Let $X$ be a non-trivial (other than singleton $x$) normed space. Prove that $X$ is a Banach space if and only if $\{x \in X \mid \|x\| = 1 \}$ is complete.
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
40319
Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But  writes that there is a converse: "If $L(V,W)$ is Banach, and if $V$ is non-trivial, then $W$ is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof.  I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in $W$, and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy)
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).
61668
Can someone give me an example of a Banach space that is not a Hilbert space? I can't think of any because I don't know how to show one space that can not have inner product structure.
36775
I am reading the proof for this theorem and it goes as follows:  To see that $B(X,Y)$ is complete when $Y$ is complete, let $\left\{ T_n \right\}$ be a Cauchy sequence (of operators) in $B(X,Y)$, that is, $$\|\ T_n - T_m \|\ \to 0 \hspace{.5cm} \text{as} \hspace{.5cm} n,m \to \infty.$$ Then, for each $x \in X$, $$\|\ T_n x - T_m x \|\ = \|\ (T_n - T_m)x \|\ \leq \|\ T_n - T_m \|\ \|\ x \|\ \to 0 \hspace{.5cm} \text{as} \space\ n,m \to \infty$$ which imples that $\left\{ T_n x \right\} $ is a Cauchy sequence in $Y$. As $Y$ is complete, $\left\{ T_n x \right\}$ converges to some vector, denoted by $Tx$ in $Y$. We have just defined a transformation $T: X \to Y$ by $x \mapsto Tx = \displaystyle\lim_{n\to\infty} T_n x$. Linearity of $T$ is clear. Next, we show that $T \in B(X,Y)$ and $T_n \to T$ with respect to the operator norm. For any $\epsilon &gt; 0$ and any $x \in X$: \begin{align*} \|\ (T_n - T)x \|\ &amp;= \|\ (T_n - T_m)x + (T_m - T)x \|\ \\ &amp;\leq \|\ (T_n - T_m)x \|\ + \|\ (T_m - T)x \|\ \\ &amp;\leq \|\ T_n - T_m \|\ \|\ x \|\ + \|\ T_m x - Tx \|\ \\ &amp;&lt; \dfrac{\epsilon}{2} \|\ x \|\ + \dfrac{\epsilon}{2} \|\ x \|\ \hspace{1cm} \exists N \in \mathbb{N} \space\ \text{when} \space\ m,n &gt; N \\ &amp;= \epsilon \|\ x \|\ \\ \end{align*} Therefore, $\|\ T_n - T \|\ &lt; \epsilon$ whenever $n \geq N$. This shows that $T_n \to T$ in operator norm, and that $\|\ T_n - T \|\ \to 0$ as $n \to \infty$, and so $T$ is continuous.   My question: Why is $\|\ T_m x - Tx \|\ &lt; \dfrac{\epsilon}{2} \|\ x \|\ $ ? That is the only part of this proof that baffles me.
50608
Let $B$ be a Banach space, $H,K$ be closed subspaces and let $K$ be finite dimensional.  Suppose $B = H\oplus K$ and $T:B\to B$ is a bounded linear operator.  How do I show that $T(B)$ is closed $\iff$ $T(H)$ is closed ?
