So, I was searching for ways of finding the dual of $l_\infty(\mathbb{R})$, (the space of bounded sequences of real numbers, under the infinity norm, i.e., $$||x|| = sup_{n\in\mathbb{N}} |x_n|.$$  I tried to approach the same way I did for $l_1, l_p, c_0$ and other similar spaces, however, I was not able to find any Schauder Basis for $l_\infty$. So I started searching, but I didn't find anything concrete about it. Does anybody knows an explicit "calculation" of $(l_\infty)^*$, or a book or paper that shows a way to obtain it?
120724
Can we identify the dual space of $l^\infty$ with another "natural space"?  If the answer is yes, what can we say about $L^\infty$? By the dual space I mean the space of all continuous linear functionals.
8283
Let $E$ be an empty set. Then, $\sup(E) = -\infty$ and $\inf(E)=+\infty$. I thought it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded?   Thank you.
42323
Let $S\subset\mathbb{R}$ be a non-empty bounded above set. Then there exists a monotone increasing sequence $\{x_n\}\subset S$ such that    $$\lim_{n\to\infty}x_n=\sup S.$$   I'm struggling with creating a formal proof for this problem. I understand that this has to be true, but I'm unsure of how to go about proving it.  I said that $\sup S = C$ and $x_n= C - y/n$ where $y$ is dependent on the size of $S$, but this is just a bunch of jumbled ideas and I can't really think of a way to formalize a proof. Any idea, even totally different from mine would be appreciated.
51335
Assume $R=(\mathcal{U}_n)_n$ is a sequence of distinct ultrafilters on some set $X$. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence $R=(\mathcal{V}_n)_n$ of $R=(\mathcal{U}_n)_n$ such that $\{\mathcal{V}_n|n \in \mathbb{N} \}$ is a discrete subspace of $\beta X$. In particular, there is a sequence $(A_n)_n$ of sets with $A_n \in \mathcal{V}_m$ if and only if $m =n$. If we let $B_n=A_n - (A_0 \cup...\cup A_{n−1})$, then the sequence $(B_n)_n$ is pairwise disjoint and $B_n \in \mathcal{V}_m$  iff $m=n$. If we let $B =\cup_n B_{2n}$, then $B \in \mathcal{V}_m$ if and only if $m$ is even. In other words, if $B =\{ \mathcal{V} \in βX|B \in \mathcal{V}\}$, then $B$ is a clopen set with $\mathcal{V}_m \in B$ iff $m$ is even. Therefore, we conclude that the sequence $(\mathcal{V}_m)_m$ cannot converge to any point, so the sequence $(\mathcal{U}_n)_n$ cannot converge to any point either.     I would like to know that:      (1) Why "$B_n \in \mathcal{V}_m$  iff $m=n$"? And " $B =\cup_n B_{2n}$, then $B \in \mathcal{V}_m$ if and only if $m$ is even "?      (2)$B$ is a clopen set with $\mathcal{V}_m \in B$ iff $m$ is even
40860
This question is from a set of optional, much harder problems from my first year analysis course, but the subject material is norms on $\mathbb R^K$.  (c) Show that there exists a constant $C &gt; 0$ such that for all $\mathbf x ∈ \mathbb R^k$, $||\mathbf x||\le C||\mathbf x||_\infty$  Hint: Use the standard unit vectors to write $\mathbf x = x_1\mathbf e_1 +. . .+x_n \mathbf e_n$, and use the triangle inequality.  (d) Show that there also exists a constant $c &gt; 0$ such that for all $x ∈ \mathbb R^k$, $||\mathbf x||_\infty \le c||\mathbf x||$  Hint: If it isn’t true, then you can find a sequence $⟨\mathbf x_n⟩$ such that $||\mathbf x_n||$ is bounded but $c_n := ||x_n||_∞ → ∞$. (Why?) Think about the sequence $⟨\mathbf x_n/c_n⟩$. The Bolzano-Weierstrass theorem may be of assistance.  Note that the unspecified norm represents any norm for $\mathbb R^k$. I've done part c), but am stuck on part d). I've had a look for proofs on other questions on MSE, but none of them seem relevant to the hint given, and frequently the terminology is too advanced for me.  I tried considering the case if it isn't true, as suggested, but I can't see how to bound any sequence using that. Any help would be much appreciated.
