Tag Info

New answers tagged

1

No. A Poincaré duality space must in particular have vanishing cohomology above some degree, but a nontrivial finite cyclic group has nonvanishing cohomology in arbitrarily high degrees.


0

Your reasoning is correct. In fact, Hahn-Banach is not needed for the isometric part either. On the other hand, it is needed to show that the dual of a subspace $M\subseteq X$ is isometric to the quotient space $X^*/M^\perp$.


0

If $V$ is finite dimensional, there is a natural isomorphism \begin{align*} V&\longrightarrow V^{**}\\ x&\mapsto x^{**} \end{align*} where $x^{**}: V^*\to \mathbb{F}$ is defined by $x^{**}(f):=f(x)$. As the dimensions of $V$ and $V^{**}$ are the same, it suffices to show that the above natural map is injective in order to conclude that it is an ...


0

We do not know a priori that $\widehat{\phi}=\phi$. We have simply evaluated $\phi$ on the basis elements and used these coefficients to define $\widehat{\phi}$ as a linear combination of elements of the dual basis. While it is not hard to see that $\widehat{\phi}$ takes on the same values as $\phi$ by linearity, it is not immediate. The fact that such ...


1

$\renewcommand{\phi}{\varphi}$The proof is really in two steps. First, one shows that in the expression for the element $$ a_1\phi_1+...+a_n\phi_n $$ of $V^{*}$, the $a_{i}$ are uniquely determined as $a_i = \phi_{i}(v_{i})$. This shows that the $\phi_{i}$ are linearly independent. To show that the $\phi_{i}$ span $V^{*}$, then note that for any $\phi \in ...


1

I assume $A$ is symmetric. Your dual manipulation is incorrect when $A$ does not have full rank. To see why, let's try to compute the dual function (using your notation): \begin{align} g(\eta, \beta) &= \sup_{\alpha \in \mathbb{R}^k} [-c\alpha^TA\alpha + d^T\alpha+\eta^T(\alpha + p) - \beta^T(\alpha -r)] \\ &= \sup_{\alpha \in \mathbb{R}^k} ...


2

The process of minimizing $\mathcal{L}$ to construct the dual results in a formula for $\alpha$ as a function of $\beta$ and $\eta$. This is exactly the connection between the optimal primal and dual variables.


1

No, it is not symmetric unless $X$ is a Hilbert space. In the conjectural formula $$\langle x, j(y)\rangle = \langle y, j(x)\rangle\quad (?)$$ the left hand side is linear in $x$, but the right hand side is not, unless $j$ is a linear map. Concrete example: in $\ell^4$, the duality map $j:\ell^4\to \ell^{4/3}$ is $$ j(x) = (x_1^3,x_2^3,x_3^3,\dots) $$ ...


1

Another proof. More elementary than Urban's proof. (But less general.) Consider the subspace $c \subseteq l^\infty$ consisting of the real sequences that converge. So $c$ is a closed linear subspace of $l^\infty$. Let $L \;:\; c \to \mathbb R$ be the "limit" linear functional. That is, for $x = (x_1,x_2,\cdots)$, define $$ L(x) := \lim_{n\to\infty} ...


1

This is not true because of the following Theorem $\textbf{Theorem:}$ If $X$ is a normed space such that its dual $X'$ is separable, then $X$ itself is separable. So, if $(l^{\infty})' =l_1$, then it will follow that $l^{\infty}$ is separable which is not true.


1

The other poster already gave a good answer explaining some of your questions, so I would just like to recommend that you pick up Aluffi's algebra text. It teaches algebra and category theory simultaneously from the very basics, and I personally love it. However, he takes a while to introduce things like natural transformations and adjunctions, and just ...


4

1) There is no definition of morphism. There is a definition of category, and a category consists of objects and morphisms. That is what morhpisms are. Just like there is no definition of what a vector is. There is a definition of vector space. So a vector is (precisely) anything which is an element of some vector space. Similarly, a morphism is (precisely) ...



Top 50 recent answers are included