# Tag Info

6

There is a theorem that if $k$ is a normed field, one of the following two things holds: 1) $k$ is a subfield of $\mathbb{C}$. 2) $k$ satisfies the non-archimedean property: $|x+y| \leq \max\{|x|,|y|\}$ (e.g. $k=\mathbb{Q}_p$). We can certainly define normed spaces over non-archimedean fields, but there are enough major differences in the theory that we ...

6

The correct equality is $\|a v\|=|a|\|v\|$ for some absolute value $|\cdot|$ on the scalar field. Absolute values can be nonarchimedean and satisfy the ultrametric inequality. In such a case, $|k|\le 1$ for all $k\in\Bbb N$. For instance, $\|(v_1,\cdots,v_n)\|:=\sqrt{|v_1|_p^2+\cdots+|v_n|_p^2}$ makes $\Bbb Q^n$ an ultrametric normed vector space over $\Bbb ... 3 I will give you a hint: Let us start with the definition, i.e. take sequences$(x_n)_n$in$M$and$(y_n)_n$in$N$with$x_n + y_n \to z$for some$z \in X$. We want to show$z \in M+N$. Now, let us write$\Gamma : X \to X/M \oplus X/N, x \mapsto (x+M) \oplus (x+N)$. We have $$\Gamma(x_n) = (0, x_n +N) = (0, x_n+y_n+N) \to (0, z + N).$$ Why/how does ... 2 A common use is in topologizing certain function spaces. For example, let$\Omega \subset \mathbb{R}$be open. We want a topology on$C^\infty(\Omega)$the collection of smooth functions. Intuitively we'd like convergence$f_n \to f$to imply the convergence locally of all derivatives. So what we do is we use a family of semi-norms to do it.$\|f\|_{K,n}$is ... 2 If$k_1,\dots,k_p$are kernels with RKHSs$\mathcal{H}_i$, then$k=\sum_i k_i$is indeed a positive definite kernel again without further assumptions. The relation$\lVert f\rVert_{\mathcal{H}}^2 = \sum_{i=1}^{p}\lVert f\rVert_{\mathcal{H}_i}^2$holds automatically for all functions that are in the intersection of all participating RKHSs. There is no ... 2 Since$\mathbb R^n$is convex (affine), and$\mathbb S^n_{++}$is a convex cone, the product$\mathbb R^n\times\mathbb S^n_{++}$is also a convex cone. This would be the shape you are looking for. For$n=1$, this is just the open right half plane.$\mathbb S^n_{++}$is also a$\frac{n(n+1)}{2}$dimensional manifold with tangent space at a point$P\in\mathbb ...

2

No, $f(S,r)$ need not be closed for closed $S$. Let $X = \ell^p(\mathbb{N})$ and $S = \left\{ \left(1+\frac{1}{n+1}\right)\cdot e_n : n \in \mathbb{N}\right\}$. $S$ is closed, but $0 \in \overline{f(S,1)} \setminus f(S,1)$.

2

It seems that you quoted the statement of Riesz's lemma from wiki with a small error. They state it in the following form. Let $X$ be a normed space, $\alpha\in (0,1)$ and let $Y$ be a proper closed subspace of $X$. Then there exists $x\in X$ with $\|x\|=1$ such that $$\|x-y\|>\alpha$$ for all $y\in Y$. Consider $Y=c_0$ inside $X=\ell_\infty$ ...

2

You need countable basis of neighboroods for any points. Indeed, suppose $E$ has countable basis $\{U_n^x\}$ of neighborhoods for any point $x$. 1) $T$ is sequence continuous implies $T$ is continuous. Proof. Let $A$ be an open set. If $B=T^{-1}(A)$ is not open there is $x\in B$ which is not interior. Thus, for any $U_{k}^x$ containing $x$ there is ...

2

The minimal condition on a space $X$ such that every function from $X$ is continuous if and only if it is sequentially continuous is that $X$ be sequential. First countable spaces are sequential. Thus a function from a first-countable space is continuous if and only if it is sequentially continuous. A space is called sequential if every sequentially open ...

1

As you write $A \cap r\cdot B_{X^*}$ is weak$^*$ly metrizable as a subspace of the metrizable space $r\cdot B_{X^*}$, now let $x_i^* \in A \cap r \cdot B_{X^*}$ be any net converging weakly$^*$ to $x^* \in X^*$. As $rB_{X^*}$ is weakly$^*$ closed (the norm is weakly$^*$ lower semicontinuous), we have $x^* \in rB_{X^*}$. Now, as $rB_{X^*}$ is metrizable in ...

1

If $k$ is the diameter of $\Gamma$, $0 < k < \infty$, then the diameters of the two sets $x+r\Gamma$ and $x'+r'\Gamma$ are $rk$ and $r'k$. So $r=r'$. We used $0<r<\infty, 0<r'<\infty$. So now we must ask, can a non-trivial translate $y + \Gamma$ of a bounded set $\Gamma \ne \varnothing$ be equal to $\Gamma$? No: let $a \in \Gamma$. ...

1

If $s>0, a \in X$ then the maps $$\lambda_s: X\to X, \lambda_s(z)=sz,\quad \tau_a: X\to X,\tau_a(z)=a+z$$ are invertible and satisfy: $$\tau_a^{-1}=\tau_{-a},\quad \lambda_s^{-1}=\lambda_{s^{-1}},$$ and $$\|\tau_a(y)-\tau_a(z)\|=\|y-z\|,\quad \|\lambda_s(z)\|=s\|z\| \quad \forall y,z \in X.$$ We have $$... 1 I think the answer is no, because l^{\infty} is not reflective. A related discussion can be found at here. The credit should be given to the answerer in the other post. 1 A proof without using a norming linear functional for f(b)-f(a) can be found in the 2nd edition of Rudin's Principles of Mathematical Analysis. It's stated for Euclidean space, but it works the same for all normed spaces. (Trivia: when the 2nd edition was translated into Russian, the translator added a footnote with the linear-functional proof; ... 1 A direct proof, in response to your comment: Let T be an isometry, and let v_1,\dots,v_n be a basis of X. We note that the vectors T(v_1),\dots,T(v_n) must be linearly independent. Why? Suppose otherwise. That is, suppose that$$ \sum_{i=1}^n a_i T(v_i) = 0 $$for some choice of a_1,\dots,a_n not all zero. It follows that$$ ...

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