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## Hot answers tagged functional-analysis

6

$p$ is not needed for the proof, but rather it's an important part of the formulation of the theorem. Not only can the linear functional be extended, but the extension is still dominated by $p$. Just an extension is not hard to produce, but the theorem actually tells you that you still have control over the extension knowing that the original functional is ...

5

No, this is not true. If $D$ is a Dedekind finite set with a Dedekind finite power set, then $\ell_1(D)$ is a Banach space which has a Hamel basis which is also a Schauder basis, and every linear operator from $\ell_1(D)$ to a normed space is continuous. But if $D$ is Dedekind finite, then $|D|^{\aleph_0}>|D|$. So it suffices to assume that an infinite ...

4

Since $X_n\xrightarrow{p} X$ there exists a subsequence $X_{n_k}\xrightarrow{a.s.}X$. However, the event $\{X\le x\}=$$\{\lim X_{n_k}\le x\} is a tail event so that X must be constant by Kolomogorov's 0-1 law. 3 Necas' book "Direct Methods in the Theory of Elliptic Problems" is a wonderful guide for the topics that you mentioned above, although the proofs are very abstract and most steps are omitted. Evans' book is more understandable and also includes the topics above, but not in the most general settings of the theorems. If you just start studying the Sobolev ... 3 Every space which has a Schauder basis is separable. We have that \ell^\infty is not separable, hence it can't have a Schauder basis. You can see a proof of this here, for example. Using the axiom of choice, we have that every vector space has a Hamel basis - in particular, \ell^\infty has a Hamel basis, but for infinite dimensional vector spaces we ... 3 We have$$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$which is compact in \mathbb{R} (since \mathrm{supp}(f) is compact and f is continuous), hence bounded. 3 Let \|\cdot\| be some C^*-norm on C_c(X). Write A=(C_0 (X),\|\cdot\|_\infty) , B=\overline {C_c(X),\|\cdot\|) }. Fix f_0\in C_c(X). Put A_0=C^*(f_0)\subset A, and \pi:A_0\to B the identity map (this works because A_0\subset C_c (X)\subset B ; since we use the supremum norm to generate it, every element has support inside that of f_0). ... 2 You have g=F(f). The Fourier transform is a continuous linear map in many good spaces (Schwartz space, its dual, or L^2). The derivative of a linear map is the map itself, and so a continuous linear map is continuously differentiable (and the second derivative vanishes). That is, \partial g/\partial f=F. (One usually denotes this by DF(f)=F or ... 2 By expanding f and then re-factorizing it in a slightly different form we find$$f(x) = \sum_{i=1}^N x^2 - 2x\cdot y_i + y_i^2 = N[x^2 - 2x\cdot \overline{y} + Y] = N(x-\overline{y})^2 + N[Y - \overline{y}^2]$$where \overline{y} = \frac{\sum_{i=1}^N y_i}{N} and Y = \frac{\sum_{i=1}^N y_i^2}{N}. From this last form, remember that x^2 \geq 0, it is ... 2 Suppose \mathcal{H}=L^{2}[0,\pi]. Suppose \mathcal{M} is a countably dense subset of compactly supported C^{\infty} functions on (0,\pi). Apply Gram-Schmidt in order to obtain an orthonormal basis \{ f_{k} \}_{k=1}^{\infty} of smooth compactly supported functions on (0,\pi). The following operators L_{\alpha,\beta} are selfadjoint$$ ... 2 Your reasoning is correct, for a continuous (bounded) everywhere defined operator$A$on a Banach (in particular on a Hilbert) space, the denseness of the range of$\lambda\mathbb{I} - A$together with the boundedness of the inverse already implies the surjectivity of$\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ... 2 As David Mitra mentioned in his comment, one such proof can be found in Morrison T.J. Functional Analysis. An Introduction to Banach Space Theory (Wiley, 2000), p.221. The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese ... 2 Let$\phi$be a state. Suppose that$\phi(a_1)=0$. Then, for some fixed$j$, $$0\leq\phi(a_ja_j^*)\leq\phi(a_1)=0,$$ so$\phi(a_ja_j^*)=0$. Now, since$\phi$is completely positive, it satisfies the Kadison-Schwarz inequality; thus $$\tag{1} 0\leq\phi(a_j)\phi(a_j)^*\leq\phi(a_ja_j^*)=0,$$ so$\phi(a_j)\phi(a_j)^*=0$, which implies$\phi(a_j)=0$. Since ... 2 I can't bring myself to write$fg$for the convolution of$f$and$g$. So I'm going to write$f\mapsto f'$for the involution, so I can write$f*g$for the convolution. Are you certain you got the definition of$f'$straight? What would make much more sense to me would be $$f'(t)=\overline{f(-t).}$$ That seems to me is the "standard" involution on$L^1$. ... 