# Tag Info

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The bra-ket notation is really an adaptation of Fourier's original ideas from more than a century before Dirac. There are ways of viewing this notation in terms of cyclic/irreducible representations associated with selfadjoint operators. However, the Laplace transform is associated with a non selfadjoint operator, which is not appropriate for study using the ...

0

Yes, this set is closed. In fact, we do not need all of your assumptions, for example the assumption $A\cap B=\emptyset$ is superflous. Let $\left(X_{n}-Y_{n}\right)_{n\in\mathbb{N}}$ be a sequence with $X_{n}\in A$, $Y_{n}\in B$ and $X_{n}-Y_{n}\xrightarrow[n\to\infty]{}Z$, where convergence is in $L^{p}$. We have to show $Z\in A-B$. To this end, observe ...

4

Let $$\sup_{n\in \mathbb N} \| f'_n\|_2 = D <\infty.$$ Using the Fundamental Theorem of Calculus, if $f = f_n$, $$|f(x)| = |f(x) - f(0)| = \left|\int_0^x f'(s) ds \right| \le \sqrt x \|f'\|_2 \le D.$$ So $\{f_n\}$ has a uniform $C^0$ bounded. Similarly, $$|f(x) - f(y)| \le \sqrt{|x-y|} \|f'\|_2 \le \sqrt{|x-y|}D.$$ Thus the family $\{f_n\}$ is ...

2

The sequence $(f_n)$ does not have a weak-* convergent subsequence. To see this, suppose $f_{n_j}$ is a subsequence converging weak-* to some $f \in (\ell^\infty)^*$ as $j \to \infty$. Thus $$x_{n_j} = f_{n_j}(x) \to f(x)$$ as $j \to \infty$ for every $x \in X = \ell^\infty$. Define $x \in \ell^\infty$ by $x_{n_j} = (-1)^j$ and $x_n = 0$ if $n$ is not equal ...

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First, we prove the fact $T(t)=T\left(\frac{t}{n+1}\right)^{n+1}$ by induction. First, we show the property hold for the case $n=1$, i.e. $T(t)=T\left(\frac{t}{2}\right)^{2}$. The right hand side is \begin{equation*} T\left(\frac{t}{2}\right)T\left(\frac{t}{2}\right)=T(t) \end{equation*} equals the left hand side. Now assume it is true for $n=k$, where ...

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No, it's not even closed. Using the interval $[-1,1]$ instead of $[0,1]$, one can see that the sequence $$f_n(x) = |x|^{2+1/n}\operatorname{sign} x$$ converges in $C^1$ norm to $f(x)=|x|^2\operatorname{sign} x$ which is not in $C^2$. Indeed, $f_n'(x) = (2+1/n)|x|^{1+1/n}$ converges uniformly to $f'(x) = 2|x|$. The second derivatives ...

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These hints can help you, provided that you know the definitions: First try to find what the limit $T$ is in the weak topology. By the Riesz representation theorem, any linear functional in $H$ is written as $\ell(x) = \langle x, \xi\rangle$ for some vector $\xi$. What happens with $\ell(T_n(x))$ when $n\rightarrow \infty$? Now consider $x = e_1$. What ...

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Of course. Because $C_r(\mathcal A,G)$ is the C$^*$-algebra generated--in the right environment--by $\mathcal A$ and $G$. You take a dense subset $\mathcal A_0$ of $\mathcal A$, and then all sums $$\sum_{j=1}^m a_jg_j,$$ are dense, where $a_j\in\mathcal A_0$ and $g_j\in G$. As you can see, you don't need $G$ finite: countable suffices.

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Assume that such functions exist. From the given condition we have $$f(0) + g(0) = 0$$ $$f(x) + g(0) = x^2 \rightarrow f(x) = x^2 + f(0)$$ $$f(0) + g(y) = y^2 \rightarrow g(y) = y^2 + g(0)$$ Assert this into the first identity we conclue $f(0) + g(0) = xy, \forall x, y \in \mathbb{R}$. The LHS is a constant, while the RHS can be an arbitrary real number. ...