32797
Consider the space of continuously differentiable functions, $$C^1([a,b]) = \{f:[a,b]\rightarrow \mathbb{R}\mid f,f' \text{ are continuous}\}$$ with the $C^1$-norm $$\lVert f\rVert := \sup_{a\leq x\leq b}|f(x)|+\sup_{a\leq x\leq b}|f'(x)|.$$ Prove that $C^1([a,b])$ is a Banach Space.   This was the proof we were given: Assuming $C^1([a,b])$ is a normed linear space all we need to show is completeness. Let $(f_n)$ be a Cauchy Sequence in $C^1([a,b])$ with respect to the $C^1$-norm. Then each $f_n,f'_n\in (C([a,b]),\|\cdot\|_{\sup})$. We know that $C([a,b])$ is complete and thus there exists $f,g\in C([a,b])$ such that $f_n\rightarrow f$, and $f'_n\rightarrow g$ (uniformly) with respect to $\|\cdot\|_{\sup}$. If we let  $$ F_n(x) = \int_a^x f_n(t)dt, \hspace{2mm}  F(x) = \int_a^x f(t)dt $$ then $F_n\rightarrow F$ uniformly because  $$\lVert F_n-F\rVert_{\sup}\leq \sup_{a\leq x\leq b}\int_a^x|f_n(t)-f(t)|dt\leq \lVert f_n-f\rVert_{\sup}&lt;\epsilon.$$ From the fundamental theorem of calculus: $$f_n(x)-f_n(a) = \int_a^x f'_n(t)dt $$ Since $f'_n\rightarrow g$ uniformly then  $$ \int_a^xf'_n(t)dt\rightarrow \int_a^x g(t)dt $$ Since we know that $f_n\rightarrow f$ uniformly, $$f(x)-f(a) = \int_a^x g(t) dt $$ which by the fundamental theorem of calculues implies $f'=g$. So we know have $f_n\rightarrow f$ and $f'_n\rightarrow g=f'$ which mean $f_n\rightarrow f\in C^1([a,b])$ with respect to $C^1$-norm. So every cauchy sequence converges. Hence $C^1([a,b])$ is a Banach Space.   So I understand most of the proof. Where I get confused is that how did we actually show this satisfies the $C^1$-norm? Maybe I don't understand what this norm actually does.   Thank you for any help, comments and advice!
106944
Problem: Show that $(C([a,b]),\|{\cdot}\|_2)$ is not a complete normed vector space.  I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question:  Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?  Attempt: In order to show that $(C([a,b]),\|{\cdot}\|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $\|\cdot\|_2$ to a discontinuous function.  Consider the following sequence of functions, $(f_n)_{n=1}^\infty\subset C([a,b])$, given by,  $$f_n(x)=\begin{cases} 0,\,\text{when}\,\,x\in[a,\frac{b-a}2-\frac{2^{-n}}{b-a}] \\[2ex]  1,\,\text{when}\,\,x\in[\frac{b-a}2+\frac{2^{-n}}{b-a},b] \\[2ex]  \frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},\,\text{when}\,\,x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}] \end{cases} $$  The definition of $f_n(x)$ when $x\in[\frac{b-a}2-\frac{2^{-n}}{b-a},\frac{b-a}2+\frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.  I claim that this is a Cauchy sequence. We consider:  $$\|f_n-f_m\|_2^2=\int_a^b|f_n(x)-f_m(x)|^2dx$$  $$=\int_a^{\frac{b-a}2-\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+\int_{\frac{b-a}2+\frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$  $$=0+\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$  The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:  $$\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=\int_{\frac{b-a}2-\frac{2^{-n}}{b-a}}^{\frac{b-a}2+\frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$  Based in the example I am following, supposing that $m\ge n$, I think that I should be able to argue that the whole thing is bound above by $\frac{2^{-n}}{b-a}$ which would mean that $\|f_n-f_m\|_2^2\le\frac{2^{-n}}{b-a}\to0$ as $n\to\infty$.  I don't quite see how I'm meant to bound it above by $\frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $m\ge n$, leading me to question whether the approach I have employed here is a fruitful one after all.
100388
I recently encountered the theorem that the sum of a closed (linear) subspace with a finite dimensional subspace is closed subspace of the Banach space in which it is contained. However, this came with the caveat that the statement does not hold for two arbitrary closed subspaces.   So, here's what I'm looking for:     Find a Banach space $X$ and closed subspaces $M,N$ such that   $$ M+N=\{m+n\mid m\in M, n\in N\} $$   Is not closed in $X$.   Any references, hints, or answers are appreciated!
22065
How to show that a Banach space $X$ is reflexive if and only if its dual $X'$ is reflexive?
103869
Problem: Let $X$ be a normed space and $f$ be a non-zero linear functional on $X$. Then prove that $f$ is discontinuous $\iff$   kernel($f$) is dense in $X$.   I have proved that if $f$ is discontinuous, then kernel($f$) is dense in $X$. How to prove the other way?
