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2

The only thing which needs explanation is that if $x$ and $y$ are in $Y$, then $x+y$ is in $Y$. In all the examples which I worked out $Y$ turned out to be a subspace. Therefore, my guess is that the answer should be affirmative. You look in the wrong direction. Take $X = \mathbb{C}^5$, and look at $T$ given by the matrix $$\begin{pmatrix}0 & 0 ...


1

Note that in step one, it is given that for any multi-index $\alpha$ with $\|\alpha\| \le k$, $D^\alpha u_n$ converges uniformly to some function $u_\alpha$. In particular, $u_n \to u $ uniformly. You do not know, a priori, how all these $u_\alpha$'s are related. In particular, you do not know $u_\alpha = D^\alpha u$. (Actually, from step $1$ alone you do ...


2

Hint: for any $p\in P$, the degree of $f(p)$ is even.


0

Here's one more answer. Every closed subspace of $\ell_1$ contains a copy of $\ell_1$ but $L_1$ contains a copy of $\ell_2$ - for example take the closed linear span of the Rademacher system and apply Khinchine's inequality.


0

First, since $V$ is finite-dimensional, the image of all transformations in $L(V, W)$ have finite rank, and are therefore bounded, so we do have the operator norm here. Otherwise I'd insist that we use $B(V, W)$ instead. Additionally, we may assume without loss of generality here that $V$ is $\mathbb{R}^n$ with the $1$-norm, for some $n$, since every two ...


2

Nothing would go wrong per se, but in an incomplete normed space, there are simply much fewer weakly compact sets. Let $X$ be a normed space, and $\tilde{X}$ its completion. Then we have a canonical isometric isomorphism $\rho \colon \tilde{X}' \to X'$, namely the restriction of a functional to $X$. Thus we may identify the two spaces (that is habitually ...


1

Approximate Unit Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$ Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$ Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$ (That is the hard part!) Quotient Norm Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$ ...


4

$\left\Vert p_{n}-p_{m}\right\Vert =\sup_{\left[ 0,1\right] }\left\vert %TCIMACRO{\dsum \limits_{i=0}^{n}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{n}} %EndExpansion \frac{x^{i}}{i!}-% %TCIMACRO{\dsum \limits_{i=0}^{m}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{m}} %EndExpansion \frac{x^{i}}{i!}\right\vert =\sup_{\left[ 0,1\right] ...


0

Special thanks to Simon B. By uniform extension: $$\varphi\in U'\implies\overline{\varphi}\in\overline{U}'$$ This happens uniquely: $$\overline{U}=E\implies\ker\pi=(0)$$ Especially one has then: $$\|\pi[\overline{\varphi}]\|=\|\overline{\varphi}\circ\iota\|=\|\varphi\|=\|\overline{\varphi}\|$$ Conversely assume: $$U=\overline{U}\neq E$$ By Riesz lemma: ...


0

Nate has already answered this perfectly well, but perhaps it is worth noting the question becomes a little more difficult if we require $X^+$ to be large enough to generate $X$, i.e. $X=X^+-X^+$. For an example in this case, let $$X^+=\{(x_n)\in c_0:\forall n(0\leq nx_{2n}\leq x_{2n+1})\},$$ which is complete w.r.t. the supremum norm. Then $X=X^+-X^+$ ...


0

You have that if $a \in l_q\big(\Bbb N\big)$ then $a \in l_q\big(\Bbb Z\big)$ since a sequence of natural numbers is a sequence of integers. You also have if $a$ and $b$ are two sequences of natural numbers then so is $ a*b$, since the convolution only involves multiplications and additions. So the equality you wrote holds for sequences of natural numbers ...


0

Since the codimension of $R$ is one, the quotient space $Y/R$ has dimension one. (For quotients of Banach spaces, see this earlier MSE question, for example.) Since $Y/R$ has dimension one, there is a linear isomorphism $f:Y/R\to\mathbb R$ (assuming you are working over the reals). The quotient map $p:Y\to Y/R$ is linear and continuous and its kernel is $R$. ...


5

No, this is not true. If $D$ is a Dedekind finite set with a Dedekind finite power set, then $\ell_1(D)$ is a Banach space which has a Hamel basis which is also a Schauder basis, and every linear operator from $\ell_1(D)$ to a normed space is continuous. But if $D$ is Dedekind finite, then $|D|^{\aleph_0}>|D|$. So it suffices to assume that an infinite ...


2

As David Mitra mentioned in his comment, one such proof can be found in Morrison T.J. Functional Analysis. An Introduction to Banach Space Theory (Wiley, 2000), p.221. The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese ...


