# Tag Info

1

As PhoemueX said: the statement follows from the Uniform Boundedness Principle applied to linear maps $\phi_n:X^*\to \mathbb{K}$ that are defined by $\phi_n(f)=f(x_n)$. ($\mathbb{K}$ stands for $\mathbb{R}$ or $\mathbb{C}$.) The principle applies here because $X^*$ is complete even when $X$ isn't.

1

Given a continuous mapping $F:X\rightarrow Y$ it induces continuous map $F^{**}:X^{**}\rightarrow Y^{**}$ given by the formula $$[F^{**}(x^{**})](y^*):=x^{**}(F^*(y^*)).$$ This assigment is functorial, i.e. it has two properties: $(id_X)^{**}=id_{X^{**}}$ and $(G\circ F)^{**}=G^{**}\circ F^{**}.$ Additionaly for every normed space $X$ we have continuos ...

2

Given a bijective linear isometry $T:X\rightarrow Y$, the dual map $T^*:Y^*\rightarrow X^*$ is also a bijective linear isometry. (This follows from the fact that $T^*$ has inverse $(T^{-1})^*$ and $\|T^*\|=\|T\|$.) From this, we have that $T^{**}:X^{**}\rightarrow Y^{**}$ is a bijective linear isometry as well. Let $J_X:X\rightarrow X^{**}$ is the canonical ...

1

Here is a theorem (paraphrased) from Ivan Singer's Bases in Banach Spaces, Vol. 1 (Springer-Verlag 1970). In what follows, "basis" means "Schauder basis". Proposition 4.3 (pg. 29): Let $E$ be a Banach space with a bounded basis $\{x_i\}$. Let $\{\alpha_i\}$ be a sequence of non-zero scalars and for each positive integer $n$, let y_n=\sum\limits_{i=1}^n ... 1 If x is analytic, \begin{align} T(t)x & = x+\int_{0}^{t}T(s)Ax ds \\ & = x + (s-t) T(s)Ax|_{s=0}^{t}-\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax -\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax +\left.\frac{t^{2}}{2!}A^{2}x\right|_{s=0}^{t}+\int_{0}^{t}\frac{(s-t)^{2}}{2!}T(s)A^{3}xds ... 2 Your proof is fine. You constructed a countable subset of E\subset X, then proved that each x\in X has a sequence (y_n)\subset E such that y_n\rightarrow x in your topology. You use the linearity of X and the separability of B_X to construct E and y_n, and the Hausdorff condition (implicitly) to prove that y_n\rightarrow x. There's ... 0 As Norbert said, this is false. When considering sequences, we don't get the accurate picture of weak topology on the unit ball of \ell^1: in fact, we can't even tell it from the norm topology. Theorem (Schur) A sequence in \ell^1 converges weakly if and only if it converges in the norm. An accessible proof of this theorem is easy to find ... 0 Let N be the null space of T. Then N is closed in X because T is bounded. So X/N is a Banach space. And T becomes \dot{T} on X/N, where \dot{T}(x+N)=T(x). Clearly \dot{T} is bounded because T is bounded. \dot{T} has inverse \dot{T}^{-1} from \mathscr{R}(T) to X/N that is also bounded because, by assumption, ...

0

You might know that a subspace of a Banach space is closed if and only if it is a Banach space. So, it is enough to show that $Im(T)$ is a Banach space. To show that $Im(T)$ is a Banach space I will use the following well known result. $\textbf{Result}:$ Let $X$ be a normed space. Then absolute convergence of a series implies convergence of the series if ...

0

You shouldn't say that the $y_n$ are convergent but that they are Cauchy in the image but otherwise I think this is fine. Though you should really fill in the details showing that $T(x_n)\to y = T(x)$. You don't have injectivity but the norm condition supplants it and lets you get away without it.

3

Yes, it's complete. First, observe that your norm is comparable to the simpler norm $$\|f\| = \sup_{n\in\mathbb{Z}} \int_n^{n+1}|f(y)|\,dy \tag{1}$$ Indeed, $\|f\|\le |f|\le 2\|f\|$ because every interval of length $1$ is contained in some interval of the form $[n,n+2]$. The space with $(1)$ is just the direct sum $\bigoplus_\infty X_n$ of Banach spaces ...

2

For completeness, I post the proof here. The first two facts are valid for all metric spaces. If a Cauchy sequence has a convergent subsequence, it converges. Every Cauchy sequence contains a subsequence (denoted $\{x_n\}$ here) such that $$d(x_n,x_{n+1})\le 2^{-n},\quad n\in\mathbb{N} \tag{1}$$ (Note that $(1)$ implies being Cauchy.) So, it suffices ...

