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The only mistake I see is where you claim that $|f|\leq|f_{n_1}|+|g|$ implies $|f|^p\leq|f_{n_1}|^p+|g|^p$ , which is not true ($1\leq1/2+1/2$ does not imply $1^2\leq (1/2)^2+(1/2)^2$). What you have to use is that the sum of two functions in $L^p$ is again in $L^p$.

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Yes. Here are two examples. 1) Let $\Omega=\mathbb{N}$, so that $L^\infty(\Omega)=\ell^\infty$. Let $\mathcal{c}$ be the subspace of convergent sequences. The linear functional $f(\{x_n\})=\lim_{n\to\infty}x_n$ is bounded on $\mathcal{c}$. By the Hann-Banach theorem it can me extended to $\ell^\infty$. This extension is not represented by any $\ell^1$ ...

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I assume $\Omega\subset\mathbb{R}^n$ with the usual Lebesgue measure. Fix $x\in\Omega$ and take for example $f(g)=g(x)$ for $g\in L^\infty(\Omega)\cap C(\Omega)$. Then $|f(g)|\le \|g\|_\infty$. Extend $f$ to $L^\infty(\Omega)$ by Hahn-Banach, so that $f\in (L^\infty(\Omega))^*$, but of course $f\not\in L^1(\Omega)$ ($f$ is the Dirac measure at $x$).

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You can use separability arguments.

