Suppose I have a sequence in a normed space.  In this space, the Cauchy-ness and convergence of the sequence is determined completely by the specific norm defined on the space, correct?  So when showing that a given sequence is Cauchy on this normed space, one would not be able to use any information about the sequence from an outside context (for example, that the sequence convergences under the standard absolute value norm in $\mathbb{R}$)?    As you can see, I am a little confused about non-standard norms on vector spaces.  If we have one, is that the ONLY norm we can have there?  Or can we have many different norms, including the standard one, on the space simultaneously?  I hope that makes sense.
121831
Let $V$ be a real vector space. Suppose that $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are two norms on $V$ which are equivalent.  I suspect the following to be true.  Let $(x_n)_{n=0}^\infty$ be a sequence in $V$. Then $(x_n)_n$ is a Cauchy sequence w.r.t.  $\Vert \cdot \Vert_1$ if and only if $(x_n)_n$ is a Cauchy sequence w.r.t. $\Vert \cdot \Vert_2$.  Is it true?   I'm looking for a good way to explain why "equivalence" of norms is the right way of "comparing" norms without mentioning anything from topology besides "convergence".
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.
41734
Consider the Banach space $X$ of null sequence whose elements are complex sequence which converges to $0$. In addition the norm is defined as $$\|(a_1, \dots, a_n)\| := \sup_n |a_n|.$$ Show this space is NOT reflexive. Recall that the dual of $X$ is $l_1$. Also use the following fact. If $X$ is reflexive space and $(x_n)$ is a sequence in $X$ and for all $\psi \in X^*$, the sequence $\psi(x_n)$ has a limit in $\mathbb C$, then $x_n$ converges weakly to some $x \in X$.   Suppose that $X$ is reflexive. Consider the sequence $$u_n = (1, \dots, 1, 0, \dots) \in X,$$ where first $n$ entries are $1$. Then for any $y \in l_1$, we have $$y(u_n) := \sum_n y_n\times u_n = y_1+\dots+y_n \to \sum_n y_n &lt; \infty,$$ since $y$ is absolutely convergent. Now it is only left to show that $u_n$ does not converges weakly to some $u \in X$ which I guess is $(1, 1, \dots)$. That is, for all $y \in l_1$, $$y(u_n) \nrightarrow y(u).$$ However, this seems correct to me. Where did I make mistake, please? Thank you!
29181
Given the normed space $C^1[0,1]$ of differentiable functions with continuous derivatives on $[0,1]$. The norm is defined as $$\|x\| = \max_{[0,1]} |x(t)|$$ I'd like to prove that the given normed space is not a Banach space. In the attempt of solving this problem, I have thought about the possibility to construct a Cauchy sequence in the space which doesn't converge in the space. However until now I haven't got any idea. Then I proceeded to think about constructing an equivalent norm to the given one where the space can be easily proven to be not a Banach space. However, I have got nothing either. Now I'm stuck without a clue. Please give me a hint to a correct direction. Any help is greatly appreciated.
69189
Can anybody please help me to prove this:  Let $p$ be greater than or equal to $1$. Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p&lt;\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $&lt;\_\,|\,\_&gt; $ s.t. $||u||^2=&lt;u\,|\,u&gt;$ if and only if $p=2$.
101636
I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function  $d(u,v) = \lVert u - v \rVert$, $u,v \in V$.   My question is whether every metric on a linear space can be induced by norm? I know answer is  no but I need proper justification.   Edit: Is there any method to check whether a given metric space is induced by norm ?  Thanks for help
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!
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: $1\leq p&lt;q &lt;\infty$
34643
Here's the proof which shows that every contractive sequence satisfies the Cauchy criterion, and I don't seem to be convinced with it.    My doubts:   Why is the fact that the RHS tends to zero on approaching larger $n$, sufficient to say that we can always find a natural number (say $K$) such that the Cauchy criterion is met for $m,n \ge K$ What exactly is the epsilon we choose, here? Is it the expression on the RHS? If so, we should be able to show that the RHS ranges over all real numbers, because we're concerned with all $\varepsilon &gt;0$
70037
Define  $\|\cdot\|\colon V  \rightarrow \mathbb{R} $ by $\|a\| = \sum 2^n |a_n|  $,  where  $V$ denotes the vector space  of all sequences  $a=(a_1,a_2,a_3,\dotsc)$ of real numbers such that $\sum 2^n |a_n|  $ converges.  Which of the following are true?  1) $V$ contains only the sequence $(0,0,\dotsc)$  2) $V$ is finite dimensional  3) $V$ has a countable linear basis  4) $V$ is a complete normed space  My answer:  it is already given that $\sum 2^n |a_n|  $ converges. That  means $V$ contain contains  only the sequence $(0,0,\dotsc)$ so option 1) is correct.  Option 2  is  incorrect because  $\mathbb{R}$ is  uncountable, as continuous image of uncountable  is  uncountable.  I'm confused  about option 3 and option 4.  Please help me.  Any hints/solution will be  appreciated.
29942
Consider $C([0, 1])$, the linear space of continuous complex-valued functions on the interval $[0, 1]$, with the norm  $$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$  I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.  Let $f_n \in C[0,1]$ be given by  $$f_n(x) := \begin{cases}  0   &amp; \text{if $0 \le x \le \frac1{2}$}\\  n(x-\frac{1}{2}) &amp; \text{if $\frac {1}{2} &lt; x \le \frac {1}{2} + \frac {1}{n}$}\\  1              &amp; \text{if $ \frac {1}{2} + \frac {1}{n} &lt;x \leq 1 $} \end{cases}$$  How to prove that $f_n$  is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$?   If I use basic definition then I have to prove that $\lVert f_n - f_m\rVert_1 &lt; \epsilon$ $\forall n, m &gt; N$. But I am finding it difficult to prove this.  Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.  Thanks
111703
If the sequence $\{a_n\}$ with $n∈\mathbb{N}$ is a Cauchy sequence,  then $\{a_n^2\}_{n∈\mathbb{N}}$ is also a Cauchy sequence. How do we prove it?
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!
75296
Note that (1) all norms on finite dimensional vector space are equivalent.  And here (2) norms on vector space are equivalent iff they have same topology.   Question : How can we prove (2) ? Please recommend reference  Thank you in anticipation
45157
Let $V$ be a vector space over  $\mathbb{C}$. If $V$ is an inner product space, then $V$  is normed (where the norm is defined as $\|x\|=\sqrt{(x,x)}\,\,$). Now if $V$ is normed, does it follow that $V$ is an inner product space ? I suspect no. I would like to see an example.  Thank you.    After reading my question again, I think it needs some clarification:  Suppose that $V$ is normed with norm $||\,||$. Can $V$ be given an inner product space structure such that $(x,x)=||x||^2$ ?
34880
As in the introduction of dual norm by Wiki, it says dual normed space $X'$ is always complete.  How to prove that? or at least explain that?  We all know the normed vector space is not always complete; if complete, all Cauchy sequences convergent. However, dual normed vector space is complete?
111704
Suppose $X$ is a possibly non-separable Banach space and let $X^*$ be its dual. Also, let $(f_n)_{n=1}^\infty$ be a sequence of [EDIT: linearly independent] norm-one functionals in $X^*$. Does there exist an element $x\in X$ such that $f_n(x) \in (0,\infty)$ for all $n$?