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Take an infinite-dimensional hilbert space with an orthonormal basis $\{e_j\}$, I'm sure you know that $e_n \xrightarrow{w} 0$... but, obviously, $\|e_n\| =1$ Check your Hahn-Banach corollary again!

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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 $C^0$ bound. Similarly, $$|f(x) - f(y)| \le \sqrt{|x-y|} \|f'\|_2 \le \sqrt{|x-y|}D.$$ Thus the family $\{f_n\}$ is ...

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From what I recall, every $C^{\ast}$-seminorm arises from a $\ast$-homomorphism. ie. $\exists$ a $\ast$-homomorphism $\varphi : A\to B$ between $C^{\ast}$-algebras such that $$p(a) = \|\varphi(a)\|$$ Your result follows from this because a $\ast$-homomorphism is norm-decreasing. To prove the result I mention, the argument (I think) is as follows: Take $N ... 3 The operator$2 \, I$is always self-adjoint. Hence,$2 \, I = I$or$2 \, I = 0$. This yields$I = 0$, hence,$\mathcal{H} = \{0\}$. Edit: Since every orthogonal projection onto a subspace is self-adjoint, it is quite easy to reconstruct$\mathcal{H}$from its self-adjoint operators. 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} $As ... 3 Let$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, $$|\langle ... 3 Let c: [0,1] \to [0,1] be the usual Cantor-Lebesgue function and let g: [0,1] \to [0,2] be given by g(x) = c(x) + x. Then g is continuous, increasing, and g'(x) = 1 almost everywhere. Since$$g(1) - g(0) > \int_0^1 g'(x) \, dx$$it follows that g is not absolutely continuous. Let f = g^{-1}(t) so that f : [0,2] \to [0,1] is continuous and ... 3 Take a sequence \{f_n\} in X such that f_n\to g and P(f_n)\to h. Then we have that f_n-P(f_n)\to g-h. Note that f_n-P(f_n)\in \ker P since P(f_n-P(f_n))=P(f_n)-P^2(f_n)=P(f_n)-P(f_n)=0. Since \mathrm{Ran}(P) and \ker(P) are closed, and P(f_n)\in\mathrm{Ran}(P), we must have that h\in\mathrm{Ran}(P) and g-h\in\ker P. But then we must ... 3 Notice that$$ \int_{-1}^{1} |f_j - 1|^2 = \| f_j \|_2^2 - 2\|f_j\|_1 + 2 = \mathcal{O}(2^{-j}). $$By the monotone convergence theorem, we have$$ \int_{-1}^{1} \sum_{j=1}^{\infty} |f_j - 1|^2 = \sum_{j=1}^{\infty} \int_{-1}^{1} |f_j - 1|^2 < \infty. $$Therefore \sum_{j=1}^{\infty} |f_j - 1|^2 is finite a.e. and hence f_j \to 1 a.e. 3 Suppose T is a bounded operator on a Banach space X. \lambda\in\rho(T) iff T-\lambda I is a linear bijection. In that case, the inverse (T-\lambda I)^{-1} is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of T-\lambda I being invertible. T-\lambda I is not injective. Equivalently, ... 2 If ||I-P||<1 \implies P is invertible \implies P=I 2 Part 2 is easier to answer: no, the way monotone operators are defined in functional analysis, -F is not in general monotone when F is. (This is unlike the concept of monotonicity in real analysis). Monotone operators correspond to (non-strictly) increasing functions. The reverse inequality defines dissipative operators. Part 1, geometric ... 2 By Sobolev embeeding, from \sqrt\rho\in H^1(\mathbb R^3) it follows that$$ \sqrt \rho \in L^6(\mathbb R^3), $$which implies$$ \rho \in L^3(\mathbb R^3). $$2 As you have defined:$$ 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) $$The linear operator p_n is an orthogonal projection operator onto the linear span of the orthonormal set \left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}. Hence, p_n^2=p_n and \|p_nf\| \le ... 2 Consider the restriction of the quotient map E \to E/A to B. This is an injective map of finite-dimensional spaces, so it has a bounded inverse. 2 No because in a Banach space weak topology has the same bounded sets as the topology induced by the norm. This relies on the Banach–Steinhaus theorem. Certainly weakly convergent sequences are bounded in the weak topology. Let X be a Banach space. Suppose that A\subset X is bounded in the weak topology. Then A regarded as a subset of X^{**} is ... 