Let $E$ be a separable Banach space and $(y_j)_{j\in \mathbb N}$ be a sequence in $B(0,1)$ which is dense in $B(0,1)$. Where $B(0,1)$ is the open ball of center $0$ and radius $1$. Show that the map \begin{align} A:~&amp;l_1~~~~~~~~~\longrightarrow E\\ &amp;(x_i)_{i\in\mathbb N}\longmapsto \sum_{i=1}^\infty x_iy_i \end{align} is linear, continuous, surjective.  I don't have any idea to show the function $A$ is surjective. Can someone please help me.
5851
Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks.
77246
Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a continuously differentiable map satisfying $\|f(x)-f(y)\|\geq\|x-y\|$ for all $x,y\in\mathbb{R}^{n}.$ Then   $A.$ $f$ is onto.  $B.$ $f(\mathbb{R}^{n})$ is a closed subset of $\mathbb{R}^{n}.$  $C.$ $f(\mathbb{R}^{n})$ is a open subset of $\mathbb{R}^{n}.$  $D.$ $f(0)=0.$  According to me $f$ is one one and onto so image is closed and open both. Please suggest me. Thanks in advance.
66211
How to prove the following mind-blowing fact?     Let $X$ be a separable Banach space and let $\ell_1$ be the space of all absolutely summable scalar sequences. Then there exists such closed subspace $A\subset \ell_1$ that factor space $\ell_1/A$ and $X$ are isomorphic as normed spaces.   Edit: So what, this is like a classification up to isomorphism of all separable Banach spaces? Each separable Banach space corresponds to some closed subspace of $\ell_1$?
107481
I wanted to show that $X' / U^\perp \cong U'$, for $U$ being a closed subspace of the Banach space $X$.  Therefore I looked at $l: X' / U^\perp \cong U' , x' + U^\perp=[x']  \mapsto x'|_U$.  It is clear that this map is onto, as we just take a $u' \in U'$. Then I can use Hahn-Banach to get an appropriate $x'$ such that $x'|_U = u'$ and in particular $||x'||=||u'||$.  Now I need to show that is isometric: since we always have $\|[x']\|\le \|x'\|$ and by Hahn-Banach we have $\|x'\|=\|x'_{|U}\|$, this shows $\|[x′]\|\le \|x'_{|D}\|$ and it should be always true that $\forall y' \in U^\perp: \|x'-y'\| \ge \|x'_{|D}\|$, hence $\|[x']\|\ge \|x'_{|D}\|$. Is this reasoning correct? The isometric property follows.
31461
Suppose that $l_1 = \{ (x_n)_{n \in \mathbb{N}} | \sum|x_n| &lt; \infty \}$ and $l_{\infty} = \{ (x_n)_{n \in \mathbb{N}}| \sup|x_n| &lt; \infty \}$.  Show that $(l_1)^* \cong l_{\infty}$, where $(l_1)^*$ is a dual space of $l_1$.  My attempt: Define a map $L: l_{\infty} \rightarrow (l_1)^*$ given by  $$L(x)(y) = \sum_{n \in \mathbb{N}}{x_ny_n}$$  where $y = (y_n)_{n \in \mathbb{N}} \in l_1$.   My aim is to show that $L$ is an isometric isomorphism.   Clearly $L$ is linear and and injective (choose $y = e_n$ for all $n \in \mathbb{N})$.  How to show that $L$ is a surjection and isometry?
2277
I want to prove that in an infinite dimensional normed space $X$, the weak closure of the unit sphere $S=\{ x\in X : \| x \| = 1 \}$ is the unit ball $B=\{ x\in X : \| x \| \leq 1 \}$.  $\\$  Here is my attempt with what I know:  I know that the weak closure of $S$ is a subset of $B$ because $B$ is norm closed and convex, so it is weakly closed, and $B$ contains $S$.  But I need to show that $B$ is a subset of the weak closure of $S$.  $\\$  for small $\epsilon &gt; 0$, and some $x^*_1,...,x^*_n \in X^*$, I let $U=\{ x : \langle x, x^*_i \rangle &lt; \epsilon , i = 1,...,n \} $  then $U$ is a weak neighbourhood of $0$  What I think I need to show now is that $U$ intersects $S$, but I don't know how.
42306
This is a problem from Rudin's Functional Analysis:  Show that there is a sequence of functions $\{ f_n \}$ in $X$ (the vector space of all complex functions) such that $f_n \to 0$ pointwise, but if any sequence $\{ \gamma_n \}$ tends to $\infty$, $\gamma_n f_n \not\to 0$.  The hint says to use the fact that the collections of all complex sequences converging to $0$ has the same cardinality as $[0,1]$. Honestly, this confused me more than it helped me.  My main issue is through the method of showing existence: I am not sure if I should attempt to construct an explicit sequence $f_n$, or if I need to appeal to other nonconstructive means such as Zorn's Lemma, etc. I have been trying to construct clever choices of sequences for a good bit now, but it always seems like I will be able to find a sequence that diverges slowly enough to allow that $\gamma_n f_n \to 0$.   Thanks!