0

The condition (11.2) says that for all sufficiently large $x$, the image $Tx$ does not lie on the half-line $\{\lambda x : \lambda \ge 1\}$. If $T$ satisfies this condition, then so does $\sigma T$ for any $\sigma\in [0,1]$. This is what the second sentence of the remark means. I found the first sentence of the remark a little misleading, because the ...


1

It holds true that $q>p\implies \ell^p\subseteq \ell^q$ Indeed, let $(a_n)_{n\in\mathbb N}$ such that $\sum_n |a_n|^p<+\infty$ Since $q>p$, as soon as $|a_n|<1$, which holds from a certain $N$ onwards, $|a_n|^q<|a_n|^p$. So $\sum_{n\in\mathbb ...


0

We have $ \lVert x \rVert_p \leq \lVert x \rVert_1$ for all $1 \leq p < \infty $. Why? Because $|x_1|^p + \ldots + |x_n|^p \leq (|x_1| + \ldots + |x_n|)^p$ (from Newton's binomial formula). So $(|x_1|^p + \ldots + |x_n|^p)^{1/p} \leq |x_1| + \ldots + |x_n|$. For given $1 \leq q < \infty $ we have: $|1 \cdot x_1| + \ldots + |x_n \cdot 1| \leq |1+ ...


2

If $\|\cdot\|_B$ is a cross norm, then yes, your argument shows that $D_1 \otimes_a D_2$ is dense in $B_1 \otimes_a B_2$ and therefore also dense in $B$. Without this assumption or something similar, the answer is no, as then there is nothing at all relating the $B$ norm to the $B_1, B_2$ norms. This really has nothing to do with tensor products, so take ...


0

There is a direct analog of closed Graph theorem of non-archimedean fields. see page 61, theorem 3.5 in Non-archimedean functional analysis. A. C. M. van Rooij.


1

Assume $k$ field and $A$ a finite dimensional $k$ algebra. Then the spectrum of any element is finite. Step 1. Every element $a$ of $A$ satisfies a polynomial equation of degree at most $n= \dim_{k} A$. Indeed, the elements $1$, $a$, $\ldots$, $a^n$ are linearly dependent. Step 2. Let $a$ satisfying a polynomial equation $P(a) = 0$, where $P \in k[X]$ , ...


2

Consider abelian C*-algebra $C^*(a)$ which is infinite dimensional ( because $\sigma(a)$ is infinite). Also $C^*(a) \subset A$ which implies that $A$ is infinite dimensional.


0

If $B\subset X'$ then that double annihilator is the weak-* closure of $B$.


1

One approach is to consider $\dot{A} : X/\mathcal{N}(A)\rightarrow Y$, which is continuous and a linear bijection between $X/\mathcal{N}(A)$ and $Y$. So the inverse of $\dot{A}$ is continuous, and the statement of that continuity gives you what you want.


0

What i know about Hann Banach Theorem is the existence of enough functionals of a dual space on a given space, and these functionals sepearate points of the space. The sufficiency of these functional guaranteed enough maps in a dual space to work with.


0

Just use the definition: If $S_{n}=\sum _{k=0}^{n}x^{n}$ converges, then it is Cauchy so for all $\epsilon >0$ there is an integer $N$ such that whenever $n,m\geq N$, $\vert S_{n}-S_{m}\vert <\epsilon$. For this $N$, take $n=m+1$ and then $\vert S_{m+1}-S_{m}\vert=\vert x_{m+1}\vert $ and this is $<\epsilon $.


0

Let the sum be $s$; then, for every $\varepsilon>0$, there exists $m$ such that, for every $n>m$, $$ \biggl\|\sum_{i=1}^n x_i-s\biggr\|<\varepsilon $$ By the triangle inequality, for $n>m$, $$ \|x_{n+1}\|= \biggl\| \biggl(\sum_{i=1}^{n+1} x_i-s\biggr)+ \biggl(s-\sum_{i=1}^{n} x_i\biggr) \biggr\| \le \biggl\|\sum_{i=1}^{n+1} x_i-s\biggr\|+ ...


0

Following your notation, let $s_n = \sum_{i=1}^n x_i$. Assuming the series converges, $s_n$ is Cauchy. Rewriting this, for any $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m \geq n > N$, we have $$\sum_{i=n}^m ||x_n|| < \varepsilon.$$ In particular, setting $m = n$ shows that $x_n < \varepsilon$ for all $n \geq N$, as ...


2

Note that $x_n \to x$ weakly means that for every $f \in X'$, we have $f(x_n) \to f(x)$. Now, note that for any $g \in Y'$, we have $g(Tx_n) = g \circ T(x_n)$, and that $g \circ T \in X'$. From there, we simply apply the definition.


2

Every Banach space is a metric space. However, there are metrics that aren't induced by norms. If this were the case, then Banach spaces and complete metric spaces would the same thing... if metric spaces had operations and an underlying field! The difference is more than just the metric/norm dichotomy. For completeness of the answer (no pun intended), a ...


