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8

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$. Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional. A ...


6

One useful example is the holomorphic functional calculus. It allows us to generalize Cauchy's integral formula from complex analysis in one variable to evaluate functions of operators. Let $V$ be a Banach space and let $T$ be a bounded linear operator on $V$. If $\Gamma$ is a positively oriented rectifiable Jordan curve such that the spectrum of $T$ is ...


6

First, there is a difference between $C([0,1])$ and $C((0,1))$ - the spaces of functions that are continuous on the closed and open interval, respectively. There is no difference between $L^2([0,1])$ and $L^2((0,1))$ though. If you define function spaces on open sets then every point of the set is an interior point, which is convenient if you want to work ...


6

Since $A$ and $B$ are closed subspaces of the Banach space $X$, they are themselves Banach spaces, and hence so is $A\times B$, endowed with the norm $\lVert (a,b)\rVert = \lVert a\rVert_X + \lVert b\rVert_X$. Now consider the map $$T \colon A\times B \to X; \quad T(a,b) = a+b.$$ We have $\lVert T(a,b)\rVert_X \leqslant \lVert (a,b)\rVert$, so $T$ is ...


5

Consider the sets $f(\Bbb I^-)$ and $f(\Bbb I^+)$, where $\Bbb I=\Bbb R\setminus\Bbb Q$. For any $x\in\Bbb I^-, y\in\Bbb I^+$, we have $x<0<y\implies f(x)<f(y)$, so each element of $f(\Bbb I^-)$ is less than each element of $f(\Bbb I^+)$. Now both sets are nonempty, so they partition $\Bbb R$ into two pieces, and there is a point $\alpha$ in the ...


5

It's something with the kernel of the functionals,right? Right. What you want is an $x$ with $f_n(x) \neq 0$ for all $n$, so $$x \in \bigcap_{n\in\mathbb{N}} \left(X\setminus \ker f_n\right) = X \setminus \bigcup_{n\in\mathbb{N}} \ker f_n.$$ The $f_n$ are continuous, so $\ker f_n$ is closed. $f_n \neq 0$, hence $\ker f_n \neq X$, and therefore $(\ker ...


4

First of all, in order to "calculate" something explicitly you need a reasonably nice Banach space $B$ as your target. For instance, suppose your Banach space is $B=C([a,b])$. Then a continuous function $f: [0,T]\to B$ is nothing but a continuos function of two variables $F(x,t)$, $x\in [a,b], t\in [0,T]$: $$ f(t)(x)= F(x,t). $$ Now, computing the Bochner ...


4

In fact, we can derive that $\,\dim S \le c^2$. Let me describe the proof as it is very elegant. Assume that $v_1,\ldots, v_n\in S$ are orthonormal functions, i.e., $\int_0^1 v_iv_j\,dx=\delta_{ij}$, and for a fixed $a=(a_1,\ldots,a_n)\in\mathbb R^n$ define $\varPhi_a :\mathbb R^n\to \mathbb R$ as $$ \varPhi_a(x)=\sum_{j=1}^n a_jv_j(x). $$ Then $$ ...


4

Both conditions imply the finite-dimensionality of $X$. If $\dim X = \kappa$ for an infinite cardinal $\kappa$, you can embed $X$ as a dense subspace into $\ell^p(\kappa) = \left\{f \colon\kappa\to\mathbb{K} : \sum_{k\in\kappa} \lvert f(k)\rvert^p < \infty\right\}$ for $1 \leqslant p < \infty$ by mapping each basis element to the "standard basis" ...


4

Intuition: this multiplier evolves $f$ by the Schrodinger equation by time one (after making Planck's constant one). And this maps the Dirac delta function to something whose absolute value is constant, which is a counterexample for $p=1$. So approximate the Dirac delta function by a narrow bell curve. This will work for $p<2$. For $p>2$, do the ...


4

Here's a way to do it that doesn't explicitly use sequences: Let the base field be $\Bbb F$, where $\Bbb F = \Bbb R \; \text{or} \; \Bbb C$. Start with the definition of $M^\bot$: $M^\bot = \{ x \in X \mid \langle m, x \rangle = 0, \forall m \in M \}; \tag{1}$ next, note that for any $m \in M$, the map $\phi_m:X \to \Bbb F, \; \phi_m(x) = \langle m, x ...


4

The statement is false, as I discovered here. Since not everybody has access to the paper, let me provide a summary of the argument: Let $R=\mathbb C[t]$, $L$ the left shift operator, and view $\ell^p$ as an $R$-module by defining $t\cdot x=Lx$. Let $X=\sum \ker L^i \subset \ell^p$ be the subspace of eventually-zero sequences. Lemma: Given a PID $P$, a ...


3

Yes, you extend $l$ to $\lambda \in C(\mathbb{C}^n)^\ast$. Then, since $\lambda$ is continuous, there is a compact $K \subset \mathbb{C}^n$ with $$\lvert \lambda(f)\rvert \leqslant M\cdot \max \left\{\lvert f(z)\rvert : z \in K\right\},$$ and you use the Riesz representation theorem for continuous linear functionals on $C(K)$ (by Tietze's extension ...


3

Let denote $T$ the two linear functional. Let the sequence $p_n(x)=nx^n$ then $$||p_n||_1=n\int_0^1 t^ndt=\frac n{n+1}<1$$ hence the sequence is bounded but $T(p_n)=n^2p_n$ and we have $$||T(p_n)||_1=\frac{n^2}{n+1}\xrightarrow{n\to\infty}\infty$$ hence $(T(p_n))$ isn't bounded and then $T$ isn't continuous. We have ...