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.
103893
Show that $\{e_{n}\}_{n=0}^{\infty}$ is a Schauder basis of $(c,\|\cdot\|_{\infty})$ under the . Here, $c$ is the collection of all convergent sequences. Also, let $e_{0} = \{1,1,1,\ldots\}$ and $e_{n} = \{0,0,\ldots,1,0,\ldots\}$ where the $1$ is in the $n$th position.    The fact that $e_{0}$ is defined this way is a source of confusion for me. Starting with the typical arguments for a Schauder basis proof, I can't find a way to reconcile this.   Does anyone have any insight into this proof?
105221
Show that $(L^{p},\|\|_{p})$ is a Banach space.  My approach: I prove the statement for $(L^{1},\|\|_{1})$, of the following way, first all, is easy show that $\|\|_{1}$ is a norm. So,  $(L^{1},\|\|_{1})$ is vector space. To show that is a Banach space, note that  Prop. 1: Let $(f_{n})_{n\in\mathbb{N}}\subset\mathcal{L}^{1}$ (where $\mathcal{L}^{1}=\mathcal{L}^{1}(X,\tau,\mu)$ is the space of all integrals function), such that $\sum_{i}{\|f_{n}\|}_{1}&lt;\infty$, then the sequences $\left(\sum_{n=1}^{N}{f_{n}}\right)_{n\in\mathbb{N}}$ converges almost everywhere to the integral function, that we called $\sum_{n\in\mathbb{N}}{f_{n}}$. Furthermore $$\sum_{n\in\mathbb{N}}{\int{f_{n}d\mu}}=\int{\sum_{n\in\mathbb{N}}{f_{n}d\mu}}\quad\lim_{N\to\infty}{\|\sum_{n=1}^{N}{f_{n}}-\sum_{n\in\mathbb{N}}{f_{n}}\|_{1}}=0$$ Prop. 2: Let $(E,\|\|)$ a normed vector space. Then $E$  is a Banach space if and only if for all $(e_{k})_{k\in\mathbb{N}}\subset E$ such that $\sum_{k\in\mathbb{N}}{\|e_{k}\|}&lt;\infty$, $\left(\sum_{k=1}^{n}{e_{k}}\right)_{n\in\mathbb{N}}$ converges in $E$  We take the proposition 2. Let $(f_{n})_{n\in\mathbb{N}}\subset L^{1}$ such that $\sum_{n}{\|\hat{f}_{n}\|_{1}}&lt;\infty$. Let $f_{n}\in\hat{f}_{n}$, then $\|f_{n}\|_{1}=\|\hat{f}_{n}\|_{1}$ and then $\sum_{n}{\|f_{n}\|_{1}}&lt;\infty$, by prop. 1, there exist $F\in\mathcal{L}^{1}$ such that $\|\sum_{n=1}^{N}{f_{n}-F\|_{1}}\to 0$, and then $\|\sum_{n=1}^{N}{\hat{f}_{n}-F\|_{1}}\to 0$. Finally note that $\hat{F}\in L^{1}$ is a Banach space. This was my answer for  $L^{1}$ space, but how I prove the general statement for $L^{p}$ (I want to do a similar response). Edit: I'm stuck in the problem, any idea or hint is appreciated.Thanks!!
121684
Suppose that $x_0 = \alpha \in \mathbb{R}$ and $x_{n+1} = x_n ^2-x_n +1$.  I am asked to study the boundedness of $(x_n)$ and then asked if $(x_n)$ converges. How can I show that $(x_n)$ is bounded?  I have noted that $$x_{n+1}-x_n = (x_n-1)^2\geq 0$$ so $x_n$ is increasing.  Suppose that $(x_n)$ converges, then $l = l^2-l+1$ so $l = 1$. This means that the sequence converges only when $\alpha \leq 1$. And so $(x_n)$ is bounded above by $1$ if $\alpha\leq 1$ (since an increasing sequence converges to its supremum) otherwise the sequence is not bounded above.  Is my reasoning correct?
