Show that norm in $ \ell_{\infty} /c_0 $ is $||\overline{x} ||= \text{lim sup}|x_i|$ for each $x=(x_i)_i \in \ell_{\infty} $.  I'm doing $$||x_n+c_0||= \text{inf}_{y_n \in c_0}||(x_n)+(y_n)||_{\infty}= \text{inf}_{y_n \in c_0 } \text{sup}_{n \in \mathbb{N}}|(x_n)+(y_n)|$$  I do not know what to do now. I thought to fix $y_n$ and use $y_n \to 0$, but I can not get anywhere.
123266
I try to show that the norm on the quotient space $\ell^{\infty}(\mathbb{N}) / c_0 (\mathbb{N})$ is given by $\limsup_{n \in \mathbb{N}} |x_n|$, where $x = (x_n)_{n \in \mathbb{N}} \in \ell^{\infty} (\mathbb{N})$. My attempt is the following:     By definition, the quotient norm $\| \cdot \|_{\ell^{\infty} / c_0}$ is defined as   $$ \| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} = d( x + c_0 (\mathbb{N}), 0 + c_0 (\mathbb{N}) ), $$   which can be re-written as    $$ \| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} = \inf \{ \| x + y \|_{\infty} \; | \; y \in c_0 (\mathbb{N}) \}. $$   Since $0 = (0,0,...) \in c_0 (\mathbb{N})$, it follows that   $$ \| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} \leq \| x \|_{\infty} = \sup_{n \in \mathbb{N}} |x_n|. $$   I thought about using Bolzano-Weiserstrass' Theorem at a certain point, but I am not sure how to obtain the wished equality from here.    Any help is appreciated. Thanks in advance.
75160
Let $(y_n)\subseteq\mathbb{C}$ a sequence such that $\sum_{n=1}^{\infty}x_n\overline{y_n}$ converges for every $(x_n)\in c_0$. ($c_0$ is the space of complex sequences converging to zero with the $\infty$-norm.) Show that  $\sum_{n=1}^{\infty}|y_i|&lt;\infty$. Hint: uniform boundedness theorem.   Since $c_0$ is a Banach space, in order to use the uniform boundedness theorem we need linear bounded operators $T_n:c_0\to \mathbb{C}$. My first guess was:  $$T_n(x)=\sum_{i=1}^nx_i\overline{y_i}$$  Clearly every $T_n$ is linear but to prove $T_n$ is continuous seems to be the same as to prove what we actually desire.   Would anyone give me a hint about how to define the family of bounded operators, in order to use the theorem?  Thank you.
81241
I need to proof that $(C_0([0,\infty)), \ \|\cdot\|_\infty)$ is complete. I think that I did it correctly (or atleast am thinking in the right direction), but I still struggle with writing it down correctly. I used that $(C_b([0,\infty)), \|\cdot\|_\infty)$ is complete, so therefore as $C_0[0,\infty) \subset C_b[0,\infty) $, I wanted to use that if $A$ is a closed subset of a complete space $(M,d)$, that $(A,d)$ is complete. Thus, it came down to showing that $C_0[0,\infty)$ is closed.  Let $f_n(k)$ be a sequence of elements out of $C_0[0,\infty)$ that converges to an element $f$ out of $C_b[0,\infty)$. I need to proof that $f(k) \in C_0[0,\infty)$.   Per the definition of convergence, we know that $\|f_n(k) - f(k) \|_\infty \rightarrow 0$ whenever $n \rightarrow \infty$. We use this to find $ |f_n(k)-f(k)| \leq\|f_n(k) - f(k)\|_\infty  \rightarrow 0$.  Then we consider $|f(k)| = |f(k) - f_n(k) +f_n(k)| \leq |f(k)-f_n(k)| + |f_n(k)|$. If we take limits on both sides, we get: $\lim_{k\to \infty} |f(k)| \leq \lim_{k\to \infty} |f(k)-f_n(k)| + \lim_{k\to \infty} |f_n(k)| = 0 + 0 = 0$ (the first term follows from what we have previously stated and the second term follows from the fact that $f_n(k)$ is an element of $C_0$. Thus, $f \in C_0$ and therefore $C_0$ is closed and therefore it is complete.  My questions are the following;   What exactly does it mean to be a sequence of functions? I used $f_n(k)$ and $f(k)$ in my proof because I saw that it was used in another exercise, but I'm not sure what it means to write that $k$ down. Is the structure of my proof sound? Is there a better way to proof such statements?   Thanks for reading,  K. Kamal
6163
How do you prove that these four definitions of the operator norm are equivalent? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &amp;= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &amp;=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ &amp;=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ &amp;=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$
120632
Is $C[0,1]$ a Banach space with respect to the norm $\|f\| = \int\limits_0^1|f(t)| \, dt$?   People keep telling me it is, but lets consider: $f_n(x) = x^n$. This function defines a Cauchy sequence, yet the limit clearly isn't a continuous function!
