defined as the denominator in lowest terms, because everything nearby enough would have a larger denominator. If we want to also define it on the irrationals this wouldn't work. However, we can compress the positive integers into $[0,1)$ in an order preserving way, then we can ...


11

I think that the function that sends any irrational number on $1$ and sends a rational $\frac{p}q$ (where the fraction is irreducible) on $1-\frac1q$ does the job. Given any rational number $\frac{p}q$, you can always find a neighborhood so that any rational number has a denominator bigger than $q$


10

I would not worry too much at first about distinguishing between "bra" and "ket". What is important is that you have an inner product $\langle x | y \rangle$ which is linear in the second coordinate and conjugate linear in the first, as opposed to the Mathematician's inner product $(x,y)$ which is linear the first coordinate and conjugate linear in the ...


7

What helped me understand it was the notion of bra and ket as vectors in Hilbert space. ket $|f\rangle$ denotes a "usual" vector, bra $\langle x|$ a "transposed" one which can be used for projection. Thus $$\langle x | f \rangle = f(x)$$ is simply the projection of $f$ into its (coordinate) representation of the $x$. This becomes more clear when you see ...


6

The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...


5

I’d attack it much more directly. HINT: Suppose that $a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty$. Then $a$ has a subsequence $\langle a_{n_k}:k\in\Bbb Z^+\rangle$ such that $|a_{n_k}|\ge k$ for each $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ let $$x_{n_k}=\frac1{ka_{n_k}}\;,$$ and let all the other terms of $x$ be $0$. Show that $x\in\ell_1$, ...


5

I'll assume $S$ not empty. Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum. If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with ...


5

Take $T_k x = k^2 x_1 -k x_2$. Then for any $x \neq 0$ we see that $|T_kx| \to \infty$. However, $T_k ({1 \over \sqrt{1 + k^2}}(1,k)) = 0$ for all $k$.


5

It fails for the sequence $1,0,0,\dots $


5

It's not quite as simple as that. The union $$ N = \bigcup_{j,k,m \in \Bbb{N}} N_{j,k,m} $$ is a countable union, so properties of measures hold. If we take the most straightforward implementation of your replacement scheme, we need $$ N = \bigcup_{\substack{j,k\in \Bbb{N} \\ 0 < \varepsilon < \varepsilon_0}} N_{j,k,\varepsilon} $$ which is no ...


5

If the $n$ real-valued continuous functions $f_1, \ldots, f_n$ separate points of $K$, then $(f_1, \ldots, f_n)$ is a homeomorphism from $K$ to a compact subset of $\mathbb R^n$. But not every compact metric space is homeomorphic to a compact subset of $\mathbb R^n$. For example, let $K$ be the Hilbert cube. For each $k$, $K$ has a subset $S_k$ ...


4

Well, functional analysis provides very natural set up for dealing with this question. We will prove the same for $$A_{c(x)}:= \left \{p(x) e^{-c(x)} \right\}$$ Where $c(x)$ is a fixed even degree polynomial with positive leading coefficient and of degree $\ge 2$ and $p(x)$ is any polynomial. Note that we need to prove that $A_{c(x)}$ is dense in $C_0 ...


4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


4

We know that for $\|A\| < 1$, $(I-A)^{-1}$ is well-defined (prove this yourself if you have not done so yet) so we can talk about the inverse. Thus: $$ I = (I-A)(I-A)^{-1}.$$ Here is a hint: $$ 1 = \|I\| = \|(I-A)(I-A)^{-1}\|.$$ Try doing some basic norm manipulations to this. You need to increase the norm, not decrease it since you want a lower ...


4

Start with an orthonormal set $\{ f_n \}_{n=1}^{\infty}$ in $L^2[1/2,1]$, and extend the functions to be $0$ on $[0,1/2]$ in order to obtain an orthonormal set $\{ \tilde{f}_n \}_{n=1}^{\infty}$ in $L^2[0,1]$. Then, for $n\ne m$, \begin{align} \|T\tilde{f}_n-T\tilde{f_m}\|^2 & =\|x\tilde{f}_{n}-x\tilde{f}_m\|^2 \\ & ...


