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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 ...


29

The whole field of partial differential equations is an application (and origin of many problems) of functional analysis. Book: Functional Analysis, Sobolev Spaces and Partial Differential Equations by Haim Brezis. Lecture notes: Applied Functional Analysis by H. T. Banks. And a (big) bit of history: On the origin and early history of functional analysis ...


29

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 $V$. ...


29

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 ...


24

Starting from von Neumann and his contribution to economic theory (1937, existence of an optimal equilibrium in the model of economic growth ) The von Neumann model and the early models of general equilibrium There are lots of applications of functional analysis in Economic theory: Functional Analysis and Economic Theory In Financial Mathematics, in the ...


19

Another real world (theoretical physics) application is the Lagrange formalism of classical and modern mechanics which relies on the Euler-Lagrange Equation - which as you properly know is a fundamental result of functional analysis. A book on the topic: Lagrangian and Hamiltonian Mechanics Particularly I find that one of the real exciting parts of this ...


17

Let's call $f$ an $n$-SOP if we can write $$f(x,y) = \sum_{k = 1}^n g_k(x)\cdot h_k(y)$$ with continuous functions $g_k, h_k \colon [0,1] \to \mathbb{R}$. If $f$ is an $n$-SOP, for every family $x_1 < x_2 < \dotsc < x_r$ of $r > n$ points in $[0,1]$, the set $$\left\{ \begin{pmatrix} f(x_1,y) \\ f(x_2,y) \\ \vdots \\ f(x_r,y)\end{pmatrix} : y \...


15

Let $\phi:\mathbb R\to\mathbb R^2$ a surjective function - for instance a suitable (continuous) relative of a Peano curve. If we define $$\pi_x:\mathbb R^2\to \mathbb R\\(x,y)\mapsto x$$ then $\pi_x\circ\phi$ is what you want. On Peano and Hilbert curves: Peano curve and Hilbert curve are examples of surjective and continuous maps $\gamma:[0,1]\to [0,1]\...


15

Limits can definitely be seen as functionals. The problem is that to consider them as a functional, you need the limit to be defined on all elements of a vector space. If you are dealing with continuous functions on a compact set (or an interval, to make things simpler), then the limit is just evaluation at a point. As soon as you are dealing with non-...


13

Note that $C_c \subset C_0$, but $C_c \neq C_0$. For example, $f(x) = \dfrac{1}{x^2+1}$ belongs to $C_0$ but not $C_c$. What you seem to be assuming is that $\lim_{|x|\to\infty}f(x) = 0$ implies that there is some $N > 0$ with $f(x) = 0$ for all $|x| > N$. This is not true, as the above example demonstrates. That is, a function can limit to zero at $\...


13

By triangle inequality \begin{align} \left\|ax + \left(1-a\right)y \right\|\le \left\|ax \right\| + \left\|\left(1-a\right)y \right\|= a\left\|x \right\| + \left(1-a\right)\left\|y\right\| \end{align}


13

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. I'...


13

You need to check if the functions are independent, as you said. A way to go about this, which that ties it in with things you likely know is to evaluate it at several points, as you did for $x=0$. You get one condition for $x=0$. You get another condition for $x=1$ and still another one for $x=2$. Each will allow more than one solution, but they'll ...


12

In my upload http://arxiv.org/abs/1509.01078 I tried to modify Sokal's idea in order to avoid the axiom of dependent choice and instead rely on the axiom of countable choice, which is strictly weaker than dependent choice. EDIT (proof sketch(EDIT2) as wished for): We first note that for a linear operator $T$, $$ \max\{\|T(x - y)\|, \|T(x + y)\|\} \ge 1/2 (\...


12

Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1. Therefore it has a finite subcover. Now reflate the subcover by a factor of 5. Et Voila! It covers the ball of radius 5.


