# Tag Info

## Hot answers tagged functional-analysis

8

Remember the definition of equivalence is that there exists numbers $c,d$ such that $c\Vert x\Vert_2 \leq \Vert x\Vert_1 \leq d\Vert x\Vert_2$. What the c and d do in the picture is to stretch or shrink the shapes you've drawn. What the inequality represents is one shape fitting inside another. What equivalence means is that I could shrink the circle for ...

7

This property is called "(H)" property and the space is called a "Radon Riesz" space. As example of this space we can say every Hilbert space as the proof given by @bartgol and $L^p$ for $1<p<\infty$, and schur space (in particular $\ell^1$).

6

Ok, assuming you are in a Hilbert space (otherwise I think we must use Hanh-Banach), you can simply consider the functional $$F(x_n) = \langle x_n,x \rangle$$ and see what happens when $n\to\infty$. Then what happens to $\|x-x_n\|^2=\langle x-x_n,x-x_n\rangle$? Edit: yes, in any Hilbert space, by definition, $\|x\|^2 = \langle x,x \rangle$.

6

Given the axiom of choice, every vector space has a basis (though it will be a very unnatural basis), and you are correct that infinite-dimensional vector spaces are exactly those where the basis is infinite. But this kind of basis (often called a Hamel basis) is rather useless and impossible to visualize. So, a more concrete way of thinking about it might ...

6

In the context of functional analysis, a functional is a function from a vector space to its base field (usually $\mathbb{R}$ or $\mathbb{C}$). In many important cases they are linear, but this is not always the case. In the context of normed vector spaces or more generally topological vector spaces, they are also often continuous, but again this is not ...

6

A function with this property does not have to be in $L^2$. Let's first note the following lemma: Lemma. Let $H$ be a separable Hilbert space and $E \subset H$ a dense linear subspace. There exists a countable set $\{e_n\} \subset E$ which is an orthonormal basis for $H$. Proof of lemma. Since $H$ is separable, so is $E$. Let $\{x_n\}$ be a ...

6

Not necessarily; this fails even in $\mathbb{R}^2$ with the usual inner product. Take $U$ to be the $x$-axis, and $V = \{(x,y):x=y\}$, the diagonal. Then $\mathbb{R}^2 = U\oplus V$, but nothing other than $0$ in $V$ is orthogonal to anything in $U$ other than $0$.

5

I'm not sure about the Laplace transform but in Joel L. Schiff's "The Laplace Transform: Theory and Applications" on page 13 the author proves that a large class of functions has a Laplace transform. I am not sure how to describe nicely the result in terms of domain and range of the operator, buy maybe that helps. As for the Fourier transform, you first ...

5

That sum does not converge in the $\ell_\infty$ norm if the sequence $x_n$ does not converge to $0$.

5

It's certainly true for $1<p<\infty$. In that case the weak topology on $B$ is the same as the weak* topology regarding $\ell^p$ as the dual of $\ell^{q}$. So $B$ is a compact Hausdorff space, and so Stone-Weierstrass implies that those polynomials are dense in $C(B)$. Note that you didn't specify what topology on $C(B)$ you're talking about when you ...

5

All normed linear spaces are metric spaces, and metric spaces are Hausdorff.