4

If $A$ is symmetric (Hermitian), then that's true (assuming you're talking about the induced Euclidean norm). However, in general, $\|A\|$ can be very large. For example, take $$ A=\pmatrix{1&t\\0&-1} $$ with $t$ as large as you want to make it. Note that for any operator on a Hilbert space, we have $\|A^*A\| = \|A\|^2$ (you should have this as a ...


4

Suppose you are trying to solve $Af=g$ where $A$ is a linear operator. As always, there are two big issues: existence and uniqueness. Start with a Hilbert space for the sake of discussion. Existence: A solution of $Af=g$ exists iff $g$ is in the range of $A$. If $A$ happens to have a closed range, then the issue is whether or not $g$ is in the range is ...


4

When dealing with conjectures about unbounded operators, it's always good to test conjectures with a differential operator. John von Neumann defined closed unbounded operators to study differential operators. Differential operators are still the best examples. For example, let $X=C[0,1]$, and let $A=\frac{d}{dx}$ on the domain $\mathcal{D}(A)$ of ...


4

Use paralleogram law for $\frac{x}{2}$ and $\frac{y}{2}$ to obtain $||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2$ and so you get $1$ in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since $x\neq y$ .


4

The accepted answer is incorrect. To show that $S$ is closed, you must show that for any sequence of points $(x_n)$ in $S$ which converges to a limit $x\in \mathbb{R}$, the limit $x$ is also in $S$. You can prove this for your $S$ by contradiction: suppose $x_n\to x$ but $x\not\in S$. Now use the definition of limit (with $\epsilon=-x$) to show that ...


4

The first thing I thought of was a "soft proof": Suppose $f \to f'$ maps $H^\infty$ to $H^\infty.$ By the closed graph theorem, this linear map is continuous. Thus there exists $C$ such that $\|f'\|_\infty \le C\|f\|_\infty$ for all $f\in H^\infty.$ The functions $z^n$ show this fails. But it's easier to just note $f(z) = \sum_{n=1}^\infty z^n/n^2$ is a ...


4

The only reason the question seems silly is that you include the answer! A locally convex TVS is one that has a basis at the origin consisting of balanced absorbing convex sets. The reason for the emphasis on "convex" is that that's what distinguishes locally convex TVSs from other TVSs: every TVS has a local base consisting of balanced absorbing sets. ...


4

All you need to do is a little algebra to substitute $f_1 + f_2$ into the new definition. To apply Minkowski inequality, let's denote the traditional $L_2$ norm of $f$: $\left(\int_0^1 |f|^2 dx\right)^{1/2}$ by $\|f\|_0$. We have: \begin{align} & \|f_1 + f_2\|^2 \\ = & \int_0^1 |f_1 + f_2|^2 dx + \int_0^1 |f_1' + f_2'|^2 dx \\ = & \|f_1 + ...


4

Suppose you have a non-trivial solution of $$ T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t) $$ Then $\lambda \ne 0$ because the above would give $f=0$ after differentiating a couple of times. For $\lambda \ne 0$, any solution of the above must satisfy $$ \lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0. $$ Any ...


3

Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...


3

This is studied in potential theory: the function $u$ is the Newtonian potential of $f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where $K(x)=c_n|x|^{2-n}$ for $n\ne 2$ and $K(x)=c_2\log|x|$ for $n=2$. In dimensions $n\ge 3$ the kernel $K$ decays at infinity, so $u(x)\to 0$ as $|x|\to\infty$ in this case, provided $f$ is reasonable (integrable and ...


3

Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$ b_1-b_2=j_2-j_1\in B\cap I, $$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto. As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, ...



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