3

The key concept here is that of annihilator. For a subset $X\subset A$, the annihilator of $X$ is the subset of $A^*$ given by $$ X^o=\{f\in A^*: f(x)=0\,\forall x\in X\}. $$ (note that almost all equal signs below mean isomorphism) It is not hard to prove that $X^{oo}=X^{**}$. Now if $A=X\oplus Y$, then $A^*=X^*\oplus Y^*$. Also $X=A/Y$, $Y=A/X$. And it is ...


3

Yes. You can factorise the identity map on $A\oplus B$ in the point-ultraweak topology taking direct sums of the respective approximations for $A$ and $B$. Then you pretend that $M_{k_A(n)}$ and $M_{k_B(n)}$ live in some bigger full matrix algebra and you are done. In other words, you can replace the matrix algebras $M_{k(n)}$ in the definition of ...


3

Let $\Omega$ be the disk of radius $r<1$ centered at the origin. Define $u(x,y) = (-\log(x^2+y^2))^{1/3}$. Clearly, this is not a continuous function. But on every line not passing through the origin, this function is Lipschitz, therefore absolutely continuous. It remains to check that its partial derivatives are in $L^2$. By the chain rule, ...


3

Consider $$f(x)=\frac{x(x^2-x+2)}{x^2-9}$$ as $x \to \pm \infty$. Note the $x^2$ terms from the numerator and denominator cancel out, so you are left with $$ f(x) = \frac{x(x^2-x+2)}{x^2-9} \approx x \text{ for very large } x. $$ As a result, $$\lim_{x \to \pm \infty} f(x) = \pm \infty$$ and it is unbounded on both ends.


3

From the inclusions of sets $$\ell_p\subset c_0\subset \ell_\infty,$$ where $c_0$ denotes the set of convergent sequences and the separability of $(c_0,\lVert \cdot\rVert_\infty)$ is separable we conclude that $(\ell_p,\lVert\cdot\rVert_\infty)$ is separable. $\ell_p$ endowed with the supremum norm is not a closed subspace of $\ell_\infty$ because its ...


3

Let us assume that the question requires $E$ to be a locally compact Hausdorff space. It is well-known that $C(E)$ with the supremum norm is a C$^*$-algebra. Sakai proved in 1971 that a C$^*$-algebra is a dual precisely when it is a von Neumann algebra. In terms of $E$, this means that it has to be compact and extremely disconnected: the closure of every ...


3

For any $f \in C([0,1])$ you have $Tf(x) = f(1-x)$ so that $$\|f\| = \int_0^1 x^2 f(x) \, dx \quad \text{and} \quad \|Tf\| = \int_0^1 x^2 |f(1-x)|\, dx.$$ If $f$ is concentrated near $0$, the first norm will be small and the second norm will be large. One way to look for counterexamples is to temporarily disregard continuity: for instance you could take ...


3

The weak topology is weaker than the norm topology. This implies that every weakly continuous map (needn't even be linear) is norm-continuous. For the other implication: The weak topology is (more or less by definition) the weakest topology making all norm-continuous linear functionals continuous. Therefore, the norm-continuous functionals are weakly ...


3

it is a topological one. In that case, $X$ must be a closed subspace. That is, because an isomorphism of topological vector spaces is not only a topological isomorphism, but also an isomorphism in the category of uniform spaces (with uniformly continuous maps as morphisms), so if $X$ (with the subspace topology induced by $L^p$) is isomorphic to $l^p$ ...


3

Define $$T_\varepsilon(\varphi):=\int_{-\infty}^{+\infty}\frac{\varepsilon}{\varepsilon^2+x^2}\varphi(x)\mathrm dx.$$ Then, after the substitution $x=\varepsilon t$, we obtain $$T_\varepsilon(\varphi)=\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\varphi(t\varepsilon)\mathrm dt.$$ Using dominated convergence, we obtain that for each $\varphi\in\mathcal D(\mathbf ...


3

Suppose that $\overline{aA_+a}\subsetneq A_+$. Use Hanh-Banach to construct a functional with $f(b)=1$ and $f(aca)=0$ for all $c\in A_+$. This functional, extended to all of $A$, is a linear combination of four states (see II.6.3.4 in Blackadar): at least one of them, say $f_1$, will satisfy that $f_1(b)>0$; by the Jordan Decomposition, we also have that ...


3

Suppose that $f=\{ f_{0},f_{1},f_{2},\cdots\} \in l^{2}$ is orthogonal to every such $f_{p}$. Let $$ F(z)=\sum_{n=0}^{\infty}f_{n}z^{n},\;\;\; |z| < 1. $$ $F$ converges absolutely in $|z| < 1$ because the Cauchy-Schwartz inequality implies $$ \left[\sum_{n=0}^{\infty}|f_{n}||z|^{n}\right]^{2} \le ...


3

We know that continuous functions are dense in $L^p([a,b])$ (where $-\infty<a<b<\infty$) and also that polynomials are dense in $C([a,b])$ and so we see that polynomials are dense in $L^p([a,b])$ by standard arguments. Thus we can see that your statement is effectively equivalent to asking if a function is orthogonal to everything in $L^2([a,b])$. ...


3

As mentioned by yoyo, the sum of the coefficients of $p$ is precisely $p(1)$. If $1 \in [a,b]$, show directly that the linear functional $p \mapsto p(1)$ is continuous. If $1 \notin [a,b]$, construct a polynomial with $p(1)=1$ but with $p$ uniformly close to 0 on $[a,b]$. The Weierstrass approximation theorem may be helpful.


3

We do not need to consider $\Omega(A)\cup \{0\}$, but it makes things easier, since that is a compact (Hausdorff, if that isn't part of your definition of compact) space. Every open subspace of a compact (Hausdorff) space is locally compact, and since the weak* topology is Hausdorff, singletons are closed, whence $$\Omega(A) = (\Omega(A)\cup \{0\}) ...



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