1

Recall that an operator is closed if a sequence $(x_{n})\subset D(T)$ converges to some $x\in X$ and $Tx_{n}$ converges to $y\in Y$ implies $x\in D(T)$. Equivalently, the graph of $T$ is closed in the direct sum $X\oplus Y$. Proof of i). Since $T$ is injective, we can define a linear operator $T^{-1}$ on the range of $T$, denoted $R(T)$, by $T^{-1}y=x$, ...

1

In your situation, you have $x_n \to x$. This can be proved by contradiction. Suppose $(x_n)$ does not converge to $x$. Then, there is $\varepsilon > 0$ and a subsequence $(x_{n_k})$ with $\|x_{n_k} - x\| \ge \varepsilon$ for all $k$. But this subsequence converges weakly to $x$ and, hence, is bounded. By compacity of $K$, a subsequence converges ...

0

In fact, we can directly conclude that $x_n \to x$ in the norm topology. Compact metric spaces are sequentially compact, so every sequence in $K$ has a convergent (in the norm topology) subsequence. Thus every subsequence $(x_{n_k})$ of $(x_n)$ has a subsequence $\bigl(x_{n_{k_m}}\bigr)$ that converges to some $y \in K$. But that subsequence also converges ...

1

The space $\ell_2$ is the space of infinity sequences square additive. You can map a vector in $\ell_2$, for example $(\xi_1, \xi_2 ,\cdots, \xi_k \, \cdots)$ to a power series $\xi_1 + \xi_2 z + \cdots, \xi_k z^k + ...$. This is known as the Z transform. think about $z=\exp(i \omega \Delta t)$ where $\omega = 2 \pi f$, and $f=1/\Delta t$ is the frequency. ...

1

First of all: since all norms on a finite-dimensional space are comparable, the condition could be simpler stated as $$C_1\|x-y\| \leq d(x,y) \leq C_2\|x-y\|$$ where $\|\cdot \|$ is a norm of our choice, e.g., Euclidean. Second: the answer is negative, for example $$d(x,y) = |x-y|+\min(|x-y|,1)$$ is a translation-invariant metric on $\mathbb{R}$ that ...

1

Let $S$ be the subspace of $c$ of convergent sequences $x = (x_1,x_2,\dotsc)$ satisfying $\sup x + \inf x = 0$. Let $y_\alpha$ denote the constant sequence $\alpha, \alpha, \dotsc$. Let $h(x)$ denote the number $\sup(x) + \inf(x) \over 2$ Then the map $c/Y \rightarrow S$ given by $[x] \mapsto x - y_{h(x)}$ is an isometry onto $S$.

3

It is well-known (and easy to see) that $c/Y$ is isomorphic to $c_0$. So we just need to renorm $c_0$ so that the natural map between them is an isometry. In fact, the norm $|||\cdot|||$ will do, where \begin{equation*}|||(x_n)|||=\frac{1}{2}\sup_{m,n}|x_n-x_m|.\end{equation*} Notice that for any $(x_n)\in c_0$ we have ...

0

This is a consequence of Riesz's lemma.

1

The following claim is not true in general. Let $X$ and $Y$ be Banach spaces and $X_1 \subset X$ a subspace. If $T: X_1 \to Y$ is a continuous bounded operator, then there exists a continuous linear extension to the whole space $X$, i.e. there exists a bounded linear operator $\hat T: X \to Y$ such that $\hat{T}\restriction_{X_1} = T$. Take for ...

1

Inspired by this question (Nonlinear function continuous but not bounded), one can construct a counterexample as follows: Let $X =c_0 (\Bbb{N})$, i.e. the space of null-sequences, equipped with the sup-norm. Define $$F((x_n)_n )= \sum_n x_n^{2n}.$$ Since the function $x \mapsto x^{2n}$ is convex, it is easy to see that every summand of the series ...

1

To answer that question, it is most convenient to solve $|f'(x)| < 1$. Since $f'(x) = \frac{1}{2} \left[ 1 - \frac{a}{x^2} \right],$ the solution for positive $x$ is $x \in \left( \sqrt{ \frac{a}{3} }, \infty \right)$. So: for every $b > \sqrt{\frac{a}{3}}$ there is such $L \in (0, 1)$ that $|f'(x)| \leqslant L$ for $x \geqslant b$ and so by the mean ...

0

The answer to question 1 is negative. For example, $L(\varphi)=\varphi(0)$ is not of this form. Neither is $L(\varphi)=\varphi'(1/2)$, etc. A continuous functional on $E$ is bounded by some finite collection of seminorms; therefore it is also a continuous functional on $C^m([0,1])$ for some $m$. The space $C^m([0,1])$ is a direct sum $P_m\oplus V_m$ where ...