102868
I can't seem to find a way of asking a sub-question in relation to , which is in itself too generic a question for my purposes. Could someone please confirm or deny my understanding that the following:   $$\left\langle \sum_{n=1}^\infty x_n, y\right\rangle =\sum_{n=1}^\infty \langle x_n, y\rangle $$ is always true in a Hilbert space equipped with the norm induced by its inner product, where the convergence on the LHS is convergence in the said norm and the convergence on the RHS is convergence in the norm of the scalar field.  Thank you very much.
55863
From the  on the dual of a norm:  $X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number $$\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.$$  At some point in the proof that $L(X,Y)$ is bounded, it seems like they use the inequality $$\|fx\| \leq \|f\|\cdot\|x\|,$$ where $\|x\|\leq 1$.  I can't see why this should holds.   $\|f|\|$ is the supremum of what $|f(x)|$ can be, given that $\|x\|\leq 1$. So, given that $\|x\| \leq 1$, $|fx|$ should be bounded by $\|f\|$. But why should it be bounded by $\|f\|\cdot\|x\|$?
111182
I have to solve this problem.      Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by      $$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} $$   Prove that $T$ is compact.    My attempt is the following.  I have considered the operator $$ T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. $$ $T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact.   Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have: $$ ||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} $$ and, given $\epsilon&gt;0$, there exists $\bar n$ such that $|\lambda_n|&lt;\epsilon$, for every $n&gt;\bar n$. So if we take $N&gt;\bar n$, we get $$ \sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon $$ Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact.   My questions are:  (1) is the proof of the compactness of the operator $T_N$ correct?  (2) is the conclusion correct?  Thanks
74001
If $X$ is a vector space over $\mathbb R$ and $B=\{x_i; i\in I\}$ is a Hamel basis for $X$, then for each $i\in I$ we have a linear functional $a_i(x)$ which assigns to $x$ the $i$-th  coordinate, i.e., the functions $a_i$ are uniquely determined by the conditions that $$x=\sum_{i\in I} a_i(x)x_i,$$ where only finitely many summands are non-zero.  If $X$ is a Banach space, then at most finitely many of them can be continuous.  I have learned the following argument from comments in .     Suppose that $\{b_i; i\in\mathbb N\}$ is an infinite subset of $B$ such that each $f_{b_i}$ is continuous. W.l.o.g. we may assume that $\lVert{b_i}\rVert=1$.      Let   $$y:=\sum_{i=1}^\infty \frac1{2^i}b_i.$$   (Since $X$ is complete, the above sum converges.)      We also denote $y_n:=\sum_{i=1}^n \frac1{2^i}b_i$. Since $y_n$ converges to $y$, we have $f_{b_k}(y)=\lim\limits_{n\to\infty} f_{b_k}(y_n)=\frac1{2^k}$ for each $k\in\mathbb N$. Thus the point $x$ has infinitely many non-zero coordinates, which contradicts the definition of Hamel basis.   I have stumbled upon  in the book Christopher Heil: A Basis Theory Primer. Springer, New York, 2011. In this exercise we are working in an infinite-dimensional space $X$. Basically the same notation as I mentioned above is introduced, $a_i$'s are called coefficient functionals and then it goes as follows:     (a) Show by example that it is possible for some particular functional $a_i$ to be continuous.      (b) Show that $a_i(x_j) = \delta_{ij}$ for $i,j\in I$.      (c) Let $J = \{i\in I : a_i \text{ is continuous}\}$. Show that $\sup \{j\in J; \lVert a_j \rVert&lt;+\infty\}$      (d) Show that at most finitely many functionals $a_i$ can be continuous, i.e., $J$ is finite.      (e) Give an example of an infinite-dimensional normed linear space that has a Hamel basis $\{x_i; i\in I\}$ such that each of the associated coefficient functionals $a_i$ for $i\in I$ is continuous.   The part (c) can be shown easily using  (a.k.a. Uniform boundedness principle). But I guess that the author of the book has in mind a different proof for part (d) from what I sketched above, since (c) is an easy consequence of (d) -- so he would probably not be put the exercises in this order. (But maybe I was just trying to read to much between the lines.)  Question: I was not able to find a proof od (d) which uses (c). Do you have some idea how to do this?  NOTE: My question is not about the parts (a), (b), (e). I've included them just for the sake of the completeness, in order to include sufficient context for the question.