98008
Show that $\mathbb R$ is cardinally larger than any separable metric space S.  I have been trying to solve this on my own. My idea was to start by mapping the open balls of positive rational radius around the points in the dense set of S to the corresponding in $\mathbb R$. Now any ${x}\subset S$ can be written as a limit of an appropriate sequence of open balls, which would correspond to a similar limit in $\mathbb R$, in case the limit exists.  Then I would need to show that different limits are obtained in $\mathbb R$ for different $x\in S$. This is a step I am not sure how to show.   Alternatively, one could show that all possible sequences of rationals have no greater cardinality than reals. What would be a good way to show this? I know that all reals can be written as some convergent sequence of rationals, but I am looking for something slightly different here.  I did not find this useful as it does not fill in the details:
61933
Let $(X,||\cdot||)$ be a reflexive Banach space. Prove that $X$ is separable if and only if $X'$ (the dual space of $X$) is separable. Does anyone have a hint for me? I have no idea where to begin
43631
This question is taken from Saxe K -Beginning Functional Analysis.   Show that the closed unit ball in $C[0,1]$ is not compact by proving that it is not sequentially compact.   (It's assumed that we are using the uniform norm).  I've been working on this for ages but I could not come up with any sequence $\{f_n\}$ in the unit ball such that there exists $N\in \mathbb{N}$ such that for all $m,n\geq N$ we have that $d(f_n,f_m)&gt;c$. Should be a nice example of this, please help me!
64894
I am reading  that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:      Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in   \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+  \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on    $\mathcal M$ (a linear map from  finite dimensional space is always    continuous) which extends to a member of $\mathcal X^*$,  by the    Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let    $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.   I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.  My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case.
4520
As the title says, the question is how to prove  Let $X,Y$ be Banach spaces. A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for some $T\in B(X,Y)$
79839
Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)&lt;\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.  The second assertion is easier since the quotient space $X/N$ has the universal mapping property.
117779
The following is problem 3.22 from Rudin's Princples of Mathematical Analysis:     Suppose $X$ is a nonempty complete metric space, and $\{G_n\}$ is a sequence of dense open subsets of $X$. Prove Baire's theorem, namely, that $\bigcap_{n=1}^\infty G_n$ is not empty. Hint: find a shrinking sequence of neighbourhoods $E_n$ such that $\overline{E}_n\subset G_n$.   Here's what I've tried so far:      Let $\{r_n\}$ be a Cauchy sequence of positive real numbers converging to $0$. Fix $x\in X$ and define $E_i=\{g\in G_i:d(g,x)&lt;r_i\}$, which is nonempty since $G_i$ is dense. I would like to show that for all $i$, $\overline{E}_i\subset G_i$ (I had convinced myself that this would be true but I am now having doubts). Let $e\in \overline{E}_i$. Then either $e\in E_i$ or $e$ is a limit point of $E_i$. If $e\in E_i$ then $e\in G_i$. Otherwise, every neighbourhood of $e$ contains a point in $E_i$.    I thought that I should be able to then choose some point $e'\in E_i$ in a neighbourhood of $e$ and, since $G_i$ is open, it'll have a neighbourhood $N\subset G_i$ which contains $e$, but this is proving to be difficult and I'm worried that it's not true. If I can show that this is true then the rest will follow from results I've already proven.  Does my approach make any sense?  Incidentally, as a secondary question, what type of a thing would $G_n$ be? A sequence of dense open subsets seems weird to me—at first I was thinking of some sequence of infinite subsets of rational numbers in the real numbers but I realized that those aren't open. Is there anything which would be familiar to my little undergrad brain which would be analogous to this problem?
88050
A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.   Prove that every separable metric space has a countable base.   Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x&gt;0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)&lt;{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.   It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)&lt;\varepsilon_x/4$ and since $d(x, d_j)&lt;{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)&lt;{\varepsilon_x}/{4}+{\varepsilon_x}/{4}&lt;{\varepsilon_x}.$$  Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.  By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r&lt;{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.  Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls  arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.  Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.  Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.
12593
Show that      A Normed Linear Space $X$ is a Banach Space iff every absolutely convergent series is convergent.   My try:  Let $X$ is a Banach Space .Let $\sum x_n$ be an absolutely convergent series .Consider $s_n=\sum_{i=1}^nx_i$. Now $\sum \|x_n\|&lt;\infty \implies \exists N$ such that $\sum_{i=N}^ \infty \|x_i\|&lt;\epsilon$ for any $\epsilon&gt;0$  Then $\|s_n-s_m\|\le \sum _{i=m+1}^n \|x_i\|&lt;\epsilon \forall n,m&gt;N$  So $s_n$ is Cauchy in $X$ and hence converges $s_n\to s$ (say).  Thus $\sum x_i$ converges.  Conversely, let $x_n$ be a Cauchy Sequence in $X$. Here I can't proceed how to use the given fact.  Any help will be great.
47578
Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, \end{align*} for all $x \in [0,1]$. Why is $E_n$ nowhere dense in $C\big([0,1]\big)$.  I have been able to show that $E_n$ is closed in $C\big([0,1]\big)$, but haven't been able to show why the interior is empty. Any advice?...