Theorem: Let $V$ be a linear space and $V_{0}$ is its subspace. Also let $p$ be a finite convex functional in $V$ and $l_{0}$ is linear functional in $V_{0}$.   While $l_{0}(v)\le p(v),v\in V_{0}$   Then exists linear functional $l$ defined in $V$, which is extension of functional $l_{0}$ and   $l(v)\le p(v),v\in V$      Stein's proof in functional analysis page$_{21}$ :     ......   Well-order all vectors in $V$ that do not belongs to $V_0$ , and denote this ordering by $\lt$ . Among these vectors we call a vector $v$ "extendable"if the linear functional $l_0$ has an extension of the kind desired to the subspace spanned by $V_0 , v$ and all vectors $\lt v$ . What we want to prove is in effect that all vectors not in $V_0$ are extendable .   ......     Since we can not apply well-ordering principle to find the largest element in $V$ . If we have proved all vectors not in $V_0$ is extendable , then how to show $V$ is extendable ?
122695
In the 4th book of Stein and Shakarchi, the statement of the Hahn-Banach theorem is as follows:     Suppose $V_0$ is a linear subspace of $V$ and $p$ is a real sub-linear function on $V$, and that we are given a linear functional $l_0$ on $V_0$ that satisfies $$l_0(v)\leq p(v) ~\forall ~v\in V_0$$   Then $l_0$ can be extended to a linear functional $l$ on $V$ that satisfies    $$l(v) \leq p(v)~ \forall ~ v \in V$$   The proof given is as follows:     We have already proved that if $l_0$ can be extended to a subspace W, then it can also be extended to the subspace $W+ w$ for any vector $w$. We well order all the vectors in $V$ that do not belong to $V_0$, and denote this ordering by $&lt;$. Among these vectors, we call a vector $v$ extendable if the linear functional $l_0$ has an extension of the kind desired to the subspace spanned by $V_0.v$ and all the vectors $&lt;v$. What we want to prove is in effect that all vectors not in $V_0$ are extendable. Assume the contrary, then because of the well-ordering we can find the smallest $v_1$ that is not extendable. Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$. The previous step, with $V_0'$ in place of $V_0$ allows us then to extend $l_0$ to the subspace spanned by $V_0'$ and $v_1$, reaching a contradiction. This proves the theorem.   Firstly, how does proving that all vectors not in $V_0$ are extendable imply that an extension exists for the whole space? We know that for any $v \notin V_0$ there exists a valid extension, say $l_v$ for the space spanned by $V_0,v$ and all vectors $&lt;v$. Say we define $l(v):=l_v(v)$. Now we take an arbitrary $w&gt;v$. Couldn't it happen that $l_w(v) \ne l_v(v)$. In this case how can one define $l$? Will it help if we change the definition of "extendable" to something like: A vector $v \notin V_0$ is called extendable if $l_0$ has an extension of the kind desired to the relevant subspace which is equal to the already defined functionals $l_w~ \forall w&lt;v$ in the relevant subspaces? Also, how does one get the statement :"Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$."
18595
Let $V$ be a Hausdorff topological vector space and $Z$ a linear subspace: $Z\leq X$  Is there a neat way to prove that: $$\dim Z\text{ finite}\implies Z\text{ closed}$$
36804
I am studying functional analysis, and in the setting of normed spaces I have seen the theorem that states that the unit ball is compact iff the space is finite dimensional.  I also saw an exercise:     Let $X$ be finite dimensional, prove that for any non-empty and closed   $C$, $x\in C$ there exist $c\in C$ s.t $||x-c||=d(x,C)$.   The proof started with:     take $c\in C$ and consider  $$ D=\overline{B(x,||x-c||)}\cap C $$      $D$ is closed and bounded hence, since $X$ is finite dimensional, is   compact.   I would be glad if someone could help me with the following two questions:  1) It seems that the theorem I read about the compactness also extends to $X$ is finite dimensional iff $\{x,||x-x_{0}||\leq d\}$ is compact for any $x_{0}\in X,d\in\mathbb{R}^{+}$. is this correct ?  2) $C$ is closed, $\overline{B(x,||x-c||)}$ is also closed, hence so is the intersection. since $\overline{B(x,||x-c||)}$ is bounded then so is the intersection.  And so I get that $D$ is closed and bounded. Why can't I conclude that $D$ is compact without the assumption that $X$ is finite dimensional ?
50168
Let $X \subset L^1(\mathbb{R})$ a closed linear subspace satisfying \begin{align} X\subset \bigcup_{p&gt;1} L^p(\mathbb{R})\end{align}    Show that $X\subset L^{p_0}(\mathbb{R})$ for some $p_0&gt;1.$   I guess the problem is that in infinite measure spaces the inclusion $L^p\subset L^q$ only holds for $p=q$. Is it maybe possbile to apply Baire's Theorem in some way?