53933
If I have a sequence $\{f_n\}_{n\in \mathbb{N}} \subset L_p\cap L_q$ that is Cauchy under both norms,  I am wondering if $f_n \to f$ in $L_p$ implies that $f_n\to f$ in $L_q$.   I have been working on this but I can't seem to figure out a definite proof.  Thank you!  PS: $1\leq p&lt;q &lt;\infty$
53561
Let $X$ be a normed vector space.   Show that if a subsequence of a Cauchy sequence converges, then the whole sequence converges. Use the part 1 to show that $S = \{x\in X : \|x\| = 1\}$ is complete if and only if $X$ is Banach.     I have found the first part in a book, but I do not how to prove part 2. Thanx regards
121959
For every $p\in [1,\infty)$, consider the set $\ell^p$ of real sequences $x=(x_n)=(x_1,x_2,\ldots)$ such that $\sum_{n=1}^\infty |x_n|^p&lt;\infty$. This is a Banach space with the norm $\Vert x\Vert_p=(\sum|x_n|^p)^{1/p}$.  I have the following problem:     Is it true that $\ell^p=\bigcup_{1\leq q&lt;p}\ell^q$ for $p&gt;1$?   I don't think it is true, but I'm having problems to show this. Clearly, the statement is true for some $p&gt;1$ iff it is true for every $p&gt;1$.  I'm trying to find a sequence $(a_n)$ such that $\sum |a_n|^s$ converges iff $s\geq 1$. I've tried many sequences (such as $(n^{1+1/n})$, involving $\log$, etc...), but none of them worked.
23096
I know the title isn't very clear. The question is the following. Let $\|\cdot\|$ be a vector norm on $\mathbb{C}^n$. Define the induced norm on $\mathbb{C}^{n\times n}$ as: $$\|A\|_{op}=\max\limits_{x\in\mathbb{C}^n\smallsetminus\{0\}}\dfrac{\|Ax\|}{\|x\|}.$$ How do I prove that:   If $\|\cdot\|_{2}$ is the Euclidean norm, then: $$\|A\|_{2}=\rho^{\frac12}(A^{*}A),$$ with $\rho(A)$ the spectral radius of $A$ and $A^{*}$ the conjugate transpose of $A$? If $\|\cdot\|_{\infty}$ is the max norm ($\|x\|_{\infty}=\max|x_i|$, $x_i$ being the components of $x$) then: $$\|A\|_{\infty}=\max_{i=1,\dots,n}\left(\sum\limits_{j=1}^n|a_{ij}|\right)?$$ And that if $\|\cdot\|_{1}$ is the sum norm ($\|x\|_{1}=\sum_{i=1}^n|x_i|$, $x_i$ being, again, the components of $x$), then: $$\|A\|_{1}=\max\limits_{j=1,\dots,n}\left(\sum\limits_{i=1}^n|a_{ij}|\right)?$$
14517
I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete.  The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in \mathbb{R}$.  I don't understand why it's not enough that $f$ is bounded. I think the limit of a sequence of continuous functions is continuous and then if $f$ is bounded, it's in $C_b(X)$. So what is this $\lim f_n \in \mathbb{R}$ about? Many thanks for your help.
