Tag Info

New answers tagged

1

The answers follow from the standard fact that $B$ has the metric $d(x,y) = \sum_{k=1}^\infty c_kp_k(x,y)/[1+p_k(x,y)]$ where $c_k$ is any sequence of scalars tending to $0$. E.g. $c_k=1/2^k$.


0

Having duly noted Martin Argerami's answer, I thought it worth mentioning that for normal $A,B$ in a $C^*$-algebra, we have $$ \|(A+B)^m\|^{1/m} \leq \|A+B\| \leq \|A\|+\|B\|=\|A^m\|^{1/m}+\|B^m\|^{1/m}. $$ Observe also that for commuting normal operators, we have the following Cauchy-Schwartz type inequality $$ \|AB x\|^2=\langle A^*Ax,B^*Bx\rangle\leq ...


3

No, let $V$ be separable and $e_n$ form an orthonormal basis in $V$, set $Ae_n=\frac1n e_n$. Then $\langle Au,u \rangle>0$ since every non-zero vector has at least one non-zero coefficient in its expansion, but there is no $\alpha$ since otherwise $\|Ae_n\|\geq\alpha$ for all $n$. The first condition is sometimes called "strictly positive definite" and ...


1

Let $v=\frac{5}{9}$ and $w=-\frac{5}{9}$. Then $(v,w)$ is clearly not in the first set, but it is in the second since 1) $|v+1w|=0\le1$ 2) $|v+(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)w|=\frac{5}{9}|\frac{3}{2}-\frac{\sqrt{3}}{2}i|=\frac{5}{9}\sqrt{3}\le1$ 3) ...


1

The part with lemma 2 isn't necessary. Any complete subspace of a metric space is closed. So the proposition follows directly from lemma 1 and you don't need to assume that $E$ is finite-dimensional. But can you prove lemma 1? Actually even if you use just “complete subspace of complete space is closed” you don't need lemma 2 since both $E$, $F$ are ...


2

I claim that $\exists M \geq 0$ such that $$ |f(x)| \leq M \quad\forall x\in E \text{ such that } \|x\| \leq 1 \qquad(\ast) $$ If not, then $\exists x_n \in E$ such that $\|x_n\| \leq 1$ and $$ |f(x_n)| > n^2 $$ Hence, take $$ y_n = x_n/n $$ Then, $y_n \to 0$ and $\{f(y_n)\}$ is unbounded. Now can you prove that $f$ is continuous at $0$ from $(\ast)$?


1

Not in general, for example the infinity norm is a norm on any vector space over a totally ordered field that does not come from an inner product. E.g. for a vector $\vec{v} = (v_1, \dots, v_n) \in \mathbb{R}^n$, defined $||\vec{v}||_\infty = \max\{v_i\}$. You can check that this is a norm, and does not come from an inner product.


1

Note that, in your case with the $2$-norm, you have $\|Ax-b\|_2^2 = \langle Ax-b,Ax-b \rangle $, where $\langle \cdot , \cdot \rangle$ is the euclidian scalar product. But this is not always possible, for example the $p$-norm with $1 \leq p < \infty$ defined by $$\|x\|_p := \left(\sum_{k=1}^n |x_k|^p\right)^{1/p},$$ is such that there is no scalar product ...


2

Yes. Because you consider the identity map $(V,\|\cdot\|_2)\longmapsto(V,\|\cdot\|_1)$. The inequality between the norms shows that it is bounded. It is not immediately obvious to me that it is a surjection in the generality of your question. But it is in the concrete case you mention, because a C$^*$-homomorphism has closed image (when the domain is a ...


0

The answer is not right. (I do not elaborate, because you only asked for confirmation.)


0

I assumed that $Y$ is a subspace and $x_0$ doesn't belong to the closure of $Y$ (hence we may furthermore assumed that $Y$ is closed). Define on $V:=Y+\operatorname{Vect}(x_0)$ the linear functional $f$ by $$f(y+\lambda x_0):=\lambda d( x_0,Y).$$ This is well defined, linear and for each $v\in V$, $|f(v)|\leqslant d(v,Y)$. The map $v\mapsto d(v,Y)$ is ...


1

If $F$ is as defined in the OP, let $M=\max{\{\|F_j\|\}}$. Since all norms on $\Bbb R$ are equivalent, take $\|x\|:=\sum\limits_j|x_j|$. Thus: \begin{align*} \|F\|&=\sup{\{\|F(x)\|\,\mid\,0\leq \|x\|\leq 1\}}=\sup{\left\{\sup{\{|\left<F(x),y\right>|\,\mid\,0\leq \|y\|\leq 1\}}\,\mid\,0\leq \|x\|\leq 1\right\}}\\ ...


2

Proof: If $A$ is closed in $Y$, then $J^{-1}(A)=A$ is closed in $X$ because $J$ is continuous. Let $(x_n)$ be a sequence in $X$ such that $x_n$ converges weakly to $x$. Now, let $f\in Y^*$. Then $f\circ J\in X^*$. Hence $f(x_n)=(f\circ J)(x_n)$ converges to $(f\circ J)(x)=f(x)$. Therefore, $(x_n)$ converges weakly to $x$ in $Y$. Since $X$ is reflexive ...


4

What you are discussing is the notion referred to as measure. While there are different notions of measure, the one that is most commonly used is that of Lebesgue measure. The $n$-dimensional volume of a set in $\mathbb{R}^n$ can be taken as the Lebesgue measure of that set. In any Euclidean space, it is indeed as you say: a set can only have infinite ...



Top 50 recent answers are included