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Oops, I misread your question, sorry. Here is a correct answer. Let $U:B\to UB\subseteq A$ be a linear isomorphism. Suppose towards a contradiction that for some $t\geq\|U^{-1}\|$ we have $\displaystyle\|b\|_B>\inf_{\tilde{b}\in B}\left(\|b-\tilde{b}\|_B+t\|U\tilde{b}\|_A\right)$ then we can find $\tilde{b}\in B$ with ...

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Yes. Namely if $Y = X \oplus U$ and $Z = Y \oplus V$, then $Z = X \oplus (U \oplus V)$.

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Because $L^2$ can also be defined as the space of square-integrable functions (up to measure zero), and one can prove both definitions are equivalent. It's not any more economical to represent $L^2$ as a space of sequences than to do so for $\mathbb{R}$: they're each only spaces of sequences up to a rather large equivalence relation, so elements don't have ...

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No, take the (incomplete) vector space $\Bbb Q$ which is a one-dimensional vector space over the rational numbers. Since it is not complete it is neither Banach nor Hilbert.

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Note: This answer is community wiki. Compactness gives rise to a cover: $$\#S_\varepsilon<\infty:\quad\bigcup_{\sigma\in S_\varepsilon}B_\varepsilon(\sigma)=C$$ Produce a separable Hilbert space: $$C\text{ compact}\implies\overline{\langle C\rangle}\text{ separable}$$ Its basis introduces increasing projections: $$\{\varepsilon_n:n\in\mathbb{N}\}:\quad ... 3 To avoid confusing yourself with sequences of sequences, I recommend thinking of the elements of C_0 as functions f:\mathbb{N} \to\mathbb{R} such that \lim_{n\to\infty}f(n)=0. Then the claim becomes: if (f_k)_{k=1}^\infty is a Cauchy sequence of functions with respect to the uniform norm, there is f\in C_0 such that f_k\to f uniformly. The ... 0 In finite (fixed) dimension all the norms are equivalent. The function$$x\longmapsto\frac{x}{\|x\|_2}$$is continuous and bijective form S_1 (compact by Heine-Borel) to S_2, so the inverse is continuous. 0 The "strong operator topology on H^*" is just the same as the weak topology on H, so you're asking whether strong and weak convergence in H is the same. No, it isn't. Standard example: an orthonormal sequence u_n converges to 0 weakly but does not converge strongly. 1 Let e_n denote the element of \ell_1 whose m'th coordinate is 0 if m\ne n and 1 if m=n. To see more directly why the bounded linear map, referenced in my comment to the main post, T:\ell_1\rightarrow\ell_1 defined by$$Te_1=e_1;\ Te_n=e_1+e_n, n>1$$is not an adjoint operator, consider the sequence (e_n) in \ell_1. Note for any ... 0 Indeed, each continuous bijection from a compact topological space into a Hausdorff space is homeomorphism(See topological books). Similarly for open functions. 0 I think I got some answer for the second part of the theorem. If f:[a,b]\rightarrow E is absolutely continuous and differentiable almost everywhere, then for \phi\in E^* we have \phi\circ f absolutely continuous hence the fundamental theorem for real valued functions gives us$$\int^b_a (\phi\circ f)'=\phi(f(b))-\phi(f(a)).$$Combining this with ... 1 The basic theorem in this regard is that the conclusion holds so long as the sequence of derivatives df_n converges locally uniformly. See 8.6.3 of Foundations of Modern Analysis by Dieudonné. 1 As mentioned in the comments, your first claim is a justly famous result of R. C. James. For the second question, you can use this result to show every continuous linear functional on the space attains its norm on the unit sphere of the space; then appeal to James' theorem. (note "attains its maximum on the unit ball" is the same as saying "attains its norm ... 