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No. Assume for contradiction that such $f(x), g(y)$ exist and let $P(x,y)$ be the statement $f(x)+g(y)=x^2+xy+y^2$. Then $P(0,0)\implies g(0)=-f(0)$. $P(x,0)\implies f(x)=x^2+f(0)$. Similarly, $g(y)=y^2+g(0)$. But then $f(x)+g(y)=x^2+(f(0)+g(0))+y^2$ and $P(x,y)$ gives $f(0)+g(0)=xy,\, \forall x,y\in\Bbb R$, contradiction.

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Let $V := \langle E\rangle$ be the span of $E$, i.e. the set of finite linear combinations of the orthonormal basis. It is easy to see $\langle x_n, v\rangle \to 0$. Now, let $y \in H$ be arbitrary and $\epsilon >0$. There is $v \in V$ with $\Vert v -y\Vert <\epsilon$, since $V$ is dense because $E$ is an orthonormal basis. Now, $$|\langle ... 3 Let ||x_n||<A Take M to be the subspace of finite linear combination of basis elements. Then it is easy to see that M is dense in H. \langle x_n,e\rangle \rightarrow 0 \forall e\implies \langle x_n,m\rangle \rightarrow 0 \forall m\in M Now fix y\in H. Take \epsilon >0 Choose m\in M such that ||m-y||<\epsilon A^{-1}  As ... 0 From comment ; Yes, they even use the notation (ℓ,x) for ℓ(x) sometimes, to emphasize this viewpoint. 1 The answer to the first question is negative. The separable Hilbert space l^2 contains an uncountable linearly independent set. For example, any Hamel basis (over \mathbb C, or over \mathbb R if you prefer real Hilbert spaces) is uncountable. For a more explicit example, recall first that \mathbb N has an uncountable family of infinite subsets A_r, ... 0 Your reasoning is correct if 'positive' is interpreted as 'positive definite'. In fact you can directly verify that (x,y)\mapsto\langle Tx,y\rangle is an inner product based on the fact that positive operators are hermitean. 2 Well, you do not quite have that \langle A \cdot, A \cdot \rangle is a inner product. T=0 would be a counterexample. The Cauchy-Schwartz inequality for \langle\cdot,\cdot\rangle suffices.$$|\langle Tx,y\rangle |=|\langle Ax,Ay \rangle |\leq \|Ax\| \cdot \|Ay\| = \sqrt{\langle Ax, Ax\rangle \langle Ay, Ay\rangle } = \sqrt{\langle A^2 x, x\rangle ...

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The answer to your question is included in the third line of the proof of the theorem you are interested in! By a classical result of Aharoni (see Theorem 7.11, p. 176 in 1), we know that there is a 3-Lipschitz-homeomorphism between $C(K)$ and some subset of $c_0$. This proof can be summarized in four steps as follows: By Sobczyk's theorem, any ...

1

As mentioned in the answer above, it is quite elementary to prove that the existence a a metric inducing the topology of a topological vector space (t.v.s.) is equivalent to the existence of a translation invariant one. However, if you then want to speak about F-spaces, i.e. complete metrizable t.v.s., the equivalence of having a complete metric or a ...

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Since $K$ is compact, $C_0(K)$. Let $A$ be a Banach space and let $B$ be a subspace. Then $B$ is dense in $A$ if and only if every linear functional $\lambda$ on $A$ such that $\lambda(b)=0$ for all $b\in B$ is the zero functional on $A$. We apply this to the Banach space $A$ consisting of all functions which are uniform limits of functions holomorphic on ...

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That proof starts with the sentence that $c_0$ is $2$-complemented in $C(K)$. And indeed $c_0$ is isometrically embedded into $C(K)$. This holds as $K$ is an infinite compact metric space, so has a convergent sequence (with all different terms) $x_n$ with limit $x$ (unequal to all $x_n$). Call $S = \{x_n: n \in \mathbb{N}\} \cup \{x\}$. Then $C(S)$ embeds ...