17835
Let $(X,B,\mu)$ be a complete measure space,Show that $$\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}, \quad \forall f \in \bigcup_{p} \bigcap_{p \leqslant q&lt;\infty} L^{q}$$ So,$\lim _{q \rightarrow \infty}\|f\|_{q}$ , $\|f\|_{\infty}$ are equal-norm with space $ L^{\infty} \cap\left(\bigcup_{p} \bigcap_{p \leqslant q} L^{q}\right)$. Case 1: $m(X)&lt;\infty $.It's easy to prove that. Case 2: $m(X)=\infty $. I have no idea about it,And I started to doubt the correctness of this conclusion. Can somebody give me a hint for this problem or just give an example to prove that this is a wrong conclusion when $m(X)=\infty $. Thanks in advance.
7724
If $\langle X_n \rangle$ is a sequence in $\mathbb R^n$ then $\lim\limits_{n\to \infty}X_n=A$. Show that $\lim\limits_{n\to \infty} \| X_n \| = \| A \|$ .   My thought process, $\| X_i \|= (\sum_{i=1}^n x_i^2)^{(1/2)}$ therefore be the multiplicative property of limits $x_i^2 = a_i^2$. Therefore the limit of norm of the components is equivalent to the limit of the corresponding limit of each component of the sequence.
60105
This is actually an exercise in Rudin's Real and Complex Analysis, $L^p$ spaces chapter. Could anyone help me out? Thanks in advance.  Motivation: It's well known that if we have a function $f$ which belongs to $L^p(0,1)$ for all $p\ge 1$. Then $\lim_{p\rightarrow \infty}\|f\|_p=\|f\|_{\infty}$ (moreover, $\|f\|_p$ is increasing in $p$). This is true even if $\|f\|_{\infty}=\infty$.  Question: How slow (fast) can $\|f\|_p$ grow when $\|f\|_{\infty}=\infty$? More precisely, given any positive increasing function $\Phi$ with $\lim_{p\rightarrow \infty}\Phi(p)=\infty$, can we always find a function $f$ which belongs to $L^p(0,1)$ for all $p\ge 1$, and $\|f\|_{\infty}=\infty$, such that $\|f\|_p\le (\ge)\Phi(p)$ for large $p$?
8578
Let $a_n \subset \mathbb C$ is a sequence Show that if $\sum_{n=1}^{\infty} a_nx_n \lt \infty$ $\forall x_n \in \ell^1$   then $a_n \in \ell^{\infty}$  What did I do : $x_n \in \ell^1 \Rightarrow \sum_{n=1}^{\infty} |x_n| \lt \infty $ Since $\mathbb C$ is a Banach space $\sum_{n=1}^{\infty} x_n \lt \infty $ and from Leibnitz rule $x_n \to 0$ $\sum_{n=1}^{\infty} a_nx_n \lt \infty \Rightarrow$ $a_nx_n \to 0$ i.e. $\forall \varepsilon \gt 0$$ |a_nx_n| \lt \varepsilon $ $|a_n||x_n|=|a_nx_n| \lt \varepsilon \Rightarrow$ $|a_n| \lt \varepsilon /|x_n| $ and $|x_n|$ is a finite number thus terms of $a_n$ are bounded. So $a_n \in \ell^{\infty}$ I’m not sure about my writtens. It seems wrong to me. Could someone fix me please? Thanks a lot