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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 ...


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 ...


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

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

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

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

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 $S$.


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 ...


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. ...



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