2 For any \varphi \in c_0^{\ast}, \exists y\in \ell^1 such that$$ \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)) $$Hence, T^{\ast}(\varphi) = \psi where \psi is as ... 2 Let g be the restriction of \Phi to the boundary of the square. Your boundary data determines g up to a constant. You can then take this g and solve the corresponding Dirichlet problem; standard theory tells that solutions exist uniquely. Therefore you can conclude that solutions are unique up to shifting by constants. (The fact that shifting a ... 2 It is always useful to have more than one answer to a problem. One learns quite a bit more. Even better, however, is to have two contradictory answers to the same problem. That is both more entertaining and more educational. (The contradiction is just about what one wants to prove, not about errors.) A friend of mine often tells an anecdote about a ... 2 The only point where the measures itself intervenes is in the definition of strong measurability, through simple convergence \mu-almost everywhere. A \sigma-finite measure is equivalent (in the sense has the same negligible sets) to a finite measure: partition \Omega into countably many sets of finite \mu-measures, and define d\nu=\rho\ d\mu where ... 2 Finally I find the solution. Proof of surjection could be done by the following 2 steps: Range of A, denoted by R(A), is a closed set. Say \{y_n\} is a converged set in R(A), and \{y_n\}\to y. We know R(A) is closed iff y\in R(A) by properties of closed sets. By definition of R(A), there exists a series \{x_n\}\in H which ... 2 Some important tricks/theorems: The spectrum is non-empty for Banach algebras (over \mathbb{C}) The spectral radius formula$$ r(a) = \lim \|a^n\|^{1/n} $$tells you that if a is nilpotent, then \sigma(a) = \{0\} If A is commutative, then$$ \sigma(a) = \{\tau(a) : \tau \in \Omega(A)\} $$where \Omega(A) denotes the set of non-zero multiplicative ... 2 The sequence (f_n) does not have a weak-* convergent subsequence. To see this, suppose f_{n_j} is a subsequence converging weak-* to some f \in (\ell^\infty)^* as j \to \infty. Thus$$x_{n_j} = f_{n_j}(x) \to f(x)$$as j \to \infty for every x \in X = \ell^\infty. Define x \in \ell^\infty by x_{n_j} = (-1)^j and x_n = 0 if n is not equal ... 2 Even more is true. Every copy of c_0 in C(K) for K compact, metric is complemented by a projection of norm at most 2. Indeed, C(K) is in this case separable (as K is second-countable we may use the Stone–Weierstrass theorem to get the claim) and then we may apply Sobczyk's theorem. 2 Yes, T_n converges strongly to zero by the Riemann-Lebesgue lemma: For any x\in H,$$ \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) - ... 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 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 ... 2 Actually$f$doesn't need to be$C^1$, or even continuous a priori. For every$x\in [0,1]$pick an open$U_x\subset [0,1]$such that$f$is constant on$U_x$. Then the family$(U_x)_{x\in [0,1]}$covers$[0,1]$. Since$[0,1]$is compact, there is a finite subcovering, say by$U_{x_1}, \dots, U_{x_n}$.$U_{x_1}$must intersect at least one of ... 2 Let$\varphi \in \mathcal D$, where$\mathcal D$is the space of$\varphi \in C^\infty_0(\mathbb R)$with the usual test function continuity notion. Consider $$\tilde \varphi(\xi) = \int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}e^{-i\xi x}\varphi(x)=\int_{\mathbb R} \frac{dx}{(2\pi)^{1/2}}\sum_{n=0}^\infty \frac{(-i\xi x)^n}{n!}\varphi(x);$$ since the integral ... 2 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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