4

Let $X = \ell^2(\mathbb{N})$, and $Y = c_{00}$ be the subspace of sequences with only finitely many nonzero terms. We know $\overline{Y} = X$. Then let $S$ be the left shift, $$(S(x))_n = x_{n+1}$$ and $T = I - 2S$. Since $S$ is bounded and $Y$ is $S$-invariant, the same holds for $T$. Furthermore, $T$ is injective on $Y$: Let $0 \neq y\in Y$ and $n_y\in ... 4 A metric space doesn't come with a vector space structure, without which there can't be much analysis going on. It woulds just be a corner of topology, then. You can imagine equipping a vector space with a separate metric, but by the time you require that the metric be compatible with the linear structure --$d(v+w,u+w)=d(v,u)$and$d(λv,λu)=|λ|d(v,u)$are ... 4 There's certainly an approximate identity in$\mathcal S$. For$\phi\in\mathcal S$and$t>0$define$\phi_t(x)=t^{-1}\phi(x/t)$. Then if$\int\phi=1$it follows that$\phi_t*f\to f$in$\mathcal S$for every$f\in\mathcal S$. Say$\psi=\hat\phi$. It's easiest to verify that you have an approximate identity if you choose$\phi$so that$\psi=1$in a ... 4 This a potential answer to the question. It is not clear whether its proof technique can be completed or not. We can obtain the result by considering convergence of the Fourier transform of the partial sums. Since$f$is smooth and compactly supported it is in$L^2(\mathbb{R})$as are all of its derivatives. Let$f_n$be the$n$th partial sum in the ... 4 For$0 < p < 1$, let $$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$ We endow it with the$p$-seminorm $$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$ and the metric$d_p(x,y) = s_p(x-y)$derived from it. Note the difference from the case$p \geqslant 1$. If ... 4 Try$A = \mathbb{Q} \subseteq \mathbb{R}$. (If a subset,$A$, of a vector space is a 2-divisible subgroup, then$A + A = 2 A$will hold, but a subgroup can only be convex if it is a subspace.) 4 Hint: choose any constant function$g$on the interval$[a,b]$. Then$g \in B$but$g$is not the zero function. 4 (a) Consider$g(t)=1$. (b) Consider$g(t)=t^2(1-t)^2$. 4 Proof by contrapositive certainly works. If$a_{n} \not \to 0$, there is an$\epsilon > 0$and subsequence$a_{n_k}$such that$|a_{n_k}| > \epsilon$for each$k$. Then, show that the sequence$\{e_{n_k}\}$is mapped to a sequence with no convergent subsequence. 4 I don't think this is the place to ask that kind of question. But since you asked, if you're having difficulty with this book and you're having trouble finding motivation to read it, why don't you just leave it for a bit, and come back to it after you're done with your other books? Or perhaps it is a too complex book for your level? I don't really know... 4 In short, the objective of the classification program is to give a (relatively) simple set of invariants that distinguish between the isomorphism classes of$C^\ast$-algebras, that is, if the invariants are isomorphic, then the$C^\ast$-algebras are isomorphic. It is thus stronger than what homology and cohomology theories yield for topological spaces: ... 4 They do not satisfy the definition of a linear map. Instead, there is a canonical bijection: $$\text{Vect}(\mathbb{K}^m,\mathbb{K}^n) \to \mathbb{K}^{n\times m}$$ where$\text{Vect}(\mathbb{K}^m,\mathbb{K}^n)$is the set of linear maps from$\mathbb{K}^m$to$\mathbb{K}^n$. It is given by mapping a linear map$f : \mathbb{K}^m \to \mathbb{K}^n$to the ... 4 You're right about the thing you say looks strange. It's simply not correct to say the step functions are dense in the continuous functions, because they're not a subset of the continuous functions. 4 Here is my favorite proof, which I think is simpler than both the one suggested by David C. Ullrich and the one I had given earlier, elaborating on David Mitra’s hint. It uses only the Hahn–Banach theorem, but not Riesz’s lemma. It is based on the hint presented in Exercise 5.25, Folland (1999, p. 160). If$X^*$is separable, let$\{f_n\}_{n\in\mathbb N}$... 4 First, note that if$0<b<T$and$u:[0,b] \to \mathbb{R}^n$satisfies $$u'(t) = F(t,u(t))$$ $$u(0) = u_0$$ then we also have that it satisfies$|u(t)| \leq C$for all$t\in [0,b]$. Now, we know that$|v(0)|=|u_0|=|u(0)| \leq C$. Suppose there is$t_1\in [0,T]$such that$|v(t_1)| > 2C$. Then there is$t_2\in [0,t_1)$such that,$|v(t_2)| = 2C$and, ... 4 Mathematics is a lot about outsmarting your definitions. Namely, "show where it says that you can't do that". This is why so many definitions in mathematics have "trivial" examples, sometimes to the point where they can be a cumbersome addition to each proof when first learning the subject. If the definition of a vector space is such that nowhere it is ... 4 You can justify the interchange of orders of integration by letting $$G(x,y) = e^{-(x-y)^{2}/2}$$ and noticing that$G(x,y)f(x)\overline{g(y)}$is jointly measurable in$x,y, and is bounded by \begin{align} |G(x,y)f(x)\overline{g(y)}| & = |G(x,y)^{1/2}f(x)||G(x,y)^{1/2}\overline{g(y)}| \\ & \le ... 4 Scalar multiplication will not be continuous unless the topology on the scalar fieldK$is also discrete (or unless$X=0$). Indeed, if$x\in X$is any nonzero vector,$a\mapsto a\cdot x$is a continuous injection$K\to X$, as the restriction of the scalar multiplication map$K\times X\to X$to the subspace$K\times\{x\}\cong K$. If$X$has the discrete ... 4 The Boolean algebra of connected components is equivalent to the projections (the elements with$p^2=p$) in the algebra of functions, with multiplication of functions representing intersection and$(p_1,p_2) \to p_1 + p_2 - p_1p_2\$ being the union of sets of components. I think there is a version of algebraic K-theory for topological algebras whose value ...

Only top voted, non community-wiki answers of a minimum length are eligible