1

Start with $M_0 = E$. Pick $y_1 \in E$, say with $\lVert y_1\rVert = 1$. By the Hahn-Banach theorem, there is an $f_1 \in E'$ with $\lVert f_1\rVert = 1$ and $f_1(y_1) = 1$. Let $M_1 = \ker f_1$. After having found $y_k$ and $M_k$ for $1 \leqslant k \leqslant m$ with $\lVert y_k\rVert = 2^{1-k}$, $y_k \in M_{k-1}\setminus M_k$ and $\dim (E/M_k) = k$, you ...

1

I suppose you are talking about some linear space. In this case, we know that the partial sums $$P_m := \sum_{n=0}^{m} K(GK)^n$$ are in $S$, because they are just finite sums of elements in $S$. If the limit $P = \lim_{m \to \infty} P_m$ exists, we know that $P \in S$ since $S$ is closed. However, I believe that there is some kind of convergence argument ...

0

Ok, here I am with the answer to my question. Let $(f_{n})\subset B(X)$ be a Cauchy sequence. Given $x \in X$ we have that $(f_{n}(x)) \subset \mathbb{R}$ is Cauchy (in $\mathbb{R}$). Lets see this. Since $(f_n) \subset B(X)$ is Cauchy, given $\epsilon > 0$ there exist $n_{0}\in \mathbb{N}$ such that $n,m \geq n_0$ implies $d_{\infty}(f_n,f_m)=sup_{x \in ... 2 Hint: given$x \in X$, the sequence of real numbers$a_n = f_n(x)$is a Cauchy sequence in$\mathbb{R}$. That should tell you all you need to know about the value of the desired limit function$f$at$x$. 1 Unbounded, convergent nets are constructed in this answer. As for what you are trying to prove, I don't think it holds. For instance let$E=c_0(\mathbb N)$, then$E^*=\ell^1(\mathbb N)$. Consider elements$f_n\in E^*$given by $$f_n(k)=\begin{cases}1,&\ k=n\\ 0,&\ k\ne n\end{cases}$$ and let$\tau:E^*\to\mathbb C$be$\tau(f)=\sum_{k=1}^\infty ...

0

We note that $$|y_k| \leq |y_k - x^n_k| + |x^n_k| = \lim_{m→∞}|x^m_k - x^n_k| + |x^n_k| \leq\lim_{m→∞} ‖x^m - x^n‖ + |x^n_k|$$ which holds for arbitrary $n$. (the superscript is which sequence, the subscript is the index of the sequence) As $x^n∈ c_0$, choose $K_n$ large such that $|x^n_k |<ε$ for $k>K_n$. Pick $N(ε)$ large such that $\|x^m - ... 1 Suppose$x^k \in c_0$and$x^k \to x$. Let$\epsilon>0$and pick$N$such that$\|x^k-x\|_\infty < {1 \over 2 } \epsilon$for$k \ge N$. Since$x^N \in c_0$, there is some$N'$such that$|x_i^N| < {1 \over 2 } \epsilon$for$i \ge N'$. Then$|x_i| \le |x_i^N| +|x_i-x_i^N| \le |x_i^N| +\|x-x^N\|_\infty <\epsilon$. Hence$x_i \to 0$and so$x \in ...

