Hot answers tagged functional-analysis
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.
1
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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