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Hot answers tagged banach-spaces

3

It's not true. Let $X$ be the space of sequences that are eventually $0$ with the sup norm. Let $X_n$ be the subspace of $X$ consisting of the sequences $x$ with $x(i)=0$ if $i\ge n$. Consider the sequence $(x_n)$ with $x_n=(1,1/2,1/3,\ldots,1/n,0,0,\ldots)$.

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This thesis contains a lot of different proofs of Hahn-Banach theorem and much more.

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The statement you say you have been able to prove is correct. In fact, it's very close to a characterization of Banach spaces. A normed linear space $X$ is Banach if and only if every absolutely summable series in convergent, i.e. if every series $\sum_{n=1}^\infty a_n$, $a_n\in X$ with $$\sum_{n=1}^\infty \Vert a_n \Vert_X < \infty$$ converges in $X$. ...

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Hints: Let $A - B = \{a-b \;|\; a \in A, b \in B\}$. Show that $A-B$ is closed, convex, and $0 \notin A - B$. Since $A-B$ is closed, there is a small $\varepsilon > 0$ so that $T := \{x\in X \;|\; \|x\| < \varepsilon\}$ and $A-B$ are disjoint. Separate them with a functional. $T$ is open, so the functional is continuous. And because of the ...

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and the latter term vanishes if the dual topology was the strong topology and the sequence $x_n$ was bounded in norm. That is the crucial thing. The sequence needs to be bounded to deduce weak convergence from pointwise convergence on a norm-dense subset. If $(x_n)$ is a bounded sequence in $X$, it is an equicontinuous sequence as a sequence of ...

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This is a corollary of the previous statement in the book: the polarization identity says that for a sesquilinear form $\sigma$, you can write $$4\sigma(x,y) = \sigma(x+y,x+y)+i\sigma(x+iy,x+iy) - \sigma(x-y,x-y) -i\sigma(x-iy,x-iy).$$ In particular, this shows that given two sesquilinear forms $\sigma, \sigma'$, if $\sigma(v,v) = \sigma'(v,v)$ for all $v$, ...

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It appears you saying that you want to assume that $1=1'$. Also, presumably $A$ and $A'$ sit together within a larger algebra (perhaps you want to think of them as subalgebras of the same $B(H)$?). Let $B$ be the $C^*$-algebra generated by $A$ and $A'$. Then the identity $1$ for $A$ and $A'$ is also the identity for $B$. Thus you know that the spectra ...

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Note that by linearity \eqalign{ \frac{h(\theta+\delta)-h(\theta)}{\delta}&=\frac{1}{\delta}\left(\xi(R(\theta+\delta))- \xi(R(\theta))\right)\cr &=\frac{\xi(R(\theta+\delta)-R(\theta))}{\delta}\cr &=\xi\left(\frac{R(\theta+\delta)-R(\theta)}{\delta}\right)\cr} Taking the limit as $\delta\to0$, we get $$h'(\theta)=\xi(R'(\theta))\tag{1'}$$ ...

2

Notice that $\|u(1-u^*u)\xi'\|^2=\|u^*u(1-u^*u)\xi'\|^2=\|(u^*u-u^*u)\xi'\|^2=0$, for every $\xi'$. Thus, $u(1-u^*u)=0$.

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A thought to get started: write a power series for $e^I$. See anything simple to do? If you don't like the idea of just jumping to a full series, start with a partial sum and see what comes out.

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A Banach space is a complete metric space with a norm, while a Hilbert space is a complete metric space with an inner product. An inner product induces a norm by $||x|| = \sqrt{<x,x>}$, which means that any Hilbert space is also a Banach space. However, the converse need not hold as the norm isn't always expressable in terms of the inner product. An ...

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The proof that $T$ is an isometry typically consists of two parts: Show that $\|Tx\|\le \|x\|$ for all $x$ Show that $\|Tx\|\ge \|x\|$ for all $x$ For the classical function and sequence spaces, 1) may involve Hölder's inequality, sometimes in its trivial $(1,\infty)$-form $\left|\sum a_nb_n\right|\le \sup_n|b_n| \sum_n |a_n|$. Step 2) then ...

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For each $x\notin \overline{M}$, as a consequence of Hahn-Banach Theorem, there is $\phi_x \in X^*$, $\|\phi_x \|=1$, $\phi_{x}|_{M}=0$, $\phi_x(x)=d(x,\overline{M})$. If $y\in \bigcap\{\ker(\phi): \phi|_M=0\} \subset \bigcap\{\ker(\phi_x)\}$, then $\phi_x(y) = 0 \quad \forall x \notin \overline{M}$. Thus $y\in \overline{M}.$