29332
Hahn Banach Theorem states that given a linear continuous functional $f$ on a subspace $N$ of a normed space $M$, it can be extended to a linear functional $F$ on the whole space $M$ and the norm of the extension is the same as the one of $f$.  Can someone tell me if something is wrong in the following lines:   Fact 1: Every vector space has a basis (use axiom of choice if it is infinite dimensional). Fact 2: A linear map is defined if we give the image of a basis.  Take $\{e_i\}_{i\in I}$ a basis of $M$ extending one of $N$, and define $F(e_i):=f(e_i)$ if $e_i$ is in $N$ and $0$ otherwise. This is clearly an extension of $f$, plus the norm of $F$ is the same as that of $f$ because sup $|F(x)|/||x||$ is $0$ for $x \in M \setminus N$ and $||f||$ if $x \in N$.   In case is correct this seems like a very simple proof of this theorem!
86732
How does one show that $L^1$  is strictly contained in $(L^\infty)^*$? Here, $(L^\infty)^*$ is the space of linear continuous functionals on $L^\infty$.
55333
Let linear $S$ be a subspace of $C([0,1])$, i.e., the continuous real-valued   functions on $[0,1]$. Assume that there exists $c&gt;0$, such that   $\|\,f\|_\infty\leq c \|\,f\|_2$, for all $f\in S$. Then show that $S$ is finite-dimensional.   This is equivalent to proving that the closed unit ball in $(S,\|\|_\infty)$ or in $(S,\|\|_2)$ is compact, but I can't derive this.   Another thought is that the $L^2$ closure of $S$ is a subset of $C([0,1])$. Indeed, if $f_n\to f\in L^2$, then the given relation $\|f_n-f\|_\infty \leq c\|f_n-f\|_2$ implies that $f_n\to f$ in $L^\infty$, so $f$ is continuous. Therefore we have the inclusion $$S\subset \overline{S}^{L^2}\subset C([0,1])\subset L^2$$ If $S$ was infinite dimensional, then $\overline{S}$ would be an infinite dimensional hilbert space, proper subset of $L^2$.   Another thought is that all the $L^p$ norms on $S$ are equivalent (by Holder). If ALL norms are equivalent, then $S$ has to be finite dimensional.
25049
I studying Brezis' book and I have somes partial solutions of the exercise $1.17.$  Let be $E$ a normed space and $f\in E^*$ be a linear functional nonzero. Consider the set  $M=[f=0]$ given by  $$M=\{x\in E \ : \ \langle f,x\rangle = 0 \}.$$  Here, $\langle f,x\rangle$ denote $f(x)$ and note that $M$ is just $\ker f$.     Dertemine $M^\perp.$   By definition   $$M^\perp=\{g\in E^* \ : \ \langle g,x\rangle=0, \forall x\in M \}.$$  So, the first doubt is: $M^\perp$ is just  $$M^\perp=\{g\in E^* \ : \ \ker f \subset \ker g  \} ?$$     Show that $\forall x\in E$, we have $d(x,M)=\inf_{y\in M}{\|x-y\|}=\frac{|\langle f,x\rangle|}{\|f\|}.$   Since $y\in M$, $|\langle f,x\rangle|=|\langle f,x-y\rangle|\le \|f\|\|x-y\|$. Then  $$\frac{|\langle f,x\rangle|}{\|f\|}\le d(x,M).$$  How can I show the equality? I know that since $\{x\}$ is a compact set and $M$ is a closed subspace of $E$, there is $y_0\in M$, such that $d(x,M)=\inf_{y\in M}{\|x-y\|}=\|x-y_0\|,$ but I don't know if that it is a help.  I tried to show that $\forall \varepsilon &gt;0$, there is $z\in M$ such that  $$\|x-z\|&lt;\frac{|\langle f,x\rangle|}{\|f\|}+\varepsilon,$$  but I could not. The last doubt is:     Let be $E=\{u\in C([0,1],\Bbb{R}) \ : \ u(0)=0\}$ and $f\in E^*$ given for each $u\in E$ by   $$\langle f,u\rangle = \int_0^1 u(t)dt.$$ Show that if $u\in E-M$ then $d(u,M)$ is never achieved.   For that, I tried something by absurd method but I found nothing. Can someone give a little help? Thanks.
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 $n$?
30893
Let be $ V $ vector space :   a) If we have an infinite chain of subspaces $ {U}_{1}\mathrm{\subseteq}{U}_{2}\mathrm{\subseteq} $  Show that $ {U}\mathrm{{=}}\mathop{\mathrm{\cup}}\limits_{{k}\mathrm{{=}}{1}}\limits^{\mathrm{\infty}}{U}_{K} $  is a subspace .   b) if  $ V $ is finite dimensional show that there is a some subspace ,call it  $ {U}_{P} $ ,such that $ {U}\mathrm{{=}}{U}_{p} $   and hence $ {U}_{P}\mathrm{{=}}{U}_{{P}\mathrm{{+}}{1}}\mathrm{{=}}{\mathrm{....}} $
105159
Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.   proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.  I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated.
