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7

Let $X$ be a normed space. You can show that if the weak topology of $X$ admits a countable base of open sets at $0$, then $X$ is finite dimensional: Prove the existence of a countable set $\{\zeta_n\}$ in $X^*$ such that every $\zeta \in X^*$ is a finite linear combination of the $\zeta_n$. Derive from this that $X^*$ is finite dimensional. Deduce that ...

5

1) This is trivially true if $x=0$ or $a=0$. So you want $x\neq 0$ and $a\neq 0$. So what you are asking is equivalent to $$\exists?\; x\neq 0, y\neq 0\qquad\Big\| \frac{x+y}{2}\Big\|=\Big\| \frac{x-y}{2}\Big\|=\frac{\|x\|+\|y\|}{2}.$$ Example: take $x=(1,1)$ and $y=(1,-1)$ in $\mathbb{R}^2$ with the $\ell^\infty$ norm $\|(x_1,x_2)\|=\max\{|x_1|,|x_2|\}$. ...

5

For me, the simplest example would be $A=M_2(\mathbb{C})$ equipped with the operator norm induced by any $\ell^p$ norm ($1\leq p\leq +\infty$, $p\neq 2$), on $\mathbb{C}^2$. In short: $A$ is naturally a $*$-algebra and an isometric embedding into $B(H)$ for the latter norms is necessarily a $*$-homomorphism, turning $A$ into a $C^*$-algebra. And a ...

4

Any point $x\in X$ gives rise to a maximal ideal $I_x=\{f\in C(X):f(x)=0\}$. Then $F^{-1}(I_x)$ is a maximal ideal of $C(Y)$, so, thanks to Gelfand-Naimark (and compactness and Hausdorffness), there is a unique $y\in Y$ such that $F^{-1}(I_x)=\{g\in C(Y):g(y)=0\}$. Define $f(x)$ to be $y$. There's a lot that needs to be checked --- continuity of $f$ and ...

4

$H$ is a Hilbert space, so is also a vector space (over $\mathbb{C}$ or $\mathbb{R}$). In addition, $H$ has an inner product $< , >$. Pick a point $h \in H$. Consider the map $t \mapsto th$, where $t \in \mathbb{R}$ and $h \in H$. $\lVert th - sh \rVert^2 =\ <th-sh, th-sh>$, since the Hilbert space norm is defined using its inner product. Then, ...

4

Every topological space can be characterized in terms of its convergent nets. In an arbitrary topological space, a point $x$ is in the closure of a subset $A$ if and only if there is a net in $A$ converging to $x$. First countable spaces, including metric spaces, have topologies that are determined by their convergent sequences. In an arbitrary first ...

3

This remark seems to be buggy. Take any sequence $u_n$ which converges towards $u$ weakly in $W^{1,p}(\Omega)$. By Rellich's theorem, the convergence is strong in $L^p(\Omega)$. By the principle of uniform boundedness, we know that $u_n$ is bounded in $W^{1,p}(\Omega)$, hence, $\nabla u_n$ is bounded in $L^p(\Omega)^N$. Now, the remark implies that $u_n$ ...

2

The two definitions are equivalent. The key point is that every neighbourhood of $0$ contains a neighbourhood $V$ of $0$ which is also $balanced$, i.e. $\alpha V\subset V$ for every scalar $\alpha$ with $\vert\alpha\vert\leq 1$ (see e.g. chapter 1 of Rudin's Functional Analysis). It follows that in any of the two definitions, one can restrict oneself to ...

