# Tag Info

105

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:

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 $V\to V$ ...

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).

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.

16

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 V be an n-dimensional vector space over the field \mathbb{F}. Given a linear map T : V \to V, there is an induced linear map \bigwedge^nT : \bigwedge^n V \to \bigwedge^n V given by \left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n). As \bigwedge^nV is one-dimensional, \bigwedge^nT = ... 13 Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space. 13 Let's take L^p space for example: how can it be in both real and functional analysis? If one uses "L^p" as notation for the set of functions that are Lebesgue integrable to the pth power, and proceeds to study the properties of individual functions in this set, this is Real Analysis. The mode of thinking here is not much different from studying ... 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 Take any constant function. Then y_n is trivially convergent for any x_n. 11 Let's go ahead and prove Eric Wofsey's claim (see his comment under the question), that if H is an infinite-dimensional Hilbert space with an orthonormal basis U of cardinality \kappa, then any Hamel basis B of H has cardinality |B| = \kappa^{\aleph_0}. Notice this implies that some (real or complex) vector spaces do not admit a Hilbert space ... 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, $$for p>1. But 0 is not an element of the set. EDIT: Here is more on the general principle involved: Your set ... 10 Make a change of variables u = e^x$$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$where g(u) = \frac{f(\log(u))}{u} is a continuous function. You can now apply Weierstrass approximation theorem. Since the above holds for all n we have that$$\int_{1}^e g(u) P(u)du = 0$$for any polynomial P(u). Now pick the polynomial to approximate ... 10 Put \displaystyle F(x)=\int_0^x f(t)^2dt, and \displaystyle g(x)=\frac{f(x)}{1+F(x)}. Clearly g is continuous on [0,1). Now you have \displaystyle F^{\prime}(x)=f(x)^2, hence \displaystyle g(x)^2=\frac{F^{\prime}(x)}{(1+F(x))^2}, and \displaystyle f(x)g(x)=\frac{F^{\prime}(x)}{1+F(x)}. Now it is easy to finish. 10 Sorry to resurrect such an old post, but I would like to supply a proof of this without invoking Jordan normal form or Schur decomposition. So we want to show the following statement. Theorem. Let T be a linear transformation on a finite-dimensional complex vector space V, say on \mathbb{C}^n for simplicity and without loss of generality; given ... 9 When d=1 it is of course possible to embed large chunks of S^d isometrically in {\mathbb R}^n. When d\geq2 it is not possible to embed even tiny pieces of S^d isometrically in {\mathbb R}^n. Proof. Take any three points x_1, x_2, x_3\in S^d forming a small equilateral triangle in the metric of S^d. This then is an "ordinary" spherical ... 9 It is certainly not dense. The linear functional$$ 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

I'm not sure if this is 100% standard, but I've always understood this notation to mean: $f$ is a function from $\Omega$ to $\mathbb{R}$ which has compact support (the subscript $0$), and is smooth/infinitely differentiable (the superscript $\infty$). I just checked, for example, that Stein and Shakarchi use this notation in their Functional Analysis. I ...

9

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 ...

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$, with uniform speed; whereever this is in the space, the path is given by the line segment; wherever it is not, this is because of a specific disk; define the path to map the interval that would have been mapped to a chord ... 8 We want to mimic$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}$, and suppose ... 8 Here's what I think the claim actually said (or at least, was meant to say): For$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.

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