0 It seems the following. Suppose that there exist points x and y in a normed space X such that \|x\|\ge 1, y\in C and$$\|x-y\|<\|x-P_C(x)\|.$$Then Triangle inequality implies$$\|x\|\le \|x-y\|+\|y\|<\|x-P_C(x)\|+1=\|x-P_C(x)\|+\|P_C(x)\| =\|x\|,$$a contradiction. 1 The answer seems to be Yes if the space Y is reflexive. Extracting a subsequence if necessary, we may assume that the sequence (\Vert Tx_n\Vert) has a limit in [0,\infty]. If \Vert Tx_n\Vert\to \infty, the inequality is trivially satisfied. So assume that \Vert Tx_n\Vert\to l<\infty and let us show that \Vert x\Vert\leq l. Since the norm of ... 2 If T: X \to Y is bounded and surjective, the Open Mapping Theorem says there is an isomorphism S:\; X/\ker(T) \mapsto Y such that T = S \circ \pi, where \pi: X/\ker(T) \to Y is the quotient map. The trouble is, a quotient of X might not be isomorphic to a closed subspace of X, so there might not be a bounded right inverse. For example, every ... 1 It is obvious that S= {~}^\perp(S^\perp) implies that S is a closed subspace. For the oposite implication, assume that S is closed and x\not\in S. Let T be the linear subspace spanned by S and x. It is closed and each y\in T is of the form y=s+\alpha x for unique s\in S and number \alpha. Define \xi_0 on T by \xi_0(s+\alpha ... 1 The following is true: If X is a Banach space, then each Hahn-Banach (norm preserving) extension is unique iff the unit ball of X^{\ast} is strictly convex. In particular, this would be true for a Hilbert space. Furthermore, there is an interesting theorem of Phelps (See this) which relates this property for a fixed subspace to a best approximation ... 2 What about the subspace$$S=\{f\in C(0,1); f(0)=f(1)\}$$and the functional T\colon f \mapsto f(0) defined on S. The you have two extensions T_0\colon f\mapsto f(0) and T_1\colon f\mapsto f(1). For these functionals we have \|T\|=\|T_1\|=\|T_2\|=1. In fact, for every a\in[0,1] we have an extension T_a\colon f\mapsto af(0)+(1-a)f(1) with ... 6 Make a change of variables u = e^x$$\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$$for any polynomial P(u). Now pick the polynomial to approximate ... 0 What you want to show is not true. To see this, note that this would imply (by taking g \equiv 0) that |Tf(x)| grows at most linearly in f. But since the integrand is quadratic in f, it is easy to see that this cannot hold (take f \equiv c for large c). 1 Almost there... You have |Sf(x)-Sg(x)| \le \int_0^x t^2 dt \|f-h\|_\infty ={ x^3 \over 3} \|f-h\|_\infty, and so \|Sf-Sg\|_\infty \le { 1\over 3} \|f-h\|_\infty. 2 Let g(x)=\frac12\sin(x). Then f(x)=g(f(1-x)) for all x\in[0,1], so in particular each point f(x) is a period-2 fixpoint of g, and f(1/2) is a period-1 fixpoint of g. But note the following: g(x)\le\frac x2 for all x>0 g(x)>0 for all x\le1 g(x) is odd. So for any x>0, if g(x)>0 then g(g(x))\le x/4, and if ... 1 Since T is continuous, we have |Tu|_Z\leq c|u|_X. Hence$$|u|_X\leq|u|_Y\leq |u|_X(1+c),$$i.e. the norms |\cdot|_X,|\cdot|_Y are equivalent. Thus the completion of X w.r.t. |\cdot|_Y is canonically isomorphic to X. 2 This question have been already answered on MO. For a more detailed discussion of the spectrum of arbitrary L_\infty-space see page 28 in Second duals of measure algebras. H. G. Dales, A. T.-M. Lau 0 Note that T is equal to the composite D\rightarrow\Gamma(T)\rightarrow Y (with \Gamma(T)\rightarrow Y the projection, which is bijective and continuous), and that since T has closed graph \Gamma(T), \Gamma(T) is a Banach space (for it is a closed subspace of the Banach space X\times Y). Then T^{-1} is the composite of continuous maps ... 