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WARNING. This is not a counterexample as it works only in real Hilbert spaces. See comments. The statement is false if $u$ and $v$ are not assumed to be symmetric. Consider the Hilbert space $\mathbb{R}^2$. The operators $$u\mathbf{x}=(-x_2,x_1)$$ and $$v\mathbf{x}=(-2x_2,2x_1)$$ are such that $$(u\mathbf{x}, \mathbf{x})=(v\mathbf{x}, ... 1 That's the easy part. The topology you need to consider on \Phi_{C(K)} is the weak* topology; that is, pointwise convergence. So, if x_j\to x in K, you want to show that \delta_{x_j}\to\delta_x. This means that \delta_{x_j}(f)\to\delta_x(f ) for all f\in C(K). But this is f(x_j)\to f(x), which is precisely the continuity of f. Conversely, ... 1 Proof Consider a bounded function \|F\|_\infty<\infty. Now, by measurability there exists a simple approximation:$$S_n\in\mathcal{S}:\quad\|F-S_n\|\to0$$After cutoff and reset it can be chosen to be bounded and decreasing:$$\|S_n\|\leq2\|F\|:\quad\|F-S_n\|\downarrow0$$Next, denote sufficiently close points by: ... 1 Yes, it holds! As it is continuous it is Bochner measurable by Pettis's criterion. As it is absolutely integrable it is also Bochner integrable. But it is bounded so on subspaces of finite measure Riemann integrable. Thus by dominated convergence also improperly Riemann integrable. 1 If the spectral radius of T is less than 1, then the unique solution of (I-T)f = 1 is$$ f = 1 + T1 + T^{2}1+T^{3}1+\cdots . $$So suppose that T1=0. Then the solution is f=1 making it unlikely that \|(I-T)^{-1}\|_{\infty}=\|1\|_{\infty}=1. For example, consider X=C[-1,1] and$$ Tf = \frac{1}{2}\int_{-1}^{1}xf(x)\,dx. $$... 1 Ok, I think I got it now... Take a slight variation of the famous example:$$F:[0,1]\to\ell[0,1]:t\mapsto\chi_t$$At least, it is bounded: \|F(t)\|\equiv1. Especially, it is absolutely integrable: \int\|F(t)\|\mathrm{d}t=1 However, it is not measurable as:$$\|\chi_s-\chi_t\|^2=2\quad(s\neq t)$$so taking a Vitali set yields:$$U:=\bigcup_{v\in ...

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The process is good. Maybe some wording could be added. For example, the first displayed inequality only holds for $m$ and $n$ large enough. The existence of the limit of the sequence $(x_n)$ has to be justified (by completeness of a Hilbert space).

1

Let $C$ be the C*-algebra generated by $\pi(A)$. Then $\pi(1)$ is a multiplicative identity for $C$, because it is one for $\pi(A)$ and multiplication is continuous. Now you can apply the proof you know to $\pi:A\to C$.

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For a subsets of topological vector space $X$ the following theorem hols true. Theorem: If $A,B,C\subset X$ and $B$ is bounded $C$ is closed and convex and $A+B\subset C+B,$ then $A\subset C$ e have $B_Y \subset T(MB_X )+ \delta B_Y ,$ hence $$\delta B_Y +(1-\delta ) B_Y \subset \overline{T(MB_X )}+ \delta B_Y ,$$ and from the theorem we obtain that ...

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And again a variant of the famous example: $$F:(0,1]\to\mathcal{H}:F(\frac{1}{n+1}<t\leq\frac{1}{n}):=e_n$$ Clearly, it is pointwise limit but can't be uniform limit.

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$z_n=(x_n,y_n)$ is a cauchy (convergent) sequence in $Z$ iff $(x_n)$ and $(y_n)$ both are cauchy (convergent).

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From comments, it sounds like by "Euclidean space" you mean "Hilbert space". Let's call it $H$. It is certainly true that for any bounded linear operator $A : H \to H$, the adjoint map exists. It's really the same as for Banach spaces: define the linear operator $A^*$ on $H^*$ via $A^* f = f \circ A$, and then use the Riesz representation theorem to ...

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Let $A=B=\Bbb R$ and $T(x)=2x$.

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There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation $$x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)}$$ Here is how we solve it: consider all the sequences (not nesessarily from $l^2$) which satisfy (*). They form a linear space (it's easy to ...

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The second statement is essentially proven few pages later, p. 487 in your link, since in the case of functions defined on the real line Gâteaux derivative coincides with Fréchet. The first one follows from the estimate (which I assume is the same you used): $$... 1 Without loss of generality we may assume that the set contains \overline{\text{conv}}\{x_n:n\in\mathbb N \} a closed unit ball B. Since the set \overline{\text{conv}}\{x_n:n\in\mathbb N \} is weakly compact thus the unit ball B is weakly compact as weakly closed subset of weakly compact set. Thus X is reflexive and every weakly compact subset of ... 1 It even suffices if just one power A^k is a contraction. For$$ \sum_n \Vert A^n\Vert = \sum_\ell \sum_{m=0}^{k-1} \Vert A^{\ell k + m}\rVert\leq \sum_{m=0}^{k-1} \sum_\ell \Vert A \Vert^m \Vert A^k\Vert^\ell,  which is finite. Now use a Neumann series argument (c.f. http://en.m.wikipedia.org/wiki/Neumann_series). Ok, to make myself more clear, the ...

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