# Tag Info

24

why do I have to restrict the measurement standard to real number? Why not any ordered field? You can use any ordered field, and the axioms will still make sense. The thing is, though, that most of our geometric intuition is built on the Archimedian property. But if we include this in our field, then it becomes a subfield of $\Bbb R$. In this case, we ...

5

You can define inner products for vector spaces and normed vector spaces over any ordered field. You can define metrics over any ordered field: they are called generalised metrics.

5

Everything boils down to just a few key facts: A non-vanishing continuous real function on the unit sphere (unit vectors) $S$ in $\mathbb{C}^{n}$ achieves its minimum (non-zero) value at some point. The Cauchy-Schwarz inequality for inner products, $|(x,y)| \le \|x\|\|y\|$, of which the following is a special case: $|\sum_j a_j b_j| \le (\sum_j ... 5 While your question could have multiple answers, perhaps the closest to what you are looking for is the notion of a non-metrizable vector space. In the general setting of topological vector spaces, we consider (as one might guess from the name) vector spaces endowed with a topology so that we can discuss ideas like the continuity of linear operators. Normed ... 3 Let$p(x)=(x-1)(x-2)\cdots(x-n-1)$. Then the polynomial$p$has degree$n$and$\|p\|=0$while$p\neq 0$so the function in 1 is not a norm. Let$p=c$,$c$is a non-zero constant. Degree of$p$is$0$and$p\neq 0$but$\|p\|=\sup_{x\in [0,1]}|p'(x)|=0$so 4 is not a norm. EDIT: Thanks for warnings. Degree of$p$is$n+1$so it is not in$P_n(\mathbb R)$. ... 3 We have $$(x,y)\in B(0,1)\iff N(x,y)=\sup_{t\in\mathbb{R}}\frac{|x+ty|}{t^2+t+1}\le1\iff |x+ty|\le t^2+t+1, \forall t\in\Bbb R\\\iff -t^2-t-1\le x+ty\le t^2+t+1,\forall t$$ the second inequality gives $$t^2+(1-y)t+1-x\ge0,\forall t\iff \Delta_2=(1-y)^2+4x-4\le0$$ and the first inequality give $$t^2+(1+y)t+x+1\ge0,\forall t\iff \Delta_1=(1+y)^2-4x-4\le0$$ ... 3 For second question, let$m=2, n=3$. We can easily draw tree circles each pair of which intersects and all three of them have empty intersection. So the answer is negative. As for the first one, a convex set$X$has, by definition, a property that for any two points$u, v \in X$point$au + (1-a)v$belongs to$X$for any$a \in [0, 1]$. In any words, whole ... 3 Let us take a closer look at$\lVert\,\cdot\,\rVert_0$. Since for$x\in X$we have$\hat{x} = \{ x-y : y \in Y\}$, we can write $$\lVert \hat{x}\rVert_0 = \inf_{z\in\hat{x}} \lVert z\rVert = \inf \{ \lVert x-y\rVert : y \in Y\}.\tag{1}$$ So we have$\lVert \hat{x}\rVert_0 = 0$if and only if for every$\varepsilon > 0$there exists a$y_\varepsilon \in ...

3

You can split your problem into $$\min \|x\| \quad {s.t.} A_1 x = b_1, \ l_1\le x\le u_1,$$ and $$\min \|y\| \quad {s.t.} A_2 y = b_2, \ l_2\le y\le u_2.$$ There is no coupling between both optimization variables.

3

It's not true. Let $X$ be the space of sequences that are eventually $0$ with the sup norm. Let $X_n$ be the subspace of $X$ consisting of the sequences $x$ with $x(i)=0$ if $i\ge n$. Consider the sequence $(x_n)$ with $x_n=(1,1/2,1/3,\ldots,1/n,0,0,\ldots)$.

2

Since the $H^k$ are descending, having $H^m = H^n$ for some $n > m$ means $H^k = H^m$ for all $k \geqslant m$, as $H^m \supset H^{m+1} \supset \dotsc \supset H^n = H^m$ forces all these spaces to be equal, and then, for $k \geqslant m$ we have $H^{k+2} = T(H^{k+1}) = T(H^k) = H^{k+1}$ by induction. Suppose that $H^1\subsetneq H$, but the $H^k$ are not ...

2

The Fourier transform is unitary on $L^{2}(\mathbb{R})$ if you choose the correct normalization: $$Ff = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ist}f(t)\,dt.$$ Some people instead absorb the constant into the exponent, but this is what I'm used to seeing because this $F$ is directly linked to the spectral resolution of the operator ...

2

The thing you're missing (I think - looks like you're almost there in your edit) is Schwarz's inequality. We have that $$|A\phi_1 - A\phi_2|_{L_2} = \lambda^2\int_a^b \left|\int_a^s K(s,t)(\phi_1(t)-\phi_2(t))\mathrm{d}\mu_t\right|^2\mathrm{d}\mu_s.$$ Then by Schwarz's inequality we can bound the inner integral squared as the ...

2

Let $D=(x_n)$ dense in $X$ and let $\sigma(X^*,D)$ the locally convex topology in $X^*$ with countable local basis at $0\in X^*$ given by: $$U_n=\{x^*\in X^*: \sup_{1\leq j\leq n}|x^*(x_j)|< 1/n \}$$ Then $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS. In fact, if $x^*\neq 0$, as $x^*$ is continuous, there is $j_0$ such that $x^*(x_{j_0})\neq 0$, so we have ...

