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6

A quick note: A 'space' is an arbitrary set of 'things' but with an additional structure or property. A metric space is a set where you can measure distances between two points, but you have to know the properties of the distance because they can get arbitrarily complicated (e.g. p-adic distance). So examples or explainations from others can help, but from ...


4

I would say it is almost impossible to get a good feeling of various kinds of spaces just by reading the definitions. Historically, those notions are not developed immediately by somebody in a dream or something like that. Instead, those concepts are formulated gradually. The concept of a manifold is a direct example. So, how these concepts are formulated? ...


4

I'm an engineering/physics student but I've also had to teach myself about certain types of spaces. I think the most important spaces to learn first to orient yourself are topological, metric, and vector spaces. Many spaces I've come across are special cases or combinations of these. Topological/metric spaces are more analytic (concerned with the ...


4

The space of continuous functions on $[0,1]$ with the norm $\|x\|= \sup\{|x(t)| : t\in[0,1] \}$ is a Banach space. To justify that claim, you would need to know that all continuous functions on $[0,1]$ are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ...


3

If that weren't the case you could find $z_n = (x_n,y_n)$ with $1 - {1 \over n} \leq |z_n| \leq 1 + {1 \over n}$ with $z_n \in (\bigcup_\alpha U_\alpha)^c$. Taking a convergent subsequence, you'd have some $z_{n_j}$ converging to a $z$ with $|z| = 1$, but it would also be in $(\bigcup_\alpha U_\alpha)^c$ since the latter set is closed. Hence you have a ...


3

(1) Define $F \colon c_0 \to \def\K{\mathbf K}\K$ by $$ F(x) = \sum_{n=0}^\infty x_n^n $$ Then $F$ is continuous, as the series converges locally uniform, but $F$ is unbounded, as the elements $x^{(n)} = (1, \ldots, 1,0, \ldots) \in \bar B_{c_0}$ have $$ \def\norm#1{\left\|#1\right\|} \norm{x^{(n)}} = 1,\qquad \def\abs#1{\left|#1\right|}\abs{F(x^{(n)})} = ...


3

You are assuming the operator $T$ is continuous at $0_X$. In a normed linear space, if an operator $T$ is continuous at $0_X$, then it is continuous (bounded) on $X$. This is a standard theorem in functional analysis.


2

Look at functions which are $-1$ on $x<- \epsilon$, $1$ on $x > \epsilon$ and go linearly from $-1$ to $1$ on the interval $[-\epsilon,\epsilon]$. Making $\epsilon$ arbitrarily small gives a function which is still in $C[-1,1]$ with norm $1$ and the value of $f$ for these is $2-\epsilon^2$. So, $||f|| \geq 2 - \epsilon^2$ for any $\epsilon >0$. ...


2

Your proof is correct. However, there are a few things you write that are strange notationally. Two aesthetic corrections: Say $u,v \in \mathcal L_c(E)$ rather than $(u,v) \in \mathcal L_c(E)^2$ Make your logical statements clearer: $\forall w \in \mathcal{L}_{c}(E): \, \Vert w \Vert \leq 1 \implies \Vert \Phi_{v}(w) \Vert \leq 2 \Vert v \Vert$ A ...


2

$$\langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle\forall \alpha\\ \langle x,x\rangle+2\alpha\langle x,y\rangle+\alpha^2\langle y,y\rangle\geq\langle x,x\rangle\forall\alpha\\ 2\alpha \langle x,y\rangle+\alpha^2\langle y,y\rangle \geq0\forall\alpha$$ But the left-hand side equals zero if $\alpha=0$ and if $\alpha=-2\langle x,y\rangle/\langle ...


2

Note: Here are two examples of partial converses which could be convenient. Partial converse in section 27 of A Hilbert Space Problem Book by Paul R. Halmos: Every bounded subset of a Hilbert Space is weakly bounded. We can read in section $27$: Uniform boundedness The celebrated principle of uniform boundedness (true for all Banach ...


2

This was going to be a comment but turned out to be a bit too long: The way I look at it is that a space is just a set that has some extra properties that give it some structure. E.g. the $L^p$ space (Lebesgue) is the set of functions that satisfy $\|f\|_p \equiv \left({\int |f|^pdx}\right)^{\frac{1}{p}}<\infty$. You can think of this as just a bag of ...


2

The category of Banach spaces is locally $\aleph_1$-presentable.


2

Let $x, y \in \def\R{\mathbf R}\R^n$. $\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$We have by Hölder as $\frac 13 + \frac 1{3/2} =1$ that $$ \abs{\sum_{i=1}^n x_i y_i} \le \norm x_3 \cdot \norm y_{3/2} $$ That proves that the map $T \colon (\R^n, \norm\cdot_{3/2})\to (\R^n, \norm{\cdot}_3)^*$ given by $$ (Ty)(x) = \sum_i x_i y_i $$ fulfills ...


2

Yes, the closure of a vector space is a vector space. Yes, $D(T)$ might not be a closed set. For example, polynomials form a non-closed subspace of $C[0,1]$ (the continuous functions on $[0,1]$).


2

To adapt and complete the OP's initial line of attack, instead of using ordinary unit balls it will make life easier to instead use "polar coordinate basis sets". For each $x = (\cos(\theta_0),\sin(\theta_0)) \in S^1$ there exists $\delta_x > 0$ such that the polar coordinate basis set $$B_x = \{(r \cos(\theta), r \sin(\theta) \bigm| 1-\delta_x < r ...


