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Notice that the question might as well be posed on $\mathbb{R}^n$: Having non-zero imaginary part only increases the norm of the $\eta_j$, so we're looking only at their real parts. In other words we may assume $\eta_j\in \mathbb{R}$. But now the second equation defines a hyperplane orthogonal to the vector $(1,\ldots, 1)$ at distance $1/\sqrt{n}$ from the ...


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I assume here the spaces are finite dimensional (as in the question), and the vector spaces are real (for convenience, one can generalize to other fields too). Recall first that a norm is induced from an inner product if and only if it satisfies the polarization identity: $$ 2\|u\|^2 + 2\|v\|^2 = \|u+v\|^2 + \|u-v\|^2 $$ for every $u,v\in V$. Let $V, W$ ...


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Let $a = (a_n)_{n\in \Bbb N}$ and $b = (b_n)_{n\in \Bbb N}$ be in $\ell^p$. If $\|a + b\|_p = 0$, the inequality is clear, so suppose $\|a + b\|_p > 0$. Since $$|a_n + b_n|^p = |a_n + b_n| |a_n + b_n|^{p-1} \le (|a_n| + |b_n|)|a_n + b_n|^{p-1}$$ for all $n\in \Bbb N$, then $$\|a + b\|_p^p = \sum_{n = 1}^\infty |a_n + b_n|^p \le \sum_{n = 1}^\infty ...


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How about this: let $\mu$ be the counting measure on $\mathbb N$. Then $$\int_{\mathbb N} |a| \, d\mu = \sum_{n=0}^\infty |a_n|$$ for any sequence $a = (a_n)_{n \in \mathbb N}$. If $a,b \in L^p(\mathbb N,\mu)$, the Minkowski inequality gives you $$ \|a+b\|_{L^p(\mathbb N,\mu)} \le \|a\|_{L^p(\mathbb N,\mu)} + \|b\|_{L^p(\mathbb N,\mu)}$$ which says that $$ ...


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If $(a_n)\in\ell_1$, then it is bounded. There is a constant $M>0$ such that $|a_n|\le M$ for all $n$. Then for any $p>1$ we have $$ \sum|a_n|^p\le M^{p-1}\sum|a_n|<\infty, $$ proving that $\ell^1\subset\ell^p$.


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First let us assume that $B$ is bounded. Let $(x_n)$ be any sequence in $B$ and $(a_n)\in c_0$. Then $\|x_n\|\leq M$ for all $n$ and given $\epsilon >0$ there exist a natural number $N$ such that $|a_n|<\frac{\epsilon}{M}$ for all $n\geq N$. So, for $n\geq N$ we have \begin{equation} \|a_nx_n\| = |a_n|\|x_n\|<\frac{\epsilon}{M}M = \epsilon. ...


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It seems there is no such formula. Vaguely speaking even for simple cases this problem is NP-hard. For details see this paper


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$L_\infty[0,1]$ is not separable PROOF Assume that $L_\infty[0, 1]$ is separable. Now consider the set $$ F= \{ f_t : f_t(x)= \chi_{(0,t)} (x)  \ \ t \in (0,1) \} $$ Since all the $f_t$ are measurable and clearly $\|f_t\|_\infty=1<\infty$, then $F\subset L_\infty[0,1]$. Of course since $(0,1)$ is uncountable, $F$ is also uncountable. Since ...


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Since $$ d:X\times X\to [0,\infty) $$ is a metric, it satisfies the triangle inequality: $$ d(a_1,a_2)\le d(a_1,a_3)+d(a_3,a_2) \quad \forall a_1,a_2,a_3\in X. $$ Given $a=(a_1,a_2)\in X\times X$, we have for every $x=(x_1,x_2)\in X\times X$: $$ d(x_1,x_2)\le d(x_1,a_1)+d(a_1,a_2)+d(a_2,x_2), $$ and therefore $$\tag{1} d(x_1,x_2)-d(a_1,a_2)\le ...


