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0

Hint: Show that the kernel of $T^*$ is the annihilator of $\operatorname{Im} T$. Show that the only closed subspace with $0$ annihilator is the whole space.


0

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 ...


0

The equality implies that if $\lvert v\rvert\xi_1 = \lvert v\rvert\xi_2$, then also $v\xi_1 = v\xi_2$, so it is immaterial which element $\xi\in \lvert v\rvert^{-1}(\eta)$ is chosen to define $u(\eta) = v(\xi)$. All choices yield the same result. Thus $u = v\circ \lvert v\rvert^{-1}$ is well-defined although in general $\lvert v\rvert^{-1}$ is not a map ...


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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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}, ...


2

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 $v$, ...


1

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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A good introduction can be found on the book Differential Equations in Abstract Spaces of G. E. LADAS and V . LAKSHMIKANTHAM.


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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): $$ ...


2

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 $$h'(\theta)=\xi(R'(\theta))\tag{$1'$}$$ ...


1

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


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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, ...


0

Hahaaa, I got it thanks to my supervisor. =D All formal calculations... Consider the abstract cauchy problem: $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\tau^t\tau_0^{-t}\right)(Z)=\tau^t(\delta-\delta_0)\tau_0^{-t}(Z)=\tau^t\imath\left[V,\tau_0^{-t}(Z)\right]=\left(\tau^t\tau_0^{-t}\right)\imath\left[\tau_0^{t}(V),Z\right]$$ so one gets the recursion relation: ...


1

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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See Corollary 6.9(1), p242 in Invitation To Operator Theory (here).


2

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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Ok, I think I got it now... Both work perfectly fine as they are always nondegenerate: $$\mathcal{A}_\text{CAR}:\quad a(f\neq0)\neq0$$ $$\mathcal{W}:\quad W(f)\neq0$$ A counterexample is provided by the angular momentum algebra: $$\mathcal{J}:\quad [J_i,J_j]=\imath\varepsilon_{ijk}J_k$$ There one has one trivial representation: $J_x=J_y=J_z=0$


0

What are the extreme points of $D = \{z \in \mathbb{C} : |z| \leq 1\}$? If $f \in C([0,1],\mathbb{C})$ has $f(x)$ an extreme point in $D$, and $f(x) = tg(x) + (1-t)h(x)$, what can we say about $g(x)$ and $h(x)$? Now what if $f(x)$ is an extreme point in $D$ for every $x \in [0,1]$?


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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 ...


1

Easy finite-dimensional counterexample: $X = \mathbb{R}$, and $X_n = [\frac{1}{n},1]$. Then each $X_n$ is complete, and $$ \bigcup_{n=1}^\infty X_n = (0,1] $$ which is quite clearly not complete. If you're looking for normed linear subspaces, consider $X = \mathbb{R}^2 = \mathbb{C}$, and $X_n = e^{i\pi/n}\mathbb{R}$, the real axis rotated by $\pi/n$. ...


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 $x_n=(1,1/2,1/3,\ldots,1/n,0,0,\ldots)$.


2

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.


1

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


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No, $\|Tx\| \ge C \|x\|$ is not equivalent. The range of a bounded linear operator $T: X \to Y$ is closed iff there is $C>0$ such that $\|Tx\| \ge C \|x + \text{ker}(T)\|$, i.e. that the operator $\widetilde{T}: X / \text{ker}(T) \to Y$ given by $\widetilde{T} \circ \pi = T$ is an isomorphism onto its range (where $\pi$ is the quotient map of $X$ to ...


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Aah, sometimes things are so simple. :) Suppose it vanishes: $a(f_0)=0$ Then one has by the CAR relations: $$0=\{a(f),a(f)^*\}=\|f\|^2\neq0$$ That is a contradiction!


2

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 ...


0

If $\sum_{j\in J} ||x_j||<\infty$, then $x_j=0$ for all but countably many $j$, so we can suppose $J=\mathbb{N}$. Then for any $\epsilon,$ let $N$ be such that $\sum_{N}^\infty||x_j||<\epsilon$. Then for every $k$, $||\sum_N^{N+k} x_j||<\epsilon$ by the triangle inequality, which shows the sequence of partial sums of the $x_j$ is Cauchy and thus ...


0

Consider the sequence $x_n = e_n/n$ in $\ell^\infty$, where $e_n$ is the $n$'th standard unit vector ($e_n(n) = 1$, $e_n(i) = 0$ otherwise). This is summable with sum $x$ where $x(j) = 1/j$ for all $j$, but $\sum_n \|x_n\| = \infty$.


1

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 ...


2

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 ...


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 ...


0

Okay, here's an attempt at an answer. First I don't think the $L^2(\partial \Omega)$ is the appropriate norm to use here, since the space $H^{1/2}(\partial \Omega)=: \text{Range}(Tr)$ is dense in $L^2(\partial \Omega)$, so continuity in this latter norm would imply a bounded extension to $L^2$, which is not the case. Instead you want to use the norm of ...


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 ...