123786
Let $X$ be a linear space, $p, q$ sublinear functionals on $X$, and $L$ a linear functional on $X$ such that $|L(x)| ≤ p(x) + q(x),$ for all $x ∈ X$. Prove that there exist linear functionals $L_1, L_2$ on $X$ such that $L(x) = L_1(x) + L_2(x),$ and $|L_1(x)| ≤ p(x), |L_2(x)| ≤ q(x),$ for all $x ∈ X.$   My Work:  First I thought to use Hahn Banach Theorem. But since there is no known subspace it was useless. Then I tried to make $L(x)$ as $L(x)=\frac{L(x+\lambda)+L(x-\lambda)}{2}$ for some scalar $\lambda$ but failed to find suitable $L_1$ and $L_2$. I think this problem is little bit tricky. I want to try it myself and I only need a hint to start. Can somebody please give me a hint?
117986
Let  $(x_n)$ be a sequence in a Banach space $E$ such that $\sum_{j=1}^{\infty} |\varphi (x_j) |&lt;\infty$, $\forall \varphi \in E'.$ Then $\sup \limits_{\|\varphi\| \leq 1} \sum_{j=1}^{\infty}|\varphi (x_j)| &lt;\infty $.  My attempt:  For all $n \in \mathbb{N}$, define $f_n: E' \to \mathbb{K}$, $f_n(\varphi) = \sum_{j=1}^{n} \varphi (x_j)$.  ($\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$)  Each $f_n$ is a continuous linear functional, since:  $$|f_n(\varphi)| = \bigg| \sum_{j=1}^{n} \varphi (x_j) \bigg| \leq  \sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{n} \|\varphi\|\|x_j\| =  (n \max \{\|x_j\|\} ) \|\varphi\|$$  For each $\varphi \in E'$, $(|f_n(\varphi)|)$ is bounded since $$|f_n(\varphi)|  \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq  \sum_{j=1}^{\infty} |\varphi (x_j)| = M_\varphi \in \mathbb{R}  $$  By Banach-Steinhaus theorem, there exists $M&gt;0$ such that $\sup \limits_{n \in \mathbb{N}}  \|f_n\|&lt;M$.  For all $n \in \mathbb{N}$, we have: $$M&gt; \|f_n\| = \sup \limits_{\|\varphi\| \leq 1} |f_n(\varphi)| =  \sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{n}\varphi (x_j)\bigg| $$  Then $$\sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{\infty}\varphi (x_j)\bigg| \leq M $$  Unfortunately, this is not  what we want to prove.   How can I fix it?
48020
From      Every linear function on a finite-dimensional space is continuous.   I was wondering what the domain and codomain of such linear function are?   Are they any two topological vector spaces (not necessarily the same), as along as the domain is finite-dimensional? Can the codomain be a different normed space (and may not be finite-dimensional)?  I asked this because I saw elsewhere the same statement except the domain is a finite-dimensional normed space, and am also not sure if the codomain can be a different normed space (and may not be finite-dimensional).  Thanks and regards!
16106
For every vector space $V$ does there exist a linear functional $f$ ( a linear map from $V$ to $F$ the underlying field ) such that for some $ \vec v \in V$ , $f(\vec v) \ne 0$ ? If it does exist , can we prove the existence without the "axiom of choice " ? Is the existence equivalent to axiom of choice ?
119612
Here's Theorem 4.3-2 (i.e. the Hahn Banach theorem for normed spaces):     Let $f$ be a bounded linear functional defined on a subspace $Z$ of a normed space $X$. Then there exists a bounbed linear functional $\tilde{f}$ on $X$ such that $$\tilde{f}(x) = f(x) \ \mbox{ for all } \ x\in Z,$$ and $$\Vert \tilde{f} \Vert_X = \Vert f \Vert_Z, $$ where $$\Vert f \Vert_Z \colon= \sup \left\{ \ \frac{ \vert f(x) \vert }{\Vert x \Vert} \ \colon \ x \in Z, \ x \neq 0 \ \right\} \ \mbox{ if } \ Z \neq \{\ 0 \ \}; \ \mbox{ otherwise } \ \Vert f \Vert_Z \colon= 0.$$ And, $$\Vert \tilde{f} \Vert_X \colon= \sup \left\{ \ \frac{ \vert \tilde{f}(x) \vert }{\Vert x \Vert} \ \colon \ x \in X, \ x \neq 0 \ \right\}.$$    I think I'm clear about the proof of this beautiful result. It uses the Hahn Banach Theorem for Complex Vector Spaces, which uses the Hahn Banach Theorem for Real Vector spaces, and the latter uses the Zorn's lemma.   Now if $X$ is a separable normed space, then is there a proof of the above result that doesn't involve the use of the Zorn's lemma?
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!!