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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 ...


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.


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 ...


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 ...


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 ...


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

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 ...


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 $$ ...


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 ...


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 ...


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. $c_k=1/2^k$.


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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