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109

This is how I used to imagine projections: If a mouse: gets run over by a steamroller: It will look like this: Now if it gets run over by a steamroller another time, it will still look like this:


27

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 ...


23

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 ...


21

It certainly cannot be "only because the unit ball becomes non-compact", because there are important algebraic differences between finite and infinite dimension that don't even depend on having a norm available (so "unit ball" would be meaningless). One of the simplest differences is that if $V$ is finite-dimensional, then a linear transformation $V\to V$ ...


18

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

$S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).


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 ...


16

A vector space is just a set in which you can add and multiply by elements of the base field. You can add polynomials together and multiply them by real numbers (in a way satisfying the axioms,) so polynomials form a vector space. A vector is nothing more or less than an element of a vector space, so polynomials can be seen as vectors.


14

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 ...


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}


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.


12

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 ...


11

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 ...


11

Take any constant function. Then $y_n$ is trivially convergent for any $x_n$.


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

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 ...


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

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

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.


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

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 ...


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 ...


8

Short answers: no, there isn't a separable solution; yes, there are eigenfunctions of the biharmonic. Unfortunately, the biharmonic isn't separable like the Laplacian. Off the top of my head, I don't know of a nice closed-form solution for the eigenfunctions of the biharmonic. The image below is an approximation of the eigenfunction for the smallest ...


8

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ ...


8

If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ...


8

The set is path-connected. Given $A$ and $B$, construct a path from $A$ to $B$ by starting with the straight line segment from $A$ to $B$, with uniform speed; whereever this is in the space, the path is given by the line segment; wherever it is not, this is because of a specific disk; define the path to map the interval that would have been mapped to a chord ...


8

Here's what I think the claim actually said (or at least, was meant to say): For $M \subset \Bbb X$ where $\Bbb X$ is a normed space, the following are equivalent: $M$ is closed For all sequences $(u_n)\subset M$, $u_n \to u$ as $n \to \infty$ implies that $u \in M$. This is quite different from what you've stated.


8

We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller. So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose ...


8

I think most of the differences can ultimately be traced back to the following properties of infinite sets (where the relevant sets would be bases of the vector space or functions of the basis vectors): There exist bijections between an infinite set and proper subsets of that infinite set. For example, there's a bijection between the integers and the even ...



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