23933
Let $(X,\|\;\|)$ be a normed vector space over $K$ and let $A,B\subseteq X$ be s.t. $A$ is compact and $B$ is closed. I want to prove that:  $$(\alpha A+\beta B)\;\text{ is closed }\;\forall \alpha,\beta\in K$$ So, given our function $\odot$ as in , my attempt goes like this: Let $\alpha,\beta\in K$. If $\alpha=\beta=0$, clearly $$\alpha A+\beta B=\{0\}\text{ is closed}$$ Now, since $\odot$ is continuous and $A$ is compact, we have that $\;\forall \alpha\neq0,\;\alpha A=\odot(\alpha,A)\;$ is compact and thus closed.  Then, let $\;y\in \beta B$ so,  $$ y=\beta b_y $$ for some $b_y\in B$. But, since $B$ is closed, there exists $\{b_n\}_{n\in\Bbb N}\subset B$ s.t. $b_n\xrightarrow[n\to\infty]{}b_y$ So let $y_n=\beta b_n\;\;\forall n\in\Bbb N$. Thus, $\{y_n\}_{n\in\Bbb N}\subset \beta B$ and $\|y_n-y\|=\|\beta b_n-\beta b_y\|=|\beta|\|b_n-b_y\|\xrightarrow[n\to\infty]{}0$ Thus, we get that  $$\forall y\in\beta B\;\exists\;\{y_n\}_{n\in\Bbb N}\subset \beta B \text{  s.t.  } y_n\xrightarrow[n\to\infty]{}y$$ so $\beta B$ is closed $\;\forall \beta\in K$. Now, $$ \text{Let }x\in (\alpha A+\beta B)\Rightarrow\;\exists\;a^*\in\alpha A\;\land\;b^*\in\beta B\text{  s.t.  } x=a^*+b^* $$ so, since $\alpha A$ and $\beta B$ are closed, $$ \exists\;\{a_n^*\}_{n\in\Bbb N}\subset\alpha A\text{  s.t.  }a_n^*\xrightarrow[n\to\infty]{}a^*\\ \exists\;\{b_n^*\}_{n\in\Bbb N}\subset\beta B\text{  s.t.  }b_n^*\xrightarrow[n\to\infty]{}b^* $$ so, taking $x_n=a_n^*+b_n^*\;\forall n\in\Bbb N$ is clear that $x_n\in (\alpha A+\beta B)\;\forall n\in\Bbb N$ and $$ \|x_n-x\|=\|a_n^*+b_n^*-a^*-b^*\|\le\|a_n^*-a^*\|+\|b_n^*-b^*\|\xrightarrow[n\to\infty]{}0 $$ so $x_n\xrightarrow[n\to\infty]{}x$ and, since this happens $\;\forall x\in (\alpha A+\beta B)$, thereby $(\alpha A+\beta B)$ is closed $\;\forall \alpha ,\beta\in K$  What do you think? Did I miss something?
41706
I know the following theorem from the lecture:     Let $X$ be a seperable Banachspace. Then $\overline{B(0)}$ is weakly* sequentially compact in $X'$.   Since it is specified that $X$ has to be separable, I want to look at an example where $\overline{B_1(0)}$ is not necessarily weakly* sequentially compact in $X'$, if we choose a Banachspace $X$ that is not separable. I found out from a book that $\overline{B_1(0)}$ is not weakly* sequentially compact in $(l^\infty)'$. $l^\infty$ is not separable ( I showed that already), but how can we now show that $\overline{B_1(0)}$ is not weakly* sequentially compact in $(l^\infty)$'?
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 $n$?
118844
I want to prove that for Banach space V there is a compact topological space $X$ so that $V$ is isometrically isomorphic to a closed subspace of $C(X)$-continuous function on a (compact) topological space X, equipped with the supremum norm $\|f\|_\infty$ = $\sup_{x \in X} |f(x)|$.
28234
I am reading a book about functional analysis and have a question:      Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be surjective?
6528
I have a homework question I can`t solve:  Let $X$ be a normed linear space, $A,B \in B(X)$. Show that there exists no nontrivial $c \in \mathbb{C} $ such that $AB-BA=cI$.  Thanks alot already guys! I seem to have a professor who gives impossible homework problems.  The hint he gave: $A^{n}B-BA^{n}=ncA^{n-1}$, $n \in \mathbb{N}$