2

By construction, the sequence $\varphi(y_n) + 2^{1-n}$ is non-increasing. It is also bounded from below (since $\varphi$ is bounded from below by assumption). It follows that $\varphi(y_n) + 2^{1-n}$ converges to some value $c\in \mathbb R$. Now $2^{1-n}$ converges to zero, thus $$c = \lim_{n\to \infty} \left(\varphi(y_n)+ 2^{1-n}\right) = \lim_{n\to \infty} ... 2 Following up on Martin's comment, if you know Riesz's lemma, you can use it, supposing that X is infinite-dimensional, to inductively create a sequence \{x_n\} of unit vectors with the property that x_n has distance at least \tfrac{1}{2} to \mathrm{span} \{x_1,\dots,x_{n-1}\} and thus show that no subsequence of \{x_n\} is Cauchy. See this blog ... 2 (I will use \omega for the ultrafilter since using a lowercase letter will improve readability) The way I see it, your algebra c_\omega is simply$$ c_\omega=\{f:\ f(\omega)=0\}\subset C(\beta \mathbb N). $$So you can make the identification c_\omega=C_0(\beta\mathbb N\setminus\{\omega\}). Note that c_\omega is an ideal in C(\beta\mathbb N), ... 2 For an inner product space the answer is no. Indeed:$$\|x+a\|^2 = (x+a,x+a) = \|x\|^2 + \|a\|^2 + 2(x,a)\stackrel{!}{=} (\|x\| + \|a\|)^2\Leftrightarrow 2(x,a) = 2\|x\|\cdot\|a\|$$and this is true only if x and a are collinear. Now, for x-a we find similarly:$$-2(x,a) = 2\|x\|\cdot\|a\|$$thus either a of x must be 0. For general ... 2 For part a) the proof by contradiction: suppose that (I+T)x=y and \|y\|<\|x\|. Then, \|(T-I)(I+T)^{-1}y\|=\|(T-I)x\|=\|y-2x\|\ge 2\|x\|-\|y\|>\|y\|, which contradicts to our norm condition. For the second part, suppose that Tx=y. The condition \|(I+T)x\|\ge \|x\| implies (x+y,x+y)\ge (x,x) or \|y\|^2+(x,y)+(y,x)\ge 0. The second condition ... 2 This answers your first question As for the second question. Consider Holder inequality$$ \sum\limits_{i=1}^n |a_ib_i|\leq \left(\sum\limits_{i=1}^n |a_i|^{s/(s-1)}\right)^{1-1/s}\left(\sum\limits_{i=1}^n |b_i|^s\right)^{1/s} $$with a_i=1, b_i=|x_i|^r, s=p/r. 2 First, note that in stating your Lemma 2, Caffarelli uses Corollary 3.3 which says that u\in C^{0,1}(\Omega). So we can assume this fact. Moreover, on the proof of this corollary, he uses your Lemma 1, and by using it he says that$$\tag{1}u(x)\leq \lim_{r\downarrow 0} \oint_{B_r(x)}u$$My conclusion is that in the statement of your Lemma 1 (Lemma 2.2 in ... 2 It seems the following. Put X=\{0;1\} and define a metric d on the set X as follows: d(x,y)=1 if x\not=y and d(x,x)=0 for each x,y\in X. Then X is not a linear space over \mathbb R. :-) PS. Less trivial are examples of linear metrizable spaces admitting no consistent norm. 2 As explained by Ben Passer, we expect K to be weak*-compact. But I think it should be noted that K is empty when \dim A\geq 2. And when A=\mathbb{C}, K is a singleton consisting of the state \omega(\lambda)=\lambda. Indeed, if \omega\in K, then for every h self-adjoint positive, we have h=k^*k whence$$ ...

2

If you know singular value decomposition, let $A=USV^H$ be a SVD, where the singular values in $S=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)$ are arranged in descending order. Then $\sqrt{\rho(A^HA)}=\sqrt{\rho(S^HS)}=\sigma_1$. If you define $\|A\|_2$ as $\sigma_1$, you are done. If you define $\|A\|_2$ as $\max_{\|x\|_2=1}\|Ax\|$, it follows immediately ...

2

It's quite true that $\|\cdot\|_\ast = \|\cdot\|_\infty$. The problem is that the norm $\|\cdot\|_\ast$ does not induce the weak-* topology. For an explicit example, take $\Omega = [0,1]$ with $R$ Lebesgue measure. Let $\phi_n = 1_{(0, 1/n)}$. It follows from the dominated convergence theorem that $E[\phi_n X] \to 0$ for any $X \in L^1(R)$, so $\phi_n ... 2 Without further assumptions this is false. If$g$is unbounded, we might not even have$g(X_n), g(X) \in L^2$. If$g$is bounded and continuous, then$g(X_n) \to g(X)$in measure by the continuous mapping theorem, and also in any$L^p$by the dominated (bounded) convergence theorem. So the statement is true in this case. It also holds when$g$is ... 2 HINT: If you understand the product topology on an arbitrary product of topological spaces, you can use that to get a better handle on the weak topology on$X$: like the weak topology in$X$, the product topology is an example of an initial topology. The weak topology on$X$is the coarsest topology on$X$that makes all$f\in X'$continuous. For each$f\in ...