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Yes, $T_n$ converges strongly to zero by the Riemann-Lebesgue lemma: For any $x\in H$, $$\lim_{n\to \infty} \langle x,e_n\rangle = 0$$ which itself follows from Bessel's inequality $$\sum_{n=1}^{\infty} |\langle x,e_n\rangle | ^2 \leq \|x\|^2$$ Furthermore, $T_n$ does not converge in norm, because if $n\neq m$ $$\|T_n - T_m\| \geq \|T_n(e_n) - ... 1 (I have withdrawn my earlier answer based on Tsang's justified criticism) Metrizable topological spaces always satisfy the first axiom of countability (take the open balls with radius 1/n). In theorem 1.24 Rudin proves that if X is a TVS with a countable local base then there is an invariant metric that is compatible with the topology. The proof ... 1 The l_p norm for (x,y)\in\mathbb{R}^2 is this:$$||(x,y)||_p=(|x|^p+|y|^p)^{1/p}$$So basically your intuition of putting zeros after the second place in an infinite dimensional vector is correct, but in fact there is no need to carry them around. 1 As you found out, the two notions are not equivalent. It happens often that two different areas of mathematics use the same word to mean two different things. Think of all the meanings of "normal", "complete", or "regular"... The map T:\ell_2\to\ell_2 defined by T(\{x_n\}) = \{x_n/n\} has a closed graph, but does not map closed sets to closed sets: ... 2 If ||I-P||<1 \implies P is invertible \implies P=I 2 For any \varphi \in c_0^{\ast}, \exists y\in \ell^1 such that$$ \varphi((x_n)) = \sum_{n=1}^{\infty} x_ny_n $$Then$$ T^{\ast}(\varphi)((x_n)) = \varphi(T(x_n)) = \sum_{n=1}^{\infty} \left(\sum_{k=n}^{\infty} x_k\right) y_n = \sum_{n=1}^{\infty} x_n \left(\sum_{k=1}^n y_k \right) =: \psi((x_n)) $$Hence, T^{\ast}(\varphi) = \psi where \psi is as ... 1 English-speaking mathematicians use the word "any" too much. Pick any projection on a normed linear space onto a subspace and then prove that it satisfies this inequality? I don't think that's what you meant, but it bears that interpretation in normal English usage. Just saying "every" instead of "any" costs nothing, except two keystrokes. I take ... 2 The deficiency indices are the same because of how \Delta commutes with complex conjugation and its domain is closed under complex conjugate. g \in \mathcal{N}(A^{\star}-iI) iff$$ ((A+iI)f,g) = 0,\;\;\; f \in \mathcal{D}(A). $$Conjugating the above gives$$ ((A-iI)\overline{f},\overline{g})=0,\;\;\; f \in \mathcal{D}(A). $$... 2 Here is a direct approach to finding all selfadjoint extensions of \Delta (sorry if it doesn't answer your questions). Let V=\mathcal S(\mathbb R^+) be the space of the smooth functions f on [0,\infty) that converge to 0 as x\to\infty, together with all its derivatives, faster than any power. For f,g\in V we have ... 0 It's a special case. If X is separable then it's not hard to show that the closed unit ball of X' is metrizable in the weak-* topology (note X' itself is not weak-* metrizable), so compactness is equivalent to sequential compactness in that case. 0 If X is separable, then the closed unit ball in X' is metrizable with respect to the weak-\ast topology. Hence compactness, given by Banach-Alaoglu, implies sequential compactness. 0 First, in a complete metric space, being totally bounded is the same as being precompact. There is a simple characterization of precompact (totally bounded) sets of \ell^1. A subset A\subset \ell^1 is precompact if and only if For all n we have \sup_{f\in A} |f(n)| < \infty, and \sup_{f\in A} \sum\limits_{n=m}^\infty |f(n)| \to 0 ... 2 Yes, this is true and you may restrict yourself to two-dimensional subspaces! That is, X is isometric to a Hilbert space if and only if every two-dimensional subspace is 1-complemented. This is due to Kakutani (1939) in the real case, and Bohnenblust (1941) in the complex case. References: P. Bohnenblust, A characterization of complex Hilbert spaces, ... 2 Marten, this is true by an observation due to Kaplansky, I believe, which asserts that an infinite-dimensional C*-algebra contains a self-adjoint element with infinite spectrum. See Ex. 4.6.12 in R. V. Kadison and J. R. Ringrose, Fundamentals of the Theory of Operator Algebras, Vol. I, Elementary Theory, Pure and Applied Math., Vol. 100 Academic Press, ... 0 I think I have a counterexample. Let r\to K(r) be a differentiable real function that takes the value 0 outside the interval (0, 3), the value 1 inside the interval [1,2] and values between 0 and 1 everywhere, and such that the absolute value of its derivative never exceeds 2. Such functions exist. Now consider$$f(x,\theta)=x.\left(\theta + ...

