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13

The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. ...


5

As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) $\mathbb R^\infty=\bigcup_n \mathbb R^n$, where $\mathbb R^n$ has its usual topology and is included in $\mathbb ...


5

By definition of bilinearity, $\langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle$. On the other hand in general there is no formula for $\langle a,b\rangle \pm \langle c,d\rangle$.


4

I think that if you want to work with norms on vector spaces over fields in general, then you have to use the concept of valuation. Valued field: Let $K$ be a field with valuation $|\cdot|:K\to\mathbb{R}$. This is, for all $x,y\in K$, $|\cdot|$ satisfies: $|x|\geq0$, $|x|=0$ iff $x=0$, $|x+y|\leq|x|+|y|$, $|xy|=|x||y|$. The set $|K|:=\{|x|:x\in ...


3

The answer for infinitely many functionals is no; there's a counterexample here. For finitely many functionals it must be yes... Right. First, there is a norm on any real vector space $X$, for example if $B$ is a (Hamel) basis define $$\left\vert\left\vert\sum_{b\in B}c_b b\right\vert\right\vert=\sum_{b\in B}|c_b|.$$Now if $||\cdot||$ is a norm on $X$ ...


3

Your answer is incorrect. $\langle f,g \rangle$ is allowed to take any value, but $\langle f,f \rangle$ must be non-negative. Try to come up with an example of two vectors whose dot-product is negative, noting that the dot-product is the prototypical inner product. The property of inner products that fails here is that $$ \|f\|^2 = \langle f,f \rangle = 0 ...


3

There are two cases: $0\in C$ and $0\notin C$. In the first case the minimum norm is $=0$, and $x_0=0$ is the unique element with this property. If $x_0$ with minimum norm has already been found (and your question indicates that this is already known to you) assume $x_1\neq x_0$ has the same norm and it is an element of $C$. It is easy to see that $x_1$ and ...


3

$y=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...) \notin c_{00}$, $y_1=(1,0,0,...)$, $y_2=(1,\frac{1}{2},0,0,...)$ and so on. $y_n \in c_{00}$ and $\|y_n -y \|_{\infty}=\frac{1}{n+1}$ which tends to $0$ when $n$ tends to $\infty$.


3

Note, that $\|\cdot\|$ is somehow the composition of the $2$-norm and the $3$-norm on $\mathbb R^2$. We have $$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$ and with this at hand the proof is easy (but a bit ugly). We have by the triangle inequality for the $2$-norm $$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:b$$ and since for all ...


3

This is just an elaboration on what others have already said in the comments. Let $V$ be a finite-dimensional normed linear space (over a subfield $\mathbb F$ of $\mathbb C$). As you say, all norms on a finite-dimensional vector space are equivalent, so we may assume that we have the usual 2-norm on $\mathbb C^n$. The definition of weak convergence is ...


3

The $\ell^p$ spaces are a special case of the $L^p$ spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on $\mathbb{N}$.


2

It is true, it is a consequence of the uniform boundness principle. https://en.wikipedia.org/wiki/Uniform_boundedness_principle Apply the contraposition. Consider the family $T^n(x) =(Tx)_n$ and suppose that $\sup_{T^n,\|x\|=1}\|T^n(x)\|=\infty$. Then there exists $x$ such that $\sup_n\|(Tx)_n\|=\infty$. This is impossible since $T$ must be defined at $x$. ...


2

Yes, the inequalities hold. Let $B$ be the closed unit ball for the norm $\|\cdot \|$. The assumptions imply that $\pm e_j$, the standard basis vectors, are in $B$. Hence, their convex hull is contained in $B$. This convex hull is the unit ball for the $\ell^1$-norm, which implies $\|\cdot \|\le \|\cdot \|_1$. Similarly, we need to prove that $B$ is ...


2

Yes, it is. In fact, we can say more. If $K$ has the trivial valuation, $X$ is a finite-dimensional $K$-vector space, and $p$ is any norm on $X$, then there exist positive constants $c_1$ and $c_2$ such that $c_1\leq p(x)\leq c_2$ for all nonzero $x\in X$. To get $c_2$, note that if $\{e_1,\dots,e_d\}$ is a basis for $X$, then writing $x=\sum a_i e_i$, we ...


2

You need to ask yourself, is $V$ is a closed subspace of $X$? Namely, given a sequence $v_n\in V$ such that for some $v\in X$, $\|v_n-v\|_\infty\to0$, does that imply that $v\in V$ as well? (You could try and show that $X\setminus V$ is open, but that's harder.)


1

It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence $(x_n)_{n=1}^\infty$ you have constructed converges, then its limit belongs to $C$. But so far you have only that $\|x_n\| \rightarrow s$. It remains to be shown that $(x_n)_{n=1}^\infty$ does converge. To this end, ...


1

I think the answer is yes. A sign- and permutation- invariant norm defined on $\mathbb C^n$ is called a symmetric gauge function. It is known that every unitarily invariant norm on $M_n(\mathbb C)$ is induced by a symmetric gauge function. See, e.g. theorem 7.4.24 on pp.438-440 of Horn and Johnson (1985), Matrix Analysis, 1/e, Cambridge University Press. To ...