2

A topological space $X$ is said to have the countable chain condition (or to be ccc) if every family of pairwise disjoint nonempty open subsets of $X$ is countable. What you are essentially demonstrating is that every separable space is ccc: Letting $A$ be a countable dense set, then if $\{ U_i : i \in I \}$ were uncountable family of pairwise disjoint ...

2

To show $T^2 = T$, just compute $T^2$. We have for $f \in L^2([0,2\pi])$: \begin{align*} (T^2 f)(x) &= T(Tf)(x)\\ &= \int_0^{2\pi} G(x-x')(Tf)(x')\, dx'\\ &= \int_0^{2\pi} G(x-x')\int_0^{2\pi} G(x'-x'')f(x'')\, dx''\, dx'\\ &= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, f(x'')\,dx'' ...

2

First, let's study the restriction of $A$ on $x\in [1/2,1]$. Clearly, $A\big|_{x\in [1/2,1]}=L^2([1/2,1])$, hence $\left(A\big|_{x\in [1/2,1]}\right)^\bot=\{0\}$. Second, as you already mentioned, if $g\in A^\bot$, then $g\big|_{x\in[0,1/2]}$ can be any function in $L^2([0,1/2])$, so we can conclude that $A^\bot=\{g\in L([0,1]):g|_{x\in[1/2,1]}=0\}$.

1

No. Take $\Omega = (0,1)$ and $u_n(x) = x^{\alpha_n}/n$ with $\alpha_n = 1/(4\,n)-1/4$. Finally, take $g(t) = t^2$. Then, $$\int_0^1 u_n^2 \, dx = \frac1{(2 \, \alpha_n + 1) \, n} \, [x^{2 \, \alpha_n + 1}]_0^1 = \frac1{(2 \, \alpha_n + 1) \, n} \to 0$$ and $$\int_0^1 g(u_n)^2 \, dx = \frac1{(4 \, \alpha_n + 1) \, n} \, [x^{4 \, \alpha_n + 1}]_0^1 = ... 1 I'm not sure if I'm missing the point of the question: First we identify V' with V and L^2(0,T,V) with L^2(0,T,V') via the Riesz Representation. Suppose that \langle f, v \rangle = 0 for all v \in L^2(0,T,V). Then in particular$$\int_0^T \! \|f(t)\|^2 \, dt = \langle f, f \rangle = 0$$Hence \|f(t)\| = 0 and thus f(t) = 0 almost everywhere. ... 1 Have you solved it? Here's a hint. Let y\in X, y\notin W. For every fixed \varepsilon >0, you can find a x\in W such that \lVert x-y\rVert\le \varepsilon. So by the triangle inequality$$\lVert T_ny-T_my\rVert\le \lVert T_n x-T_mx\rVert+\lVert T_n(x-y)\rVert+\lVert T_m(x-y)\rVert.$$The sequence (T_nx) is Cauchy. Moreover, we have a uniform ... 1 Hints: We have$$w\in U^\perp\;\wedge v\in A^{-1}(U)\;(\text{so}\;\;v=A^{-1}u\;\;\text{for some}\;\;u\in U\;):\;\langle A^*w,v\rangle=\langle w,Av\rangle=\langle w,u\rangle=0 and thus we get $\;A^*(U^\perp)\subset \left(A^{-1}(U)\right)^\perp\;$ . Now you try to prove the other direction inclusion.

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Hint: Let $x \in X - \{0\}$ and $E_x := \mathbb K\cdot x$ denote the one-dimensional subspace generated by Define a functional $\phi \colon E_x \to \mathbb K$ by $\phi(\lambda x) = \lambda \def\norm#1{\left\|#1\right\|}\norm x$. Then $\phi \in E_x'$ with $\norm{\phi}_{E_x'} = 1$ (Can you see why?). Choose an Hahn-Banach extension $\psi \in E'$ with ...

1

Could I answer this way ? (half inspired by an answer i got on MathOverflow) We know that a C$^\star$-algebra is Arens Regular. We know that for a locally compact $\textit{infinite}$ group $G$, $L^{1}(G)$ is not Arens Regular. Therefore, if there was a representation of $L^{1}(G)$ in some $H$, we would have a contradiction, since a closed subalgebra of a ...

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