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You need to show: Let $(x_n,y_n)$ be any Cauchy sequence in $X\times Y$, then $(x_n)$ is Cauchy in X, and $(y_n)$ is Cauchy in Y. If you can show 1, since $X,Y$ are Banach space, $\exists x\in X,y\in Y$, s.t. $x_n\to x,y_n\to y$. Then you need to show $||(x_n,y_n)-(x,y)||\to 0$ in $X\times Y$. Since $x_n\to x,y_n\to y$, $\forall \epsilon >0, ... 2 Try with $$f(x) = \frac 12 \left( x + \frac 2x \right)$$ This is not Lipschitz on$(0,\infty)$but on any$[a,\infty)$. It remains to prove that$x$is bounded from below to conclude with the theorem. 2 This follows directly from the Orlicz–Pettis theorem (applied to the$p^{{\rm th}}$power of your expression). You can however avoid using this theorem by showing that the function $$f(\phi) = \left( \sum_{n=1}^{\infty} \lvert \phi(x(n))\rvert^p \right)^{1/p}\quad (\phi\in B_{E^*})$$ is continuous with respect to the weak*-topology. Then it will be ... 1 If one knows the value at one point and has an inequality, one can actually derive an inequality for the derivative as follows: Let us assume for simplicity that$f : [0,\infty) \to \Bbb{R}$fulfils$f(x) \leq 0$for all$x$and$f(0) = 0$. Then (if$f$is (one-sidedly) differentiable at$0$), $$f'(0) = \lim_{h \downarrow 0} \frac{f(h) - f(0)}{h} \leq 0, ... 0 No. For example, on \ell^1 let (T x)(n) = n x(n), with A = \{x \in \ell^1 \mid \sum_n n |x(n)| < \infty\}. Take x = 0 and x_n = e_n/n, where e_n is the sequence with e_n(n) = 1, e_n(m) = 0 otherwise. 2 If 1<p<q<\infty, and f\in L^q(0,1), then$$ \int_0^1 \lvert\, f(x)\rvert^p\,dx=\int_0^1 \lvert\, f(x)\rvert^p\cdot 1\,dx \le \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}\left(\int_0^1 1^r\,dx\right)^r= \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}, $$where$$ \frac{p}{q}+\frac{1}{r}=1\quad\text{or}\quad r=\frac{q}{q-p}. $$Hence ... 0 Hint: apply the Holder inequality with the parameters \frac qp > 1 and$$ \frac{\frac qp}{1-\frac qp} $$0 We have that$$ S+T=T(I+T^{-1}S). $$Define$$ K=\left(\sum_{n=0}^\infty (-1)^n (T^{-1}S)^n\right)T^{-1}. $$Then, the series above converges as \|T^{-1}S\|<1, and hence K is a bounded linear operator, and is readily shown that$$ K(S+T)=I.$$2 Yes, your argument is correct. 1 I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition. In Wikipedia, a core (of a closed operator) is defined as a subset D of the domain of A such that A is the closure of A|D. If we replace this by requiring that D be a subspace, I can prove your claim. By symmetry, it ... 1 The isomorphism H \to H^* is always there, whether you pay attention to it or not. Making it explicit would be annoying; it would be akin to disallowing yourself to treat a rational number as if it were a real number, and having to add extra notation whenever you want to convert a rational number to its corresponding real number. 0 This is a result of Varopoulos. You'll find it in R. S. Doran and V. A. Belfi, Characterizations of C{}^*-algebras: The Gelfand–Naimark theorems. Monographs Textbooks Pure Appl. Math. 101. on p. 69. 1 This is what is often called Helly's theorem. Let X be a Banach space. Let also f_1, \ldots, f_n\in X^* and scalars \alpha_1, \ldots \alpha_n be given. Then the following conditions are equivalent: there exists x\in X such that \langle x, f_i \rangle = \alpha_i (i\leqslant n) there exists \gamma \geqslant 0 such that for any scalars ... 1 A set X can have two metrics d and d' (and thus Hausdorff topologies) and a sequence (x_n) so that the limits x and x' are different. For example, take X=[0,1] and let d be the usual metric. Let f:X\to X be the bijection that interchanges 0 and 1 but fixes all other points. Let d'(a,b)=d(f(a),f(b)). If x_n=1/n, then the limit points ... 1 If these two norms would be equivalent, there would exist a constant c>0 such that$$ |||x||| = \|x\| + |\phi(x)|\le c\|x\|\quad \forall x\in V. $$This would imply boundedness of \phi a contradiction. So continuity goes wrong. 0 Sorry but i don't understand well. 0) What does mean the symbol \oplus ? (i never saw it. (neither \oplus_p)) 1) First when you write (X \times Y)'=X' \times Y' you mean that you have a linear isometric homeomorphism ? If you change of norms on the product you can loose the isometric property i think, so is the isometric property really necessary ? ... 0 here is my proof. (I used the biadjoint) My proof 0 Theorem 9 of Chapter 3 of Peetre's New Thoughts on Besov Spaces'': For any multi-index \alpha, we have the continuous map \begin{equation*} D^{\alpha}: B_p^{s,q}(\mathbb{R}^n) \rightarrow B_p^{s-|\alpha|,q}(\mathbb{R}^n) \end{equation*} Proof Sketch: The B_{p,q}^s(\mathbb{R}^n) norm of a function u consists of terms of the form \| ... 1 i'm a bit out of my depth here, so apologies if this is nonsense, but i think your idea works. suppose \|x\|=N and define f_n=\sum_{j=0}^n \frac{x^j}{j!}. this sequence converges absolutely for all N, but is not a polynomial. 5 Take$$ f_n(x)=\left\{\begin{array}{ccc} \frac{1}{x} & \text{if} & \frac{1}{n}\le x\le 1,\\ n^2x & \text{if} & 0\le x\le \frac{1}{n}. \end{array}\right. $$Then \{f_n\}_{n\in\mathbb N}\subset C[0,1], and for m\ge n,$$ \|f_m-f_n\|=\sup_{x\in [0,1]}x^2\lvert\,f_m(x)-f_n(x)\lvert \overset{\ast}=\sup_{x\in ... 0 Here is a proof when$X$is a F-space(this is needed to apply the closed graph theorem) If$T$is continuous, then$g \circ T$is continuous as composition of continuous mappings. If$g\circ T$is continuous, we are going to apply closed graph theorem to prove$T$is continuous: Suppose$x_n \to x$in$X$and$Tx_n \to y$in$Y$, then we have$g\circ ...