1

Let $x\in X$ and $r>0$, and consider the open ball $B(x;r)=\{y\in X:\|x-y\|<r\}$. For any $y,z\in B(x;r)$ and $t\in(0,1)$, we have \begin{align} \|ty + (1-t)z - x\| &= \|t(y-x) + (1-t)(z-x)\|\\ &\leqslant t\|y-x\| (1-t)\|z-x\|\\ &<r, \end{align} so $B(x;r)$ is convex. Write $f(t) = t(a-b) + t$, then it is clear that $f$ is an affine ...


1

Using the Axiom of Choice, one can construct in a given infinite dimensional normed linear space $X$, a discontinuous linear functional (see this, e.g.). The kernel of such a functional is a proper and dense subset of $X$ (see this). This would provide a subspace $Y$ of $X$ that is proper and such that for every $x\in X$, $\text{dist}\,(x,Y)=0$. From ...


1

You have $M=\mathcal{N}(f)$. Then $g\in M^{\perp}$ iff $\mathcal{N}(f)\subseteq\mathcal{N}(g)$. Suppose $f(u)\ne 0$ for some $u\in X$.Then $f\left(y-\frac{f(y)}{f(u)}u\right)=0$ holds for all $y\in X$, which gives $$ 0= g\left(y-\frac{f(y)}{f(u)}u\right)=g(y)-\frac{f(y)}{f(u)}g(u),\;\;\;y\in X \\ \implies g = ...


1

No, not even if $X=Y=H$, a Hilbert space. Say $H$ is an infinite-dimensional Hilbert space and let $B$ be a Hamel basis for $H$. Say $m:B\to(0,\infty)$ is unbounded and define $T:H\to H$ by $$T\left(\sum_{b\in B}c_b b\right)=\sum_{b\in B}m(b)c_b b$$(where $c_b=0$ except for finitely many $b\in B$.) Then the kernel of $T$ is $\{0\}$, but $T$ is unbounded ...


1

Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have $$ ...


1

Suppose $F$ is such an extension. Then $F(x) = x$ for $x \in Y$. But since $Y$ is dense, that implies $F(x) = x$ for all $x \in X$. So we must have $Y = X$.


1

An infinite-dimensional vector $x = (x_1, x_2, x_3, \ldots)$ (or I guess we can just say sequence) is in $\ell^2$ iff $\displaystyle \sum_{k=1}^{+\infty} |x_k|^2 < +\infty$. Well, no matter what the value of $n$ is in $x = (1, 0, 0, \dots, 0, -n^2, 0, 0, 0, \dots)$, we'll always have $\displaystyle \sum_{k=1}^{+\infty} |x_k|^2 = 1 + n^4$, and this is ...


1

Suppose $w$ is in the image of $T$ then $w=(b_1/1,b_2/2,\dots$) for some bounded sequence $(b_n)$, but then $$\frac{b_n}{n}=\frac{1}{\sqrt n}$$ hence $b_n=\sqrt n$ which is not bounded. An accumulation point is the same as a limit point it just means that for any neighbourhood $U$ of $w$ there is a point $x\in \mathrm{Im}(T)$ such that $x\in U$.


1

Yes. First $f$ is surjective. Indeed, let $y \in Y$. $\frac{y}{||y||} \in B_Y = f(B_X)$ hence there is $x \in B_X$ s.t $f(x)=\frac{y}{||y||}$ and so $f(||y||x)=y$ by linearity. We conclude that $f$ is surjective and hence bijective (we already know it is injective by assumption) Because of the facts that $B_Y = f(B_X)$ and that $f$ is bijective, we have ...


1

If $a=\lim a_n$ and $b=\lim b_n $, then you have to show that $a+b=\lim (a_n+b_n) $.


1

Notice that we can write $$c_1^2 + c_2^2 = |\nabla v|^2 = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2,$$ where $\tau$ is the direction of the tangent line. By assumption, we know that the second term on the RHS is $0$ on $\Gamma_0$. What is left to show is that also the tangential part of the gradient is $0$. Do you see why this is true? EDIT: take ...


1

(i) is correct. (ii) Show that $\{ f \in X : f(0)=0\}$ and $\{ f \in X : f(1)=0\}$ are closed subspaces of $X$: from this, you will have that $V$ is an intersection of closed sets, hence closed. This can be done using sequences, and (i). Pick a sequence $f_n \subset \{ f \in X : f(0)=0\}$ converging to some $f$ in $X$. But (i) shows that $$f(0) = \lim_n ...


1

This is perhaps best understood from a topological perspective. A base for the weak topology is formed by finite intersections of sets of the form, $$\{u:a < \phi(u) < b\},$$ for any continuous linear functionals $\phi$. Geoometrically, one of these sets looks like the infinite slab between two parallel hyperplanes. In finite dimensions, you can ...



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