11

You could also go the more numerical route with functional analysis. In the study of Finite Element Methods (abbr. FEM) for example, some results of functional analysis that have applications in PDE's have natural applications as well, the Riesz representation theorem and the Lax-Milgram lemma being two popular ones. The applications in question tend to be ...


11

Write $$\alpha e^x + \beta e^{2x} + \gamma e^{3x} = 0$$ You can go ahead and cancel out a positive number like $e^x$ so: $$\alpha + \beta e^{x} + \gamma e^{2x} = 0$$ Suppose you have some solution for this with $\alpha$, $\beta$, $\gamma$ not all zero. Then, as you say $$ \alpha + \beta + \gamma = 0\qquad \qquad (1)$$ Because this must be true at $x = 0$ but ...


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

Pick an enumeration of the rationals $r_1,r_2,r_3,...$. Let $d(f,g) = 0$ if $f = g$. Else let $r_n$ be the first rational in the enumeration for which $f(r_n) \neq g(r_n)$. Define $d(f,g) = f(r_n) - g(r_n)$. This is well defined since if $f \neq g$, there must be some rational number for which $f(r) \neq g(r)$, since $f$ and $g$ are continuous, and the ...


10

Generally, without completeness, you can't deduce that a weak$^\ast$ convergent sequence is bounded. Let $X = c_{00}$ be the space of sequences with only finitely many nonzero terms, endowed with the $\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is $c_0$, the space of sequences converging to $0$, and its dual therefore isometrically isomorphic to $\...


10

For $p>1$, the set is not closed. To see this, note that $$ x_n = (1/n, \dots ,1/n,0\dots) $$ is an element of $S$ (the number $1/n$ appears $n$ times). Now, we have $$ \Vert x_n \Vert_{\ell^p} = 1/n \cdot n^{1/p} = n ^{1/p -1}\to 0, $$ for $p>1$. But $0$ is not an element of the set. EDIT: Here is more on the general principle involved: Your set ...


10

You will find what you are looking for in Chapter 3.1 of Gazzola, F., Grunau, H.-Ch., Sweers, G.: Polyharmonic Boundary Value Problems. Lecture Notes 1991. Springer, Berlin (2010). EDIT: To be more specific, you are looking for Theorem 3.8 at page 69. The original proof goes back to Friedrichs, K. Die Randwert- und Eigenwertprobleme aus der Theorie der ...


10

Put $\displaystyle F(x)=\int_0^x f(t)^2dt$, and $\displaystyle g(x)=\frac{f(x)}{1+F(x)}$. Clearly $g$ is continuous on $[0,1)$. Now you have $\displaystyle F^{\prime}(x)=f(x)^2$, hence $\displaystyle g(x)^2=\frac{F^{\prime}(x)}{(1+F(x))^2}$, and $\displaystyle f(x)g(x)=\frac{F^{\prime}(x)}{1+F(x)}$. Now it is easy to finish.


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 ...


10

Let $k_n$ such that $a_{k_n}\geq 2^n$. Define $x_{k_n}= a_{k_n}^{-1}$ and $0$ elsewhere.


9

When $d=1$ it is of course possible to embed large chunks of $S^d$ isometrically in ${\mathbb R}^n$. When $d\geq2$ it is not possible to embed even tiny pieces of $S^d$ isometrically in ${\mathbb R}^n$. Proof. Take any three points $x_1$, $x_2$, $x_3\in S^d$ forming a small equilateral triangle in the metric of $S^d$. This then is an "ordinary" spherical ...


9

It is certainly not dense. The linear functional $$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n} $$ Is continuous on $l^2$. Thus, the set $T^{-1}(0)$ must be a closed subspace of $l^2$. If it were dense, we'd have $l^2 = T^{-1}(0)$. But the sequence ${\frac{1}{n}}$ obviously isn't in this space.


9

In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same (...


9

Much of the theory of probability can be considered as a branch of functional analysis (although some probabilists might object to this statement). For example, the Strong Law of Large Numbers can be considered as a special case of Birkhoff's Ergodic Theorem.



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