98519
Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on  $\mathcal{C}^0(\mathbb{R},\mathbb{R})$?  A related question is: Can we proved that $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ has a norm without the axiom of choice?
55721
Let $\ell^1$ be the space of absolutely summable real or complex sequences. Let us say that a sequence $(x_1, x_2, \ldots)$ of vectors in $\ell^1$ converges weakly to $x \in \ell^1$ if for every bounded linear functional $\varphi \in (\ell^1)^*$, $\varphi(x_n) \rightarrow \varphi(x)$ as $n \to \infty$. How may I show that weak convergence, in this sense, is the same as the usual convergence-in-norm? It's clear the weak convergence implies pointwise convergence, but that's not good enough to conclude strong convergence...  By linearity, it suffices to prove that if $\varphi(x_n) \longrightarrow 0$ for every $\varphi \in (\ell^1)^*$, then $\| x_n \| \longrightarrow 0$. Let $x_n(k)$ be the $k$-th component of the vector $x_n$. Then, $x_n(k) \longrightarrow 0$ for every $k$, so $\sup_n |x_n(k)| &lt; \infty$ for each $k$, and this implies $$\lim_{N \to \infty} \lim_{n \to \infty} \sum_{k=1}^{N} |x_n(k)| = 0$$ This is almost what I want, but the limits are the wrong way around. The obvious next thing to try is to construct some clever functional, or even a family of clever functionals, but I can't think of anything useful here. I can see that pointwise convergence alone is not good enough — if $x_n$ is the standard basis vector, then $x_n \longrightarrow 0$ pointwise, but $\| x_n \| = 1$ for all $n$. The fact that it doesn't converge strongly can be detected by the linear functional $\varphi(x_n) = \sum_k x_n(k)$, but I'm at a loss as to how to generalise this.
57796
$C[a,b]$ is a normed vector space of all continuous complex valued functions on $[a,b]$, with supremum norm   $$\|f\|_\infty=\sup_{t\in [a,b]}|f(t)|.$$  The metric induced by the norm is  $$d(f,g)=\|f-g\|_\infty = \sup_{t\in [a,b]}|f(t)-g(t)|.$$  Show that $C[a,b]$ is a complete metric space under this metric induced by the supremum norm (i.e. show $C[a,b]$ is a Banach space).
22963
If we are in a sequence space, then the $ l^{p} $-norm of the sequence $ \mathbf{x} = (x_{i})_{i \in \mathbb{N}} $ is $ \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} $.  The $ l^{\infty} $-norm of $ \mathbf{x} $ is $ \displaystyle \sup_{i \in \mathbb{N}} |x_{i}| $.  Prove that the limit of the $ l^{p} $-norms is the $ l^{\infty} $-norm.  I saw an answer for $ L^{p} $-spaces, but I need one for $ l^{p} $-spaces. Besides, I didn’t really understand the $ L^{p} $-answer either.  Thanks for your help!
91772
(Euclidean Norm)  Proof that $\|X\|_2\leq \sqrt{n}\|X\|_\infty$, if $x \in \mathbb{R^n}$.   With that, proof $\|X\|_\infty \to 0 \implies \|X\|_2\to 0$  I'm stuck:  $\|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{n}\sqrt{x^2_1+x^2_2+...+x^2_n}$  I don't know how to use the fact: $\|X\|_\infty=\max(|x_1|,|x_2|,...,|x_n|)$
74618
Consider the vector spaces $C([0, 1])$ and $C^1([0, 1])$ with norm, \begin{align*} \displaystyle \|f\|_\infty=\sup_{x\in [0, 1]}|f(x)|, \end{align*} and let $T:C^1([0, 1])\to C([0, 1])$ the operator given by, \begin{align*} \displaystyle Tf=f'. \end{align*} Is $T$ defined this way an open mapping? If it's, how to prove it?
90252
Let $(V,|\cdot|)$ be a normed space. Let $x, y\in V.$ If $|x|,|y|\geq 1$, then does it follow that   $$\left|\frac{x}{|x|}-\frac{y}{|y|}\right|\leq |x-y|\;?$$   Actually I got a hint in form of a picture, but I failed to use it.