0

I think what is used here is the following version of the Schauder theorem: Let $V$ be a separable reflexive Banach space. Let $K \subset V$ be nonempty, convex, bounded, and compact for the weak topology. Let $S : K \to K$ be weak-to-weak continuous. Then $S$ has a fixed point. I couldn't find a reference for this. However, it seems to me that the usual ...


2

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


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

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.


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Bochner Since it is separable valued: $$\varphi\in\mathcal{C}(\mathbb{R},E):\quad\mathbb{R}\text{ separable}\implies (\alpha\varphi)(\mathbb{R})\text{ separable}$$ and weakly measurable: $$l\in E':\quad(\alpha\varphi)\text{ continuous}\implies l\circ(\alpha\varphi)\text{ measurable}$$ so by Pettis' criterion strongly measurable: $$\varphi\text{ Bochner ...


0

Consider the Banach space $\ell^\infty(\mathbb{N})$. Define the group on the basis: $$T(t)\chi_{\{n\}}:=e^{it/n!}\chi_{\{n\}}$$ and extended by linearity and continuity. Then it is a group: $$T(s+t)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)=\sum_{k\in\mathbb{N}}e^{i(s+t)/k!}a_k\chi_{\{k\}}=T(t)T(s)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)$$ ...


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

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.


1

The answer to the general question is no. It is a result of Banach that a basis for the weak topology is also a basis for the norm topology. Now take a reflexive space $X$ without the approximation property (for instance, any subspace of $\ell_p$ where $p\neq 1,2,\infty$ which lacks the approximation property). The ball $B$ of $X$ is weakly* (=weakly) ...


-1

One can choose an appropriate sequence: $$\|F-S_n\|\downarrow0$$ This way one obtains a dominant: $$\|F-S_n\|\leq\|F-S_0\|:\quad\int\|F-S_0\|\mathrm{d}\mu\leq\int\|F\|\mathrm{d}\mu+\int\|S_0\|\mathrm{d}\mu<\infty+\infty$$ Thus by dominated convergence and linearity: $$\int F\mathrm{d}\mu-\int S_n\mathrm{d}\mu=\int(F-S_n)\mathrm{d}\mu\to0$$ (For more ...


0

Fix $x\in X$. If $f\in X'$ then $|f(x)|=|[(T')^{-1}T'f](x)|=|(T')^{-1}f(Tx)|\leq\|(T')^{-1}\|\|f\|\|Tx\|$. Hence $$\|Tx\|\geq\frac{|f(x)|}{\|f\|}\cdot\frac1{\|(T')^{-1}\|}.$$ By a corollary to the Hahn-Banach theorem, there exists $f\in X'$ with $\|f\|=1$ and $f(x)=\|x\|$, completing the proof. This method manages to avoid any spectral theory.


0

Since $T'$ is invertible, $\operatorname{spec}(T')=\operatorname{spec}(T)$ is bounded away from zero, and we can define: $$\lambda^- = \min_{\lambda\in\operatorname{spec}(T)}|\lambda|.$$ Obviously we have: $$ \| T x\| \geq \lambda^{-}\|x\|,$$ hence it is sufficient to check that: $$ \lambda^{-} = \frac{1}{\|T'^{-1}\|} $$ that follows from: $$ \|T^{-1}\| = ...


0

First of all, check integrability: $$\mathcal{P},\mathcal{P}'\geq\mathcal{P}_{N(\varepsilon)}:\quad\|\sum_{\mathcal{P}}F(a)\mu(A)-\sum_{\mathcal{P}'}F(a')\mu(A')\|\\\leq\|F-F_N\|_\infty\mu(\Omega) +\|\sum_{\mathcal{P}}F_N(a)\mu(A)-\sum_{\mathcal{P}'}F_N(a')\mu(A')\|+\|F-F_N\|_\infty\mu(\Omega)<3\left(\frac{\varepsilon}{3}\right)$$ Next, check ...


0

The Riemann integral for vector functions with values in a Banach space $X$ is essentially the same as for scalar functions. $$ \int_{a}^{b} F(t)\,dt = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}} F(t_{j}^{\star})\Delta_{j}t, $$ where $\mathcal{P}$ is a partition with partition points $$ a = t_{0} < t_{1} ...


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 ...


1

The nub of it is that total boundedness is an intrinsic property of a metric [more generally, uniform] space. It does not depend on whether the space is a subspace of some larger space, and if so, what that larger space is, all that matters is the metric. We have two metric spaces, $X = A^\ast(S^\ast)$, with the metric induced by the norm on $E^\ast$, and ...


1

You can use the Rayleigh quotient (see here: http://en.wikipedia.org/wiki/Rayleigh_quotient). This works as follows. Let $A$ be a real symmetric matrix with minimal eigenvalue $\mu_\min$ and maximal eigenvalue $\mu_\max$. Then we have $$ \mu_\min = \min_{v^Tv=1}v^TAv $$ $$ \mu_\max = \max_{v^Tv=1}v^TAv $$ This is easy to see if we transform $A$ into ...



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