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I feel like the other answer is a bit sketchy, and possibly is begging the question (by assuming that the diagram commutes). So here's my own proof: If we have a bounded operator $f:X \to Y$ then $f^*:Y^* \to X^*$ and $f^{**} = (f^*)^*: X^{**} \to Y^{**}$ then $$f^{**}(i_X x) = i_Y f(x)$$ because $$f^{**}(i_X x)(\psi) = i_X x (f^*(\psi)) = f^* (\psi)(x) = ... 2 The answer to your question is no. No copy of B(H) is complemented in \ell_\infty(I, H). This is because the latter space is a Banach lattice and B(H) lacks the local unconditional structure. Y. Gordon and D. R. Lewis, Absolutely summing operators and local unconditional structures, Acta Mathematica, 133 (1974), 27–48. See also Y. Gordon and ... 11 Preduals of \ell_1 are very interesting creatures. The question is sligthly ill-posed but let me comment on that anyway. We have quite a lot isometric preduals of \ell_1. Indeed, for any countably infinite ordinal number \alpha (endowed with the order topology) we have C_0(\alpha)^* \cong \ell_1. This essentially follows from the ... 2 In finite dimensional complex spaces, you can find a branch of complex log function that doesn't intersect any of the eigenvalues. And that offers one explanation of why everything works so nicely. You can see it by using Cauchy's integral formula$$ \log(A) = \frac{1}{2\pi i}\oint_{C} \frac{1}{\lambda I-A}\log(\lambda)\,d\lambda. $$You can prove that ... 1 A neighbourhood basis of a point x in a topological space is a family \mathscr{B}_x of neighbourhoods of x such that every neighbourhood of x contains some member of \mathscr{B}_x. In particular, since the intersection of finitely many neighbourhoods of x is again a neighbourhood of x, we must for every pair V,W\in \mathscr{B}_x have an U\in ... 1 The solutions are unique if the norm of the Banach space is strictly convex:$$ \|\lambda x + (1-\lambda)y\| = \lambda \|x\|+ (1-\lambda)\|y\| $$only if x=y or \lambda\in\{0,1\}. Or equivalently,$$ \|\lambda x + (1-\lambda)y\| < \lambda \|x\|+ (1-\lambda)\|y\| $$for all x\ne y, \lambda\in(0,1). Now assume that the norm is strictly convex. ... 2 Let \{f_n\} \in (\ell_\infty(M), ||\cdot||_\infty) be a Cauchy sequence. Then \forall x\in M,f_n(x) is a Cauchy sequence of \mathbb{R}. Since \mathbb{R} is complete, the sequence converges, say to f(x). This defines a f for each x\in M. Then you just need to show \{f_n\} \rightarrow f in ||\cdot||_\infty (by using the definition \{f_n\} ... 2 Hint: Let \{f_n\}_{n \in \mathbb{N}} be a cauchy sequence in \ell^\infty(M) For any x \in M lets define$$ f(x) = \lim_{n \to \infty} f_n(x) $$3 You should try to prove directly that the set is a Banach space by showing it is complete. That is, show that for any Cauchy sequence of bounded real-valued functions on M, it has a limit that is also a bounded real-valued function on M. One big hint I can give you is that you already know that \mathbb{R} is complete. 0 Let Q be a countable dense subset of your Banach space X. In particular, every element of X may b written as the limit of a sequence of points in Q, so there is an onto map Q^{\mathbb{N}} \to X, hence$$|X| \leq |Q^{\mathbb{N}}| = \mathfrak{c}.$$On the other hand,$$|X| = |\mathbb{K}| \cdot \dim (X)$$where \mathbb{K}= \mathbb{R} or \mathbb{C} ... 2 This is a community wiki answer posted with the goal of getting this question off the unanswered list This question was crossposted to mathOverflow and received an answer there by GMark: http://mathoverflow.net/a/76019/19965 I'm reproducing the solution here, in case this link is somehow broken: 1 Takesaki (and most authors) do this because it is technically simpler. It reduces directly to standard complex analysis. Otherwise, it would be necessary to develop a little of the theory of complex analysis with values in a Banach space and a little integration theory so as to express Cauchy's formula, or use a special method. The extension of complex ... 1 Extended discussion... (Find a draft of the solution below!) Query T.A.E. nicely showed Hadamard's criterion saying that the series:$$\sum_{k=0}^\infty A_k$$converges for \limsup_{k\to\infty}\|A_k\|^\frac{1}{k}<1 and certainly diverges for \limsup_{k\to\infty}\|A_k\|^\frac{1}{k}>1. Now, why does one need bounded linear functionals anyway? ... 0 The value infinity is not allowed. Your example functional$$ f(x) = \sum_{i=1}^\infty x_i $$is not defined on the whole l^2, D(f)\subsetneq l^2. Unbounded functionals are somewhat different: they are defined on the whole space, but are unbounded. In every neighborhood of the origin there are points, where the unbounded functional admits arbitrarily ... 1 Take \ell_1(\mathbb{Z}) with the *-operatorion given by \delta_n^* = \delta_{-n}. Consider the element \tfrac{1}{2}\delta_1 + \tfrac{1}{2}\delta_2. It has norm one but its spectral radius is 1/2. Instead of \mathbb{Z} you can take your favourite finite, non-trivial group to have a finite-dimensional example. 0 Your series are convergence in this norm indeed, and you have already proven it. For an example to show the noncompleteness, consider fn(x) to be a function such that it equal to 1 when x < 1/2, it equal to 0 when x > 1/2 + 1/n, and it is linear when 1/2 < x < 1/2 + 1/n. If such fn converges to a function f in the norm above, then view it as L1 we ... 0 Notice that, if g(t)=\sum_1^\infty x_n(t), then, for t\in (0,1)$$ g(t)= \frac{\sqrt{1-t}}{1-t}. $$You can prove this as you did, using the fact that we have a geometric series. On the other hand, since x_n(0)=0 for all n, we have g(0)=0, which is impossible since the sum representation above forces g(0)=1. Therefore the series doesn't converge ... 0$$ x_n(x):=1-nx,\ x\in \bigg[0,\frac{1}{n}\bigg],\ x_n(x):=0,\ x\in \bigg[\frac{1}{n},1\bigg] $$Note that$$ \parallel x_n-x_0 \parallel\rightarrow 0 $$where$$ x_0(0):=1,\ x_0(x):=0,\ x\in (0,1]$$x_0 is not continuous so that it is not Banach 0 You are correct in your observation about proving boundedness without applying linear functionals. In fact, you can take a direct approach using only vector arguments to prove the existence and value of the radius of convergence of a vector power series: Theorem (Radius of Convergence): Suppose X is a complex Banach space and consider$$ ...