2

For example, note that $f(2x) = f(x + x) = f(x) + f(x) = 2 f(x)$, and hence $f(x) = \frac{1}{2} f(2x)$, which means $f(\frac{1}{2} x) = \frac{1}{2} f(x)$. Similarly one can prove that, for every $q \in \mathbb{Q}$, we have $f(qx) = q f(x)$. Now suppose $\lambda \in \mathbb{R}$. Choose a sequence $q_n \in \mathbb{Q}$ such that $q_n \rightarrow \lambda$. This ...


2

Since $$ f(0)=f(0+0)=f(0)+f(0)=2f(0), $$ we have $f(0)=0$. It follows that $$ f(-x)=f(-x)+f(x)-f(x)=f(-x+x)-f(x)=f(0)-f(x)=-f(x) \quad \forall x\in E. $$ Also for every $x\in E$ we have \begin{eqnarray} f(2x)&=&f(x)+f(x)=2f(x)\\ f(3x)&=&f(2x)+f(x)=3f(x)\\ &\vdots&\\ f(nx)&=&nx \quad \forall n\in \mathbb{N}. \end{eqnarray} ...


2

First, as a side remark, note that $\bigcup_{\beta < \alpha_0} \operatorname{span}(V_\beta)= \operatorname{span}(\bigcup_{\beta < \alpha_0} V_\beta)$. The statement "Then clearly $\operatorname{span}(V_\gamma)=X$ " is flawed. I'll explain why it is, and how to fix it. Say $B$ is a basis for $X$. One may easily come up with examples where $X$ is ...


1

It suffices to show that at least one $x$ with $\|x\|_{\mathbb{R}^n}=1$ achieves the specific bound. Let $$i^*=arg\max_{i=1,2,\ldots,m}\sum_{j=1}^n{|a_{ij}|}$$ i.e. $i^*$ is the index of the row of $A$ with the largest sum of absolute values of its elements. Then, if we select $\xi_j=sign(a_{i^*j}$) we have that $\|x\|_{\mathbb{R}^n}=1$ and ...


1

Suppose the maximum $\max_{k=1,\ldots,n}\sum_{i=1}^{m}|\alpha_{ik}| = M$(say) is attained for $ k = k_0$, i.e, $M = \sum_{i=1}^{m}|\alpha_{ik_0}|$. Then take the vector $x= (0,\ldots,1,\ldots,0)$ with $1$ at the $k_0$ position. Notice that $\|Tx\|_{\mathbb{R^m}} = M$. Since there is a vector such that this supremum is attained, it follows that $\|T\| = M$.


1

To show $V$ is connected use the following facts: The set in question $V = \displaystyle\bigcup_{a \in A}B(a, \varepsilon)$ where $B(a, \varepsilon)$ is the ball in Euclidean space with centre $a$ and radius $\varepsilon$. Each $B(a, \varepsilon)$ is connected. If two connected sets (for example $A$ and some $B(a, \varepsilon)$) overlap their union is ...


1

Norms with infinite values are discussed here, where they are called extended norms. As pointed out in Anguepa's answer, extended norms $||\cdot||$ define extended metrics $d(x,y)=||x-y||$ which define topologies generated by balls $B_x^r=\{y\in V:d(x,y)<r\}$ in the usual way. However, this topology will be disconnected if $||x||=\infty$ for any $x\in ...


1

If the system is unstable, the $H_\infty$ norm is infinite. You have a discrete time system with a pole greater than 1 in absolute value hence the answer inf. First one doesn't check for stability and converts it to state space system. The discrepancy is due to the fact that norm() command only checks for $L_\infty$ not necessarily $H_\infty$ (for stable ...


1

Suppose \begin{equation} T(x_1)(t) = T(x_2)(t) \end{equation}for all $t \in[0,1]$. As you have already figured out that the range of the operator is the set of all continuously differentiable functions (this follows from Fundamental Theorem of Calculus). In light of your observation just differentiate the two sides of the equation to get $x_1(t) = x_2(t)$, ...


1

At each point $x$ on the sphere, let $\rho(x)$ be the largest number for which the open ball of radius $\rho(x)$ centered at $x$ is a subset of $\bigcup_\alpha U_\alpha$. The $\rho(x)$ must be finite unless $\bigcup_\alpha U_\alpha=\mathbb R^n$, and if that happens then it's easy to answer the question. If I'm not mistaken $\rho$ is continuous. I'd try to ...


1

First of all you should understand that what is meant when one says that "$\textit{Not all normed linear spaces are inner product spaces.}$" Here is the explanation. An $\textit{inner product space}$ is a vector space with an inner product defined on it. An inner product on $X$ defines a norm on $X$ given by \begin{equation} \|x\| = \sqrt{\langle x,x ...


1

$\mathbb{R}$ is an infinite dimensional vector spaces over $\mathbb{Q}$, but it is not an Hilbert space. You can see that all the axioms for a vector space are verified if you define the sum of two ''vectors" as the usual sum of real numbers and the product for a scalar $q \in \mathbb{Q}$ as the usual product. This space has an infinite dimensional Hamel ...


1

In fact all normed spaces are subspaces of some function spaces. This could be the reason why functional analysis have its name.


1

One example might be $M(K)$, the space of all regular Borel measures on $K$ of finite variation, where $K$ is compact space. This space arises as the dual of $C(K)$.



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