New answers tagged

0

The conclusion of the theorem you want to prove holds if and only if $X^*$ has the Radon–Nikodym property with respect to the Lebesgue measure on $(0,1)$. This is precisely what you need to recover Rademacher's theorem about differentiation of Lipschitz maps from $(0,1)$ to $X^*$. Separability in your proof is redundant but it makes it easier as you may ...


3

One need not to know the whole dual space of $C[0,1]$ to see the lack of reflexivity. Simply note that $C[0,1]^*$, opposed to $C[0,1]$, is non-separable because for each $t\in [0,1]$ the map $$\langle f, \delta_t\rangle = f(t)\quad (f\in C[0,1])$$ is a norm-one linear functional on $C[0,1]$ and $\|\delta_t - \delta_s\| =2$ for distinct $s,t$. To see this, ...


2

No it is not. The dual of $C[0,1]$ is the space $\mathfrak{M}([0,1])$ of complex (or signed) regular Borel measures on $[0,1]$. The dual of $\mathfrak{M}([0,1])$ is the set $\mathcal L^\infty([0,1])$ of bounded Borel functions on $[0,1]$. For example, $\chi_{\{0\}}$, the function which vanishes everywhere, except at $x=0$, where it is equal to $1$, ...


0

In this lecture there is a proof of this proposition, using arguments from convex analysis. The "meat" of the proof starts in page 28, but you should read all the lecture to understand the notation, definitions and such.


0

In page 147 of (Sierksma, 2001), an example (Example 2) is given for this case. That is, "In case of (a), a dual optimal solution can either be multiple or non-multiple ...". Primal Optimal Solution Dual Optimal Solution (a) Multiple implies Degenerate (b) Unique and nondegenerate implies Unique and ...


1

Suppose that $W \nsubseteq V$, i.e. that there exists some $w \in W$ with $w \notin V$. Then clearly $w \neq 0$. So if we start with a basis of $(v_1, \dotsc, v_n)$ of $V$ the familiy $(v_1, \dotsc, v_n, w)$ is still linearly independent. We extend this to a basis $(v_1, \dotsc, v_n, w, u_{n+2}, \dotsc, u_m)$ of $U$. Now define $f \in U^*$ by $f(v_i) = 0$ ...


5

Nobody claims that $A \times B = A + B$ holds in $\mathcal{C}$. But it is true that $A \times^{\mathcal{C}} B = A +^{\mathcal{C}^{op}} B$, where the supscript denotes the category in which we take the (co)product. More generally, $\lim^{\mathcal{C}}=\mathrm{colim}^{\mathcal{C}^{op}}$. If you want to rename your objects in the dual category - that's ok. In ...


2

There's no problem. The exact definition of $\mathcal{C}^{op}$ is as a quintuple $(\text{ob}\mathcal{C},\text{mor}(\mathcal{C}),t,s,\circ^{op})$ where $\circ^{op}(f,g)=\circ(g,f)$. That is, $\mathcal{C}^{op}$ has the same objects and morphisms, the source and target functions of $\mathcal{C}$ serve as the target and source functions, respectively, of ...


2

The answer to both questions is yes. Let's adress the second question first. Let $A : X \rightarrow Y$ is a bijective bounded linear operator with bounded inverse $B=A^{-1}: Y \rightarrow X$. We have: $$ \text{id}_{X} = BA, \quad \text{id}_{Y} = AB $$ and therefore $$ \text{id}_{X^\ast} = A^\ast B^\ast, \quad \text{id}_{Y^\ast} = B^\ast A^\ast $$ Hence ...


0

Let $f\in L^p(\Omega)$, $1<p<\infty$. $f$ is completely determined by the values of $f(x)$ for almost all $x\in\Omega$; The values of $f(x)$ for almost all $x\in\Omega$ are completely determined by the values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$; The values of $\int_{\Omega} f(x)h(x)\;dx$ for all $h\in L^q(\Omega)$ are ...


1

I do not agree. It looks correct to me. First of all, if variables $x$ and $y$ are unconstrained, it is correct to have $=$ type constraints in the dual. Authors are not required to specify that a variable is unconstrained, they only have to specify which ones are not, and how. This makes sense. If your primal is a minimization problem and constraints are ...


0

It means that if for $f,g \in L^p$ we have $$ \int f h = \int gh$$ for all $h \in L^q$, then $f = g$ (in the usual $L^p$ sense).