0

The triangle inequality is obvious, if you use the fact that the euclidean distance is a metric together with "monotonicity" of the Riemann-integral. For the completeness i think this function might work as a counterexample $$f_{n}(x) = \left\{ \begin{array}{lr} 0 & : x \in (-1,-\frac{1}{2})\\ (x+\frac{1}{2})^{n} & : x \in ... 1 "Is it always the case that"? No, why should it? Let X = Y = \Bbb R, let U = (0, 1)\subseteq \Bbb R, and let y = 0. The (open) line segment from (0, 0) to (1, 0) is not open in \Bbb R^2. Whenever you have an idea of some property, you should always check it out in the most familiar settings (in the case of vector spaces, and much of topology and ... 1 This would only be the case if \{y\} is open in Y, which would imply that Y = \{0\}. 2 Yes. More generally this holds for separable Banach lattices that are dual spaces (spaces with unconditional bases are Banach lattices with the positive cone consisting of vectors having non-negative coefficients with respect to a given unconditional basis). Indeed, we have the following generalisation of a theorem of James (This is Theorem 1.4.c in the ... 2 It's just the English that is messing up with you. The plural is used like when you say "crossing streets without paying attention is dangerous"; where's the "second street"? And interchange refers to exchanging the position of \lim and \langle. 1 From Minkowski,  \| \sum_1^n |f_k|\|_p \le \sum_1^n \|f_k\|_p \le \sum_1^\infty \|f_k\|_p. Now raise to the pth power. 1 Triangle inequality then dominated convergence theorem. Edit: By triangle inequality, we have,$$\int G^p= \lim \int G_k^p=\lim \int (\sum_{k=1}^n |f_k|)^p=\lim ||(\sum_{k=1}^n |f_k|) ||^2\leq \lim \sum_{k=1}^n ||f_k||^2$$By DCT, we then conclude the given inequality. 0 Your idea of proving the remaining direction by contradiction seems to be a sensible approach. A couple of hints to help you along: Assuming M+N is complete, the function f:M+N\to \mathbb{R}: m+n\mapsto ||m||+||n|| is continuous and well-defined (why?). Now, take (m_k,n_k) such that ||m_k||+||n_k||\geq k ||m_k+n_k||. What can you say about ... 1 If you look at this answer, we need to check whether your operator satisfies or not the equality \text{ess nul}(A)=\text{ess nul}(A^*). We have |A|=(A^*A)^{1/2}=I-E_{11}, where E_{11}(x_1,x_2,\ldots)=(x_1,0,0,\ldots), and |A^*|=(AA^*)^{1/2}=I. For A, E_A[0,\epsilon]=E_{11} for all \epsilon\in(0,1), so \text{ess nul}(A)=1. For A^*, ... 2 I think you are making your estimate too convoluted, probably because you try to calculate \|f_n-f_m\|_1 exactly. What I would do: assume m>n; then f_m=f_n=1 for x>1/n, and so$$ \|f_m-f_n\|_1=\int_0^{1/n}|f_m-f_n|\leq\frac2n $$since |f_n|\leq1, |f_m|\leq1. This shows that the sequence is Cauchy, because given \varepsilon>0, if ... 1 Hint: consider m>n, m = n + h, and prove that the expression converges to 0 at a speed independent of h. In other words: prove that$$ \sup_{h>0} \left\| f_{n+h} - f_n \right\| \to 0 when $n\to\infty$.

3

Hint: The determinant is a polynomial in the matrix entries. If it was identically zero in any open subset of $\mathbb R^{n\times n}$, it would need to be the zero polynomial, but it isn't.

0

Let $X$ be a Banach space. A sequence $\{X_n\}_{n=1}^\infty$ of finite-dimensional subspaces of $X$ is called a finite-dimensional decomposition of $X$ if every $x\in X$ has a unique representation of the form $x=\sum_{n=1}^\infty x_n$ with $x_n\in X_n$ for every $n\in \mathbb{N}$. This is Definition 4.31 on page 198 of Banach Space Theory: The Basis ...

1

A few comments: 1) you don't need "contradiction". Your argument (if right) shows that if $g\in X^*$, then $g$ is in the span of $F_N$. 2) I don't see why you claim that $\cap_{f\in F_N}\ker f\subset\ker g$ implies $g\in \text{span}\,F_N$. 3) You seem to prove that $X=\text{span}\,\{F_N:\ N\}$ without closure, which looks suspicious to me. 4) Here is ...

2

In this answer, I will use $x_n$ as a sequence in $l^2$ and write $x_n(k)$ as the $k$-th member of that sequence. The norm in the Hilbert space is given by $\|x\| = \sqrt{\langle x, x \rangle}$. We wish to show that if a sequence $\{ x_n \} \subset l^2$ is Cauchy, then it converges in $l^2$. Suppose that $\{x_n\}$ is such a Cauchy sequence. Let $\{ e_k \}$ ...

0

Hint : Let $(x^n)$ be a Cauchy sequence. We can write $x^n=\sum _k x_k^ne_k$ where $(e_k)$ is the canonical basis. Show $|x_k^n-x_k^m|\leq \|x^n-x^m\|$ so $(x_k^n)_n$ is a Cauchy sequence in $\mathbb{C}$, so there exist $x_k$ such that $\lim x_k^n=x_k$ for every $k$. Define $x=\sum x_ke_k$ and prove that $\|x^n-x\|\to 0$.

0

Let $(\mathbf{x_n})$ be a Cauchy sequence in $l^2$, where $\mathbf{x_n} = (x_1^{(n)},x_2^{(n)},\ldots)$, i.e., given $\epsilon >0$ there exists a natural number $N$ such that for all $m,n\geq N$ $$\|\mathbf{x_n}-\mathbf{x_m}\| = \left(\sum_\limits{j=1}^{\infty}|x_j^{(n)}-x_j^{(m})|^2\right)^{\frac{1}{2}} <\epsilon$$ In ...

3

You know that $\|S\| \le 1$ which gives $\|S^{\star}\|=\|S\| \le 1$ and, hence, $\|S+S^{\star}\| \le 2$. However $S+S^{\star}$ is selfadjoint and, so, its norm and spectral radius are the same, and its spectrum is real. That means $\sigma(S+S^{\star})\subseteq[-2,2]$. So that part is fairly straightforward. Suppose $\lambda \in [-2,2]$. You want to show ...

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