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0

This result was proved in 1935 by A. Kolmogoroff: The topology of a locally convex topological vector space is generated by a norm iff there exists a bounded neighborhood of $0$.


2

Any induced norm satisfies $\|A\|\ge \rho(A)$, where $\rho(A)$ is the spectral radius of $A$, i.e. the largest absolute value of its eigenvalues. This fact is stated without proof in Wikipedia; here is a proof. Any induced norm is of the form $$\|A\|=\max_x\frac{\|Ax\|}{\|x\|}.$$ Let $x^*$ be an eigenvector corresponding to the eigenvalue $\lambda^*$ ...


0

Note the expression for the norm can be altered to; $$ ||(x_1, x_2)|| = \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} $$ Hence the graph you look for is points $(x_1, x_2)$ such that, $$ \max \{|x_1|, |x_2|\} + \frac{\min \{|x_1|, |x_2|\}}{3} \lt 1 \iff 3 \max \{|x_1|, |x_2|\} + \min \{|x_1|, |x_2|\}\lt 3 $$ I'm not sure about this part. ...


0

You are either implicitly assuming the series already converges (to argue that some tail of it has norm as small as you desire), or you are essentially showing that the partial sums form a Cauchy sequence. The latter only suffices to obtain convergence under completeness, which we assumed not to hold.


0

What needs to be shown is that given the subsequence $(\delta_{h_k})$ there is a functional $f\in X''$ such that $(f(\delta_{h_k}))$ does not converge. Now let $x$ be the sequence defined under $(2)$. Then $\hat x:X'\rightarrow \mathbb{K},x'\mapsto x'(x)$ is a bounded linear functional and $\hat x(\delta_{h_k})$ doesn't converge, so $(\delta_{h_k})$ doesn't ...


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 ...


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

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.


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

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 ...


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 ...


0

If $A$ has the norm $N$ and $N'$ is an equivalent norm, $N'$ is continuous. Since $N'(x)\neq 0$ for all $x\in A, f$ is continuous as well. A similar argument can be used to show the inverse is continuous. There is an easier way however. Since $E$ is finite-dimensional, there is an isomorphism $L:E\rightarrow \mathbb{R}^n$ for some $n$. If $N$ is the norm on ...


0

The two properties are equivalent as long as you only consider sequences: $RR\implies KK$ because on the unit sphere, by definition, all norms are 1, therefore the norm obviously converges. For the other direction, just scale your sequences, then you get $$ \frac{f_n}{\|f_n\|}\rightarrow \frac{f}{\|f\|} $$ if the norms converge, this converges strongly ...


0

Hint: Since $T$ is finite rank, you can pick an orthonormal basis $g_1,\ldots,g_n$ of the range of $T$, $$ Tv=\sum_{j=1}^n \langle Tv,g_j\rangle g_j. $$ Show that $\mbox{ker } T\supset \mbox{span }\{T^*g_1,\ldots,T^*g_n\}^\perp$. Pick and orthonormal basis of $\mbox{span }\{g_1,\ldots,g_n,T^*g_1,\ldots,T^*g_n\}$, say $e_1,\ldots,e_m$. Show that $m\leq 2n$. ...


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 ...


0

The linearity is not needed. In fact, what you have is a continuous function $T$ between topological spaces $X$ and $Y$ (they're normed but that's not relevant) and a convergent net (or sequence) $x_h \rightarrow x$ in $X$. Then show that $T(x_h)$ converges to $T(x)$. (weak topology not needed here.) And this is quite easy: take an open set $O \subset Y$ ...


0

You can easily prove this by using nets. Let $(x_i)$ be a convergent net and $x=\lim_i x_i$. Note that $x_i \rightarrow x$ iff $x'(x_i)\rightarrow x'(x)$ for all $x'\in X'$. Now, since $T$ is continuous, we have for all $y'\in Y'$ that $y'\circ T \in X'$. It then follows for all $y'\in Y'$ that $y'(Tx_i)=(y'\circ T)(x_i)\rightarrow (y'\circ T)(x)=y'(Tx)$. ...


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

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 $a \in \Gamma$. ...


0

No ! For exemple, take $\Gamma=\{e^{i\theta}\mid\theta\in[0,2\pi[\}$. You have $\Gamma=e^{i\pi}\Gamma$ but $1\neq e^{i\pi}$


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 ...


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 $0 \in \overline{f(S,1)} \setminus f(S,1)$.


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 ...


0

By the Holder inequality, $\lVert Af \rVert \leq ||\sum_{m \geq 1} \alpha_{mn} ||_{\infty} ||f||_1$. By (a) and the definition of sup norm, $M=||\sum_{m \geq 1} \alpha_{mn} ||_{\infty}$. Thus, $A$ is a bounded operator with operator norm at most $M$. Can you get the second part now?


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 ...



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