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You could prove that the metric $d$ is a continuous function from $A\times A$ to $\mathbb{R}$ by considering the product topology on $A\times A$. Let $M = \left({A, d}\right)$ be a metric space. Let $\tau$ be the topology on $A$ induced by $d$. Let $\left({A \times A, T}\right)$ be the topological product of $(A, \tau)$ and $(A, \tau)$. ...


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1/ It seems the following. Yes, both the necessity and the sufficiency can be easily checked. Let $\{x_n\}$ be a sequence of vectors of the space $V$. Necessity. Assume that the sequence $\{x_n\}$ converges to a vector $x\in V$, $k\in\Bbb N$ be an arbitrary index, and $\varepsilon>0$ be arbitrary. Choose a number $\delta>0$ such that ...


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Could anyone help me with solution for the following question. Let X be a normed space. Then Riesz lemma for r=1 holds iff for every proper closed subspace of X, there exist x in X and y in Y such that the norm of x-y equals dist(X,Y).


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I think there is a much easier proof if you know some linear algebra. As in the text, it's enough if we show that $$||\beta_1 x_1 + \beta_2 x_2 + \ldots + \beta_n x_n|| \geq c $$ where $\sum_{j=1}^{j=n} |\beta_j| = 1 $ or $|\beta|_1 = 1$. Let $X$ be the matrix whose nth column is $x_n$. Let $\beta$ be the vector whose nth entry is $\beta_n$ Then we want ...


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


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Not every normed space will provide an inner product, because there are normed spaces which are not prehilbert. The additional point to be satisfied is parallelogram law. The result you need is just the Jordan-Von Neumann Theorem.


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


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For your second question the above answer is good.... For the first one let $x,y \in \overline{D(T)}$, i.e., there exist sequences $(x_n),(y_n)$ in $D(T)$ such that $x_n \longrightarrow x$ and $y_n \longrightarrow y$, then $(x_n+y_n)$ is a sequence in $D(T)$ (as $D(T)$ is a vector space) such that $x_n+y_n \longrightarrow x+y$ and if $\alpha$ is any scalar ...


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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 ...


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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 ...


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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 ...


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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 $x_1(t) = x_2(t)$, ...


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If $\parallel x\parallel =M$ then $$ |T(x)(t)|=|\int_0^t x(s)ds | \leq \int_0^t M \leq M $$ Hence bounded. And $\frac{d}{dt} T(x)(t)=x(t)$ is continuous. And $T(x)(0)=0$. If $T(x)=T(y)$ then $ \parallel T(x)- T(y)\parallel =0$ So $$ \forall t,\ \int_0^t (x-y)(s) ds =0 $$ Assume that $t_0\in (0,1)$ with $(x-y)(t_0) >0$. Then $x-y \geq c> 0 $ on $ ...


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Yes. Your answer about range is correct. The operator is injective. To see it, since $x(t)$ is continuous at $[0,1]$, you can apply Fundamental theorem of calculus, given $F(t)\in R(T)$, $x(t)$ is uniquely determined by $x(t)=F'(t)$. Or you can argue directly that if $x_1(t_0)\not =x_2(t_0)$, for some $t_0\in[0,1]$, by continuity $T(x_1)(1)\not =T(x_2)(1)$ ...


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You can define $G$ as follows: if $x_n \in D(T)$ and $w \in W$ with $x_n \to w$, then $G(y) = \lim_{n \to \infty} T(x_n)$ (use boundedness of $T$ to show that this is well-defined). Then if $x_n \to w$ and $y_n \to v$, and $a,b$ are scalars, you want to show that $G(aw + bv) = a G(w) + b G(v)$. Well, what sequence (defined in terms of $x_n$, $y_n$, $a$, ...


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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 $[0,1]$).


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This question Limit of convolution answers the question for general convolution operator on $L^p$.