1 Since your comment to my first answer is tricky to read, let me restate it. The follow-up question is: how does$$ \frac{|v(x)-v(y)|}{|x-y|^\gamma} \le M_\alpha < \infty \qquad independent\ of\ x\neq y \qquad\qquad\qquad(*)$$for every multi-index \alpha of order k imply$$ u \in C^{k,\gamma}(\bar{U})?  The answer is as follows: in order to show $u ... 5 It seems that Nigel Kalton gave an answer to your question. He proved in [1], theorem 3.5, that for every compact metric space$K$,$C(K)$is a$2$-absolute Lipschitz retract. [1]: Kalton, N. J. Extending Lipschitz maps into C(K)-spaces. Israel J. Math. 162 (2007), 275–315. Link to article Added by OP: For completeness, I'll outline the steps of ... 1 The problem does not originate from the weak-* topology, it also is present within the weak topology on Hilbert spaces. On$\ell^2$, consider the set \begin{equation*} B := \{ \sqrt{n} \, e_n : n \in \mathbb N\}, \end{equation*} where$e_n$are the canonical unit sequences. Now, it is possible to show that$0$belongs to the weak closure of$B$, but no ... 1 We have to show the following: given a Cauchy sequence$(u_n)_{n\in \mathbb N}$in$C^{k,\gamma}(\overline U),$there is a$u \in C^{k,\gamma}(\overline U)$such that$\lim_{n\rightarrow\infty}u_n = u$in$C^{k,\gamma}(\overline U).$Note that the Hölder norm is the "sum" of the$C^k$norm (i.e. sup-norm up to the$k$-th derivatives) and the Hölder ... 0 Let$L$not be a maximal chain in the lattice of all subspaces of$X$.$M \notin $Lat K for some M subspace of X, i.e.$Kx \notin M$for some$x \in M$. Suppose that$\{x_n\}$is a sequence in$M$such that tends to$x$in$M$weakly. By Prop. VI.3.3 [A course in Functional Analysis, John B. Conway]$K$is completely cont., so that$\{ K{x_n}\}$converges ... 1 Take the self-adjoint diagonal operator$Te_k=\frac1k e_k$on a Hilbert space with orthonormal base$\{e_k, k\in\mathbb N\}.$The kernel of$T$is$\{0\}$but$T$is not surjective, since e.g.$\sum_k \frac1k e_k$is not in the range. 0 I finally figured it out; both separately... Suppose it holds:$\omega(A^*A)\geq0$Positivity can be proven by Kadison's inequality: ... 0 Disclaimer There was another answer giving the motivation for this one. Sadly, it got deleted. Crucial Point It boils down to the existence of an isomorphism:$\pi[A]=A'$&$\pi[1]=1'$This is ill-defined only for the case:$1\propto A$&$1'!\propto A'$Prework Suppose proportionality holds:$1=\lambda A$Then the original algebra was ... 1 K. Davidson, "C*-Algebras by Example", Fields Institute Monographs, 1996. Update. Added a more complete list. I am also studying C * algebras. A more complete list can be this. O. Bratteli and D. W. Robinson, Operator algebras and quantum statistical mechanics. 1", Texts and Monographs in Physics, Springer-Verlag, 1987. K. R. Davidson,$C^*$-algebras by ... 0 Just if somebody is interested... I found the book by Zhu: An Introduction to Operator Algebras. It just fits quite nicely when trying to get a first touch with the subject. 2 The first inequality is the triangle (Minkowski) inequality for the$L^p$norm, the second inequality is Minkowski inequality for the counting measure, http://en.wikipedia.org/wiki/Minkowski_inequality. 2 It is not too hard to show that if$V$is a complete$\Bbb{Q}$vector space, one can extend the scalar multiplication uniquely continuously to$\Bbb{R}\times V\to V$, so that$V$is also a$\Bbb{R}\$ vector space. Hence, we assume this to begin with. Also, if the vector space is complete, it is natural to assume that the underlying field is complete too. ...

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