2

The statement you say you have been able to prove is correct. In fact, it's very close to a characterization of Banach spaces. A normed linear space $X$ is Banach if and only if every absolutely summable series in convergent, i.e. if every series $\sum_{n=1}^\infty a_n$, $a_n\in X$ with $$\sum_{n=1}^\infty \Vert a_n \Vert_X < \infty$$ converges in $X$. ...

2

The usual definition of completion of a normed space is obtained by considering equivalence classes $[\{ x_{n} \}]$ of Cauchy sequences, grouped according to the equivalence relation that $$\{ x_{n} \} \sim \{ y_{n} \} \iff \lim_{n}(x_{n}-y_{n})=0.$$ This $\sim$ is an equivalence relation because it is (a) reflexive (b) symmetric and (c) transitive: (a) ...

2

If $V$ and $W$ are normed vector spaces of the same finite dimension $n$, then $V$ and $W$ are isomorphic as topological vector spaces, i.e., there is a a vector space isomorphism $f : V \rightarrow W$ such that both $f$ and $f^{-1}$ are continuous. However, if $n \ge 2$, $V$ and $W$ may not be isometrically equivalent, i.e., it may be impossible to choose ...

2

I like Valeriy's answer, but let me also point out that the first question is a special case of Helly's theorem, which says that in $\mathbb R^d$, if you have at least $d+1$ sets such that any $d+1$ of them have a non-empty intersection, then the intersection of all the sets is non-empty.

2

A thought to get started: write a power series for $e^I$. See anything simple to do? If you don't like the idea of just jumping to a full series, start with a partial sum and see what comes out.

2

Let $e_n$ be the element of $\ell_\infty$ whose $m$'th coordinate is $1$ if $m=n$ and $0$ otherwise. The closed linear span of $\{e_n\mid n\in \Bbb N\}$ in $\ell_\infty$ is the space $c_0$ of sequences that tend to $0$. For your operator, we have $T(je_j)=e_j$; so the range of $T$ contains each $e_j$. Since the range of a linear operator is a linear space, ...

2

You can use the same idea that works for $+$: $$\|(\alpha,x)\|_{K\times X}=|\alpha|+\|x\|.$$ In any case, you can endow $K\times X$ with the product topology and no explicit norm is required. EDIT: \eqalign{\|\alpha x - \alpha_0 x_0\|_X & = \|\alpha x - \alpha_0 x + \alpha_0 x - \alpha_0 x_0\|_X\cr &\le\|\alpha x - \alpha_0 x\|_X + \|\alpha_0 ...

2

The condition says: for every $k$, the $k$th coordinate of $x\in M$ is uniformly bounded (independently of $x$) by some number $\gamma_k$. Geometrically, this means $M$ is contained in an infinite-dimensional rectangular box with sidelengths $2\gamma_1, 2\gamma_2,\dots$. The necessity of this condition follows from the fact that the projection onto $k$th ...

2

Every (real or complex) vector space admits a norm. Indeed, every vector space has a basis you can consider the corresponding «$\ell^1$» norm.

2

Of course there are. :-) First of all, there are linear spaces which are not endowed by a topology, and there are topological vector spaces which are not endowed by a norm. So a wise question is: when a topology of a topological vector space $X$ is generated by a norm? It seems it is iff $X$ is a locally convex Hausdorff space containing such a neighborhood ...

2

Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space.

2

If $V$ is finite dimensional, it is normable, in the sense that you can use an isomorphism into $\mathbb R^n$ to pull back the $\mathbb R^n$-norm . And then this norm is equivalent to any other norm topology-wise, i.e., in a finite- dimensional space, all norms are equivalent in this sense.

1

Let $\xi_n$ a sequence in $M$, we note $(\xi_n)_1$ its first coordinate, now because $|(\xi_n)_1| \leq \gamma_1$ for all $n$, by Bolzano-Weierstrass there is some subsequence $(\xi_{n_k})_1 \to \xi_1$, now take the sequence $(\xi_{n_k})$, by repeating the process there is some subsequence on the second coordinate $(\xi_{{n_k}_j})_2 \to \xi_2, we can repeat ... 1 The fields$\mathbb R$and$\mathbb C$are equipped with a standard topology (derived from the standard metric and the standard absolute value). 1 The proof that$T$is an isometry typically consists of two parts: Show that$\|Tx\|\le \|x\|$for all$x$Show that$\|Tx\|\ge \|x\|$for all$x$For the classical function and sequence spaces, 1) may involve Hölder's inequality, sometimes in its trivial$(1,\infty)$-form$\left|\sum a_nb_n\right|\le \sup_n|b_n| \sum_n |a_n|$. Step 2) then ... 1 For each$x\notin \overline{M}$, as a consequence of Hahn-Banach Theorem, there is$\phi_x \in X^*$,$\|\phi_x \|=1$,$\phi_{x}|_{M}=0$,$\phi_x(x)=d(x,\overline{M})$. If$y\in \bigcap\{\ker(\phi): \phi|_M=0\} \subset \bigcap\{\ker(\phi_x)\}$, then$\phi_x(y) = 0 \quad \forall x \notin \overline{M}$. Thus$y\in \overline{M}.\$

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