Let $X$ be a normed space. Is it always possible to have a linear injection $T:X\rightarrow X^\ast$ which associates to every vector in $X$ a unique functional in $X^\ast$?  I suspect this is not the case, but the space $X$ with which I am working has no more structure than a normed space, yet I find myself wanting to produce functionals from a specific family of vectors I have on hand.
121415
Today our teacher said that dual spaces are "big" and told us that this is a consequence by Hahn-Banach's theorem. So I was wondering whether the dual space of a space is always "bigger" or equal compared with the space itself? I thought a good way to check the adjective "big" by mathematical methods is to ask whether there is an injective map in the dual space.   I am very interested in any comment on this.
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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 $(x,x)=||x||^2$ ?
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On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?  How about $T:V\rightarrow W$ linear surjection in general?
47642
If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm &lt; $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$.   Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?
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.
66521
I realize that every inner product defined on a vector space can give rise to a norm on that space.  The converse, apparently is not true.  I'd like an example of a norm which no inner product can generate.
17657
What are the benefits of using a conjugate linear inner product in a complex vector space vs a simple linear inner product?  That is, why do we demand that $(y,x) = \overline{(x,y)}$ as opposed to $(y,x)=(x,y)$?  Of course, this ensures that $(x,x)$ is real and thus makes an easy definition of norm, but is that necessary?
74589
Let $\mathrm{C}[0,1]$ be the space of continuous functions $[0,1]\rightarrow \mathbb{R}$ endowed with the norm $||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for $x(t)=t$ and $y(t)=1$). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties...
38525
Let $X$ and $Y$ be normed spaces, $T:X \rightarrow Y$ a linear operator. Show that $T$ is continuous if $y' \circ T$ is continuous for every $y' \in Y'$. My idea is the following: Suppose $T$ is not continuous, then there exists a sequence $x_n$ with $\|x_n\|=1$ such that $\|Tx_n\|&gt;n$. If I could assure there is a subsequence $x_{n_k}$ such that $Tx_{n_k}$ are linearly independent I think I could construct an element $z'$ of $Y'$ such that $|(z' \circ T)x_{n_k}|&gt;n_k$.  The first thing is, can I assure the existence of that subsequence? And if I can, how do I conclude applying H-B?  I'd like a solution with this same approach, but if there is a simpler proof i'd like to see it too.
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Let $X$ be a separable normed space. Is it true that every subspace is separable? If it was Hilbert space I would take the dense set and then their projections.  It sounds trivial but I cannot prove or disprove it...  Thank you.
43056
I want to prove the following statement:   If the unit sphere in a normed space contains a segment, then there exists two vectors $x,y$ such that $\Vert x + y \Vert = \Vert x \Vert + \Vert y \Vert$ with $x$ and $y$ linearly independent.   This is what I have done: Let $x$ and $y$ be the end point of the segment contained in the unit sphere. Then $\forall \lambda \in (0,1)$ I can write:   $$ \Vert \lambda x + (1-\lambda)y \Vert \leq 1 $$  From here I am struggling on getting the relationship on $x+y$. Any pointers?
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My lecturer gave us this result today in class, but he didn't give a proof, he said we can prove it ourselves, only I'm really struggling to see how to do it.  Let $E$ be a normed space with dual $E'$. Then $E$ is finite dimensional if and only if $E'$ is finite dimensional, and in fact $\dim{E} =\dim{E'}$.  Would appreciate any input!
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I would like to show that a continuous functional on a subspace $W$ of a normed space $(V,\|\cdot \|)$ has a unique continuous extension to $V$ iff $W$ is dense in $V$.  I have proved $(\Leftarrow$). But the converse is currently eluding me! Help would be much appreciated!
84189
There is a conjecture: "The only subspace of a normed vector space $V$ that has a non-empty interior, is $V$ itself." (here, the topology is the obvious set of all open sets generated by the metric $||\cdot||$).  I have a proof for the case $V$ is finite dimensional. Because, let $V$ have dimension $n$ and a subspace $S$ of $V$ have dimension $m &lt; n$. Let $\{v_1,v_2,\ldots,v_m\}$ be a basis for $S$, extended to the basis $\{v_1,\ldots,v_m,\ldots,v_n\}$ of $V$. Now, suppose that $p = b_1v_1+\cdots+b_mv_m$ is an interior point of $S$. Now, consider the norm $N(a_1v_1 +\cdots+a_nv_n)=\max(|a_1|,\ldots,|a_n|)$. Then, there is an $r&gt;0$, such that $||x-p||&lt;r$ and $x$ is in $V$ $\implies$ $x$ is in $S$, since on a finite dim. space, all norms are equivalent. Let $m&lt;k\leq n$,and chose $v$ in $V$ as:  $$v=\left(b_1+\frac{r}{2}\right)v_1+\left(b_2+\frac{r}{2}\right)v_2+\cdots+\left(b_m+\frac{r}{m}\right)v_m+\frac{r}{2}v_k$$  Then, $N(v-p)\leq\frac{r}{2}&lt;r$, so $v$ is in $S$ and by the subspace property of $S$, $v_k$ is in $S$ too, a contradiction to $m&lt;n$.    I have primarily 2 questions:  (1) Is there a simpler method to proof the conjecture for the finite dimensional case?  (2) Is the conjecture true for the infinite dimensional case?  Sorry, if the question admits a very trivial answer. The motivation behind my question , is the fact that an open interval in $\mathbb{R}$ is not open in $\mathbb{R}\times\mathbb{R}$, etc.
22535
I'm trying to find a discontinuous linear functional into $\mathbb{R}$ as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since $\ell_2$ is infinite-dimensional, there must exist a linear functional from $\ell_2$ into $\mathbb{R}$. However, I'm having trouble actually coming up with it.   I believe I'm supposed to find an unbounded function (although I'm not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors $e^i$, which have all entries equal to zero, except for the $i$-th one. Then, you can define $f(e^i)=i$.  That'd be unbounded, but I'm not sure if it'd be linear, and even if it is, I'm not sure how to define it for all the other vectors in $\ell_2$.   A friend mentioned that at some point the question of whether the set $E=\{e^i:i\in\mathbb{Z}^+\}$ is a basis would come up, but I'm not sure what a basis has to do with continuity of $f$.   I'm just learning this topic for the first time, so bear with me please.   The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed.  It also helped to cement the notions of Hamel basis, not continuous, etc.
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?
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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?