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The spectrum in a C*-algebra is always nonempty. (Read below.) This remains true for unital(!) Banach algebras. It fails, however, to hold for unital algebras in general. (See the thread on Existence of Norm for C-Algebras.)* Every C*-algebra $\mathcal{A}$ is densely embeddable into a unital C*-algebra $\mathcal{A}_e$. (An interesting view can be found ...

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I will assume that you are interested in $L^p$ spaces, and not spaces of continuous functions. The answer to your questions depends mostly on the underlying measure space. 1) If $\Omega$ is an open subset of $\mathbb{R}^n$ with Lebesgue measure, then $L^p(\Omega)$ contains unbounded functions for all $p<\infty$. On the other hand all sequances in ...

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Suppose that $xx^{\star}=x^{\star}x$ for some $x \in B$. Then $\|x^{2}\|=\|x\|^{2}$ holds because $$\|x^{2}\|^{2}=\|x^{2}(x^{2})^{\star}\|=\|(xx^{\star})(xx^{\star})\|=\|(xx^{\star})(xx^{\star})^{\star}\|=\|xx^{\star}\|^{2}=\|x\|^{4}.$$ The second part of showing that $\overline{f(x)}=f(x^{\star})$ for all $x$ is equivalent to showing that $f(x)$ is ...

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You have $$\lvert F_n(x) - F(x)\rvert \leqslant \int_{-\infty}^x \lvert f_n(t) - f(t)\rvert\phi(t)\,dt \leqslant \int_{-\infty}^\infty \lvert f_n(t)-f(t)\rvert\phi(t)\,dt.$$ Use the Cauchy-Schwarz inequality and $\lVert f_n-f\rVert\to 0$.

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Consider $\mathbf{R}^2$ with the sup norm. Define $a=(1,0)$, $a^{*}(x,y) = x$, $b=(1,1)$, $b^{*}(x,y) = y$. Then $a$ and $a^{*}$ are aligned, as well as $b$ and $b^{*}$. However $\langle a,b^{*} \rangle = 0$ and $\langle b,a^{*} \rangle = 1$

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B*-algebra is just an old-fashioned name for an abstract C*-algebra. Can you use the Gelfand–Naimark theorem? If so, then the first clause follows directly from it as in this case your algebra is isometrically *-isomorphic to the space of continuous functions vanishing at infinity $C_0(X)$ defined on some locally compact Hausdorff space $X$. You can use it ...

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