0

I assume that you are working over $\mathbb{C}$, otherwise it does not make sense to talk about DeRham cohomology. By hodge decomposition theory, $H^2(X,\mathbb{C} )=H^{0,2}\oplus H^{1,1}\oplus H^{2,0}$. By GAGA principal, $H^{0,2}$ and $H^{2,0}$ are isomorphic to cohomology of sheaf of algebraic differentials. $H^2(X,\mathcal{O}_X)=0$ by Grothendieck ...


2

This is not a complete answer to the question, but only my try so far, as requested by the question author. In particular I do not arrive at the given answer, but at one having different signs. As I am not sure about the used identifications I will try to make all identifications explicit, albeit at the costs of elegance. We may also interpret $H$ as a ...


0

Let's look like this. 1) A functional $f$ is in $W^\circ$ if and only if $f(x)=0$ whenever $x_1+\cdots+x_n=0$. Define a functional $G$ by $$g(x)=x_1+\cdots+x_n.$$ Then the kernel of the functional $g$ is contained in the kernel of the functional $f$. With symbols this means $\ker g\subseteq \ker f$. We claim that there exists a scalar $c$ such that ...


2

Put $V = F^n$. By definition, $W^0 = \{f\in V^*:\, f\vert W = 0\}$. The inclusion $W \to W\oplus W^\perp = V$ induces an isomorphism $W^0 \oplus W^* \to V^*$, given by $(f, g) \to f + g$. Hence $$\dim W^0 = \dim V - \dim W^* = \dim V - \dim W = 1.$$ The function $f(x) = x_1 + \cdots + x_n$ clearly lies in $W^0$, so it spans it. Part (2) is just the ...


2

Hint. First prove that $$ \dim W^0=\dim F^n-\dim W $$ Then prove that $\dim W=n-1$ so $$ \dim W^0=1 $$ Do you see how this proves (1)?


0

If we have a Banach space $X$ with norm $\|\cdot\|$ and a (proper) closed subspace $Y\subset X$, then their duals are related via $X' \subset Y'$: This is due to the fact that the dual of $X$ is the space of bounded linear functionals $f$ on $X$, i.e. the quantity $$\sup_{x \in X, \|x\|=1}|f(x)|$$ has to be finite in order for $f\in X'$ (assuming ...


0

\begin{align} \min_{x} c^Tx \\ s.t.~Ax=b \\ Unrestricted \end{align} Take $x=x_1-x_2$ \begin{align} \min c^T(x_1-x_2) \\ s.t.~A(x_1-x_2)=b \\ ~ x_1, x_2 \ge 0 \end{align} This is can be written as \begin{align} \min {\begin {pmatrix}c \\ -c \\ \end {pmatrix} }^T \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix} \\ s.t.~\begin {pmatrix} A, & -A \end ...


0

I will provide an intuitive solution using visualization (and I do not remember the method you are using - so will not be able to help with that). We are given: $$\mbox{min} ~-x_1-2x_2, ~~\mbox{s.t.} \\~ 3x_1+2x_2-6\leq 0 \\ -x_1+2x_2 -4\leq 0 \\ 0 \leq x_1 \leq 3/2, 0 \leq x_2 \leq 3/2$$ We can draw a region plot for the first function, $3x_1+2x_2-6\leq ...


0

Let's assume that such $q$ exists for the sake of finding one. Since now for all $p\in V$ we have $\langle p,q\rangle=p(2)$, we know, for p(x)=1: $\int_{-1}^{1}{q(x)dx}=1$. For $p(x)=x$ we find $\int_{-1}^{1}{xq(x)dx}=2$ and for $p(x)=x^2$ we get $\int_{-1}^{1}{x^2q(x)dx}=4$. We know $q$ is of the form $ax^2+bx+c$, so we can substitute this, and we'll start ...


2

Here's another way, if you don't need to find $q$: define $\varphi:\mathbb{R}_2\left[x\right]\to\mathbb{R}$ by $\varphi(p)=p(2)$. This is a linear functional on a finite dimensional space, and thus we may apply Riesz representation theorem to prove that such $q$ exists.


1

It suffices to find $q$ that works for $1, x, x^2$, i.e., find $q$ such that $$\int_{-1}^1q(x)\,\mathrm dx = 1 $$ $$\int_{-1}^1xq