0

A complete metric space need not be a complete normed space. A complete normed space is also called a Banach space! Since every norm induces a metric, these Banach spaces reside in the collection of all complete metric spaces. Thus, complete metric space and complete normed space - two different notions but related indeed.


2

It was already covered in comments, but a complete normed vector space has a lot more structure than just a complete metric space. Every normed vector space is a metric space but not the converse. With a normed vector space, you have to have some idea of adding the elements, a field of scalars, an idea of scalar multiplication. In arbitrary metric spaces ...


2

Doesn't this operator has norm 1 ? $$\| Tp \| = \sup \{ |p(t^2)| : t \in [0,1] \} = \sup \{ |p(x)| : x \in [0,1] \} = \|p\|$$


1

Continuous linear functionals on $C$ correspond to signed measures on $[0,1]$: $\Gamma(f) = \int_[0,1] f \; d\mu$. Consider an absolutely continuous measure $d\mu = \rho(x)\; dx$. Then $\|\Gamma\| = \int_0^1 |\rho(x)|\; dx$. In order to have $\Gamma(f) = \omega |\Gamma\|$ with $\|f\| \le 1$ and $|\omega|=1$ you would need $f \rho = \omega |\rho|$ a.e. If ...


0

$c_0$ is a vector subspace of the closed space $c$, so you just have to prove: If $x_n \rightarrow x$ where $x_n \in c_0$ and $x\in c$, then $x\in c_0$.


1

Yes, you're looking at a sequence of sequences. The “distance” between two bounded sequences $(x_1,x_2,\ldots)$ and $(y_1,y_2,\ldots)$ is given by the so-called $\ell^{\infty}$-norm, which is defined as $$\|(x_1,x_2,\ldots)-(y_1,y_2,\ldots)\|_{\infty}\equiv\sup_{n\in\mathbb N}|x_n-y_n|.$$ The Cauchy property and convergence of a sequence of sequences should ...


1

(2.) If $g$ is injective then $g'$ is injective. Notice $g(x_1)=g(x_2)$ implies $(x_1,f(x_1)) = (x_2,f(x_2))$ hence $x_1=x_2$ which shows $g$ is injective. (my notation below may be non-standard, but, I believe when you clean-up the solution it's something like what follows:) (3.) Suppose $s(x,z)=z-f(x)$. Observe $g(x) = (x,f(x))$ implies $g'(x) = Id ...


1

The topology on $X\times Y$ is given by the norm $\|(x,y)\|=\max\{\|x\|,\|y\|\}$. So if $(f_n,f_n')\to(f,g)$, this means that $\|f_n-f\|\to0$, $\|f_n'-g_n\|\to0$, i.e. $f_n\to f$, $f_n'\to g$ uniformly. Fix $x\in X$. Then \begin{align} \left|\frac{f(x+h)-f(x)}h-g(x)\right| ...


1

The example mentioned in the question you quoted is what you are looking for. Let us take $X=C^\infty[0,1]$ with the supremum norm, so we don't have codomain issues, and let $D:X\to X$ be the derivative. This is an unbounded operator, and it's graph is closed. Indeed, if $(x_n,x_n')\to (x,y)$ in $X\times X$, this means that $x_n\to x$, $x_n'\to y$ uniformly. ...


-1

As $C(X)$ is a Banach space, you only need to prove as you noticed that $C_0(X)$ is closed in $C(X)$. Suppose that $f_n \to f$. $f$ is continuous because $C(X)$ is a Banach space. You only need to prove that $f$ vanishes at infinity. Take $\epsilon > 0$. By hypothesis you can find $N \in \mathbb{N}$ with $\Vert f-f_N \Vert_\infty < \epsilon/2$. As ...


1

Let $\epsilon >0$ and $n_0$ such that for all $n\geq n_0$, $\| f_n-f\|_\infty \leq \epsilon $. Since $f_n\in C_0(X)$ there is compact $K_0$ such that $|f_n(x)|<\epsilon $ for $x\in X\setminus K_0$. So $\sup _{X\setminus K_0} | f(x)|\leq \sup _{X\setminus K_0} | f(x)|+\|f_n-f\|_\infty \leq 2\epsilon $.


2

This is a typical $\varepsilon/2$-argument. Fix $\varepsilon>0$. Then there exists $n$ such that $\|f-f_n\|<\varepsilon/2$. Since $f_n\in C_0 (X) $, there exists $K\subset X $, compact, with $|f_n (x)|<\varepsilon/2$ for all $x\in X\setminus K $. So, outside $K $, $$ |f (x)|\leq|f(x)-f_n (x)|+|f_n (x)|<\frac\varepsilon2+\frac\varepsilon ...



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