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1

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

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 ...


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 ...


0

It's good enough to have $E$ be first countable, i.e. at each $x\in E$ we need a countable basis $\{U_i\}_{i\in\mathbb{N}}$ of open neighborhoods of $x$. The easiest proof uses that a function $f$ is continuous just if $f\left( \overline{A}\right)\subset \overline{f\left(A\right)}$ for every $A\subset E$. In a first countable space $\bar A$ is exactly the ...


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 $$ ...


5

Consider $\newcommand{\inner}[2]{\langle #1 \mspace{-3mu}\mid\mspace{-3mu} #2\rangle}$ $$\inner{u}{v} = \int_0^\infty u'(t)v'(t)p(t)\,dt.\tag{1}$$ $\inner{\cdot}{\cdot}$ is a bilinear form, and $\inner{u}{u} = \lVert u\rVert^2$ is zero only for $u \equiv 0$. Hence $\inner{\cdot}{\cdot}$ is an inner product inducing the norm. Every space whose norm is ...


0

This was sorted out in comments, so I'll add a slightly different point of view. Every bounded linear operator $T:X\to V$ has the adjoint operator $T^*:V^*\to X^*$ defined by $T^*(f)=f\circ T$. The definition shows that $\|T^*\|\le \|T\|$. (In fact, equality holds by Hahn-Banach, but we do not need this here). In the special case when $T$ is an isometric ...


1

Perhaps an example will help. Let $A = \{7, 13\}$. Then $E$ is the vector space of complex-valued functions on $\{7, 13\}$. Let $a = 7$; then $N_7$ is a seminorm on $E$. Letting $a=13$, it's also true that $N_{13}$ is a seminorm on $E$. For instance, letting $f(7) = i, f(13) = 0$, we have $N_7(f) = |f(7)| = |i| = 1$, whereas $N_{13}(f) = |f(13)| = 0$. Since ...


0

$N_a(f)=0$ does not imply $f(x)=0$ for all $x\in A$. It only implies $f(a)=0$.


0

If $V,W$ are vector spaces and $|\cdot|$ is a norm (ort seminorm) on $W$, and $T\colon V\to W$ is linear, then $v\mapsto |Tv|$ is a seminorm on $V$


4

Take $X=c_{00}$---the space of all sequences which are almost everywhere $0$ and as $x_n$---the sequence having $\frac{1}{2^n}$ on $n$-th place and $0$ elsewhere.


0

The FOIL rule with column vectors $u, v \in \mathbb R^n$ tells us that \begin{align} \|u - v\|^2 &= (u - v)^T (u - v) \\ &= u^T u - u^T v - v^T u - v^T v\\ &= \|u\|^2 - 2 u^T v + \|v\|^2. \end{align} Applying this to our particular problem, we find that \begin{align} \|Y - X \beta \|^2 &= \|Y\|^2 - 2 Y^T X \beta + \| X \beta \|^2. ...


0

It is clear that the completeness of the whole space implies the completeness of the unit sphere. Suppose we have that the unit sphere is complete, we want to show that the whole space is also complete. $\{x_n\}$, not necessarily on the unit sphere, is a Cauchy Sequence, so is $\{\|x_n\|\}$. We have that $\|x_n\|\to a$. If $a=0$, i.e. $\|x_n\|\to 0$ which ...


0

Here is a 'constructive' proof of boundedness of $T^{-1}$: Define $\|x\|_* = \|Tx\|$. Then $\|\cdot\|_*$ is a norm on $\mathbb{R}^n$ and so is equivalent to, say, $\|\cdot\|_1$. In particular, for some $c>0$, you have $\|x\|_* \ge c \|x\|_1$, or in other words, $\|Tx\| \ge c \|x\|_1$. Now let $v=Tx$, then you have ${1 \over c} |v\| \ge \|T^{-1} v\|_1$, ...


3

You gave a nice proof but the contradiction isn't (at least for me) clear. To explain it we can do as this: since the sequence $(x_n)$ is bounded ($\|x_n\|=1$) in finite dimensional space (and here we used this assumption) then by the Weierstrass theorem there's a convergent sub-sequence to say $x$ and by the continuity of the norm we have $\|x\|=1$ but by ...


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) ...



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