1

In this case you have $$\|p_q(x)\| = \max_{s\ne q} \|x_s\| = \|x_r\| = \|x_q\|.$$ Hence, $$1-\frac{\|p_q(x)\|}{\|x_q\|} = 0$$ and similarly for $r$.

2

Part 2 is easier to answer: no, the way monotone operators are defined in functional analysis, $-F$ is not in general monotone when $F$ is. (This is unlike the concept of monotonicity in real analysis). Monotone operators correspond to (non-strictly) increasing functions. The reverse inequality defines dissipative operators. Part 1, geometric ...

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It is not true, take $\mathbb{R}^2$ with the $l_1$ norm. Let $x=e_1,y=e_2$, and $z=0$. Then $\|x-y\| = 2$, $\|x-z\| = 1, \|z-y\| =1$, but clearly $0$ is not on the line through $x,y$.

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The operator $-\Delta$ is densely defined on $\mathcal{D}(-\Delta)=\mathcal{C}_{0}^{\infty}(0,\infty)$. If $f,g \in \mathcal{D}(-\Delta)$, then the evaluation terms vanish when integrating by parts in the following: $$(-\Delta f,g)=(f,-\Delta g)$$ So $-\Delta$ symmetric on its domain, which is enough to guarantee that $-\Delta$ is ...

2

I believe it is not true. Let $X$ be $R^2$ and $u,v$ a basis of $X$, endows $X$ with the norm sup in the basis $u,v$: $\|xu+yv\|=sup(\mid x\mid,\mid y\mid)$ consider $c=5u, d=10u$ $\| 10u-5u\|=5$. Let $z=8u+1v$, $\| z-c\|=3$ $\| z-d\|=2$, 2+3=5

2

As you have defined: $$p_nf = \sum_{k=1}^{n}\left[n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(y)dy\right]\chi_{[\frac{k-1}{n},\frac{k}{n}]}(x)$$ The linear operator $p_n$ is an orthogonal projection operator onto the linear span of the orthonormal set $\left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}$. Hence, $p_n^2=p_n$ and $\|p_nf\| \le ... 0 Hint: your infinite sum is a geometric series. The sum of a geometric series can be written in closed form. 0 Hint: What can you say about$\inf_{a \in \bar D}\|a\|$? 0 Assume$M=\overline{M}$If$d(x,M)=0$then$x\in \overline{M}$. So$x \in M$. Now if$x\in M$.$d(x,M)=inf\{ |x-w|; w \in M|\} \leq |x-x|=0$so d(x,M)=0 0 No. If the norm were induced by an inner product it would satisfy the parallelogram law $$||f+g||^2+||f-g||^2=2(||f||^2+||g||^2).$$Simple examples show this is not so (for example, the characteristic functions of two disjoint sets). 0 Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||, ||f+g|| and ||f-g||. 1 Hint: take a look at the parallelogram law. 1 No, it is not a problem. What you are doing is to extend$f$by setting it equal to$0$outside of$\Omega$and then doing the convolution with the extension. 2 Even more is true. Every copy of$c_0$in$C(K)$for$K$compact, metric is complemented by a projection of norm at most 2. Indeed,$C(K)$is in this case separable (as$K$is second-countable we may use the Stone–Weierstrass theorem to get the claim) and then we may apply Sobczyk's theorem. 0 Start with the one dimensional subspace$W$spanned by$x_0$and define$f:W\to k$by $$f(\alpha x_0) = \alpha p(x_0)$$ This is a well-defined linear functional that satisfies$|f(x)| \leq p(x)$for all$x\in W$. Now simply apply Hahn-Banach to get$\overline{f}\$ defined on the whole space.

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