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First, suppose $T$ is bounded. Let $\{x_n\}$ be a sequence in $X$ converging to zero. Then, $\{\lVert x_n\rVert\}$ also converges to zero and hence is bounded (by some $M$). Thus, $\lVert Tx_n \rVert \leq \lVert T \rVert \lVert x_n \rVert \leq \lVert T \rVert M < \infty$. Conversely, suppose $T$ maps zero-convergent sequences to bounded sequences. If $T$ ...


0

Your proof is virtually complete as it stands (apologies for writing it out again with slightly different notation in an earlier version of this answer). To show that $X \cup Y \cup Z$ is connected where $X$, $Y$ and $Z$ are connected, you do not need to show that $X \cap Y \cap Z$ is non-empty. All you need is to show that two of $X\cap Y$, $X \cap Z$ and ...


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 ...


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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 ...


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 ...


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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 ...


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

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 ...


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As $f(\alpha x) = |f(x)|$, $f(\alpha x)$ is real, so it is equal to its real part. That is, $f(\alpha x) = u(\alpha x)$.


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By definition of the contraction, you need to show that there exists such $0\leq k < 1$ that $$d(f(x), f(y))\leq k d(x, y)$$ for all $x, y\in\mathbb{R}^N$ Now, here the distance $d$ is a normal euclidian distance on $\mathbb{R}^N$, so this boils down to proving that $$\sqrt{\sum_{i=1}^n (f_i(x) - f_i(y))^2}\leq k\sqrt{\sum_{i=1}^n (x_i - y_i)^2}.$$ Try ...


0

Yes, you can write Holder's inequality for random vectors (for $\mathbb{R}^n$-valued functions more generally). $$\mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} |X^i(\omega)Y^i(\omega)|\mathrm{d}\mathbb{P}\right]\leq \mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} |X^i(\omega)Y^i(\omega)|^p\mathrm{d}\mathbb{P}\right]^{1/p}\mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} ...


1

I think you should review the notations in your question and from what I understood I don't think that you even need to use the fact that $J$ is an isometry for that passage as you can see that for any $f \in Y^*$, \begin{align*} \Psi_n(f) &= \varphi_{\Psi_n}(f) & & (\text{ if this is what you meant })\\ & = f(\Psi_n)\\ &= f(T(x_n)). ...


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 ...


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)})} = ...


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 ...


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 $\epsilon >0$. ...


1

Since no one provided the answer, I write mine. So, it seems the following. Parallelogram law implies that it suffices to verify the claim for two-dimensional vector spaces. So, let on the plane $\Bbb R^2$ is defined a norm $\|\cdot\|$ satisfying the proposed equality. Let $K=\{x\in\Bbb R^2:\|x\|\le 1\}$ be the unit ball of the norm $\|\cdot\|$. Since the ...


0

Such example does not always exist. For example, if $\dim X<\infty$, $X$ is isomorphic to $\mathbb{R}^n$ and every convex function $f:\mathbb{R}^n\to \mathbb{R}$ is continuous. For $\dim X=\infty$, recall that every infinite-dimensional normed space contains a discontinuous linear functional, say $f$. Let $g=f+c$, with constant $c\not=0$. Then $g$ is ...


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 $C(K)$.


0

The result that you are trying to use will not help as it is valid only for finite-dimensional spaces and clearly the setup in your problem is for infinite-dimensional spaces because any proper subspace of a finite-dimensional normed space is closed and hence it can not be dense. For showing the first part what you can do is as follows. We need to show that ...


0

Given $\varepsilon > 0$, there is $n_0 \in \mathbb{N}$ such that: $$ m,n \geq n_0 \implies \Vert f_m - f_n \Vert < \frac{\varepsilon}{4}$$ Then, we have: $|f_m(0) - f_n(0)|< \varepsilon /4$ $\sup \left\{ \frac{|(f_m-f_n)(t)-(f_m-f_n)(s)|}{|t-s|^p}: t\neq s, \quad t,s\in [0,1] \right\} < \varepsilon /4$ Given $t\in]0,1]$, we have for all ...


1

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


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 ...


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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 ...



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