2 Since$U$is a bounded domain, its closure$F=\bar U$is compact. Therefore any (smooth) function$F\to\mathbb R$is compactly supported and so$C_0^\infty(F)=C^\infty(F)$. Consequently$W^{1,2}_0(F)=W^{1,2}(F)$. It is possible that someone prefers to work with a closed set instead of an open one (as in the case of manifolds with boundary) and means ... 2 If I understood you well, your$\ell^0$is the space of complex sequences that are eventually vanishing. If that is right, consider the sequence$(c_n) \in (\ell^0)^{\mathbb N}$defined by$c_n(k) = \frac{1}{k}$for$1 \le k \le n$and$c_n(k) = 0$for$k > n$.$(c_n)$is a Cauchy sequence as for$n < m$$$\Vert c_n - c_m \Vert^ 2=\sum_{k=n+1}^m ... 2 In the complex case, you have (Su,v)=\overline {(Sv,u)} for all u,v. Then$$ (Su,v)=\overline {(Sv,u)}=(u,Sv)=(S^*u,v). $$So (\, (S-S^*)u,v)=0 for all v , which implies (S-S^*)u=0. As this occurs for all u , S-S^*=0. For the real case, just remove the bars. 2 It should be closed. Let T:V\to W be a bounded linear operator between normed linear spaces. We have that$$ \ker(T) = \{x\in X: Tx =0\} = T^{-1}(\{0\}). $$Since bounded operators are continuous and \{0\} is a closed set in the norm topology, we see that the kernel is closed. The kernel is not open unless T is the zero operator. Assume T\neq 0. We ... 2 From Hölder:$$ I=\int_{a}^{b}(x-x_1)^2\ldots(x-x_n)^2 dx \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^4 \ldots (x-x_n)^4dx \right)^{1/2} \leq \\ \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^8 dx \right)^{1/4}\left(\int_{a}^{b}(x-x_3)^{8}\ldots(x-x_n)^8 dx\right)^{1/4} \leq \ldots \\ \leq ... 2 There are many equivalent ways to construct the topology on$C^\infty_c(\Bbb{R^d})$that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive. First, we set our goal to be to find a topology on$C^\infty_c(\Bbb{R^d})$that makes it a locally ... 2 You may observe that, using a standard integral representation of the gamma function, we have $$\int _0^{\infty }x^{s/2-1}e^{-\pi n^2x}dx=\frac{1}{n^s}\pi^{-s/2}\Gamma(s/2)$$ then your series rewrites $$\sum_{n=1}^{\infty} \int_{0}^\infty x^{\frac{s}{2}-1}e^{-\pi n^{2}x}dx=\pi^{-s/2}\Gamma(s/2)\sum_{n=1}^{\infty} ... 2 Let x = (\xi_n) and y = (\omega_n). We note that$$ p(x + y) = \lim_{n \to \infty} \sup_{m \geq n} (\xi_m + \omega_m) \leq \lim_{n \to \infty} \left(\sup_{m \geq n}\xi_m + \sup_{m \geq n}\omega_m\right) = p(x) + p(y) $$To further expand on the inequality: we note that A \subset B \implies \sup A \leq \sup B. This allows us to state that$$ ... 2 The spectrum of the$C^*$-algebra$C_b(\mathbb R)$of bounded continuous functions on$\mathbb R$corresponds to the Stone-Čech compactification$\beta \mathbb R$of$\mathbb R$. Any finite positive Borel measure on$\beta \mathbb R$gives you a positive linear functional on$C_b(\mathbb R)$. However, the points of$\beta \mathbb R \backslash \mathbb R$... 1 The reason that$0$is in the spectrum of$S$is that$S$is not invertible. Indeed, this is clear by seeing that$S$is not bijective. The element$(1,0,0,…)$is not in the range of$S$, for example. However the range of$S$has (Hilbert space dimension) equal to$\ell^2$. If$e_0,e_1,…$is the canonical orthonormal basis for$\ell^2$(where$e_i$is ... 1 For any Banach space of dimension$>2$,$X \backslash \{0\}$is simply connected. 1 Remember the definition of the norm of an operator from$A$to$B$is:$\|f\|_{op} = \sup\limits_{\|t\|_A=1}\{\|f(t)\|_B\}$In this case,$t$is a$3$-tuple,$(x,y,z)$, and we take the sup norm of the 3-tuple. When$\|(x,y,z)\|_\infty=1$is whenever$|x|,|y|$or$|z|=1$and the others are$\leq 1$. It is plain to see that this is maximized when ... 1 Yes, since$B(H)$is a Banach algebra it makes perfect sense to talk about questions of convergence. Unfortunately, the sum in question need not converge. Let$v$be some unit vector in$H$. We can construct a counterexample as follows: Let$P_v$denote the orthogonal projection onto the subspace spanned by$v$. Let$a = - P_v$. Then$a^{n} = (-1)^{n} P_v$, ... 1 You can use the Lagrange method such as $$\frac{\partial f}{\partial x_1}=0 \quad \frac{\partial f}{\partial x_2}=0 \quad \frac{\partial f}{\partial x_3}=0$$ and solve the system of equations. 1 Each$f_n(s)= \int_0^\infty x^{s/2-1}e^{-\pi n^2x}\,dx$is defined on$(0,\infty).$Each$f_n \to \infty$at$0$and$\infty.$You make the change of variables$x=y/(\pi n^2)$and you get the identity $$\int _0^{\infty }x^{s/2-1}e^{-\pi n^2x}dx=\frac{1}{n^s}\pi^{-s/2}\Gamma(s/2)$$ -as observed by Olivier Oloa. It follows that we need$s>1$for even ... 1 Note that your statement can be written as$A$surjective$\LeftarrowR(A)_\perp =\{ w\in W: \ \langle Av,w\rangle = 0 \ \forall v\in V\}= \{0\}$The desired implication fails if$A$has dense range but is not surjective (provided$V,W$are normed spaces,$A$continuous,$W^*$continuous dual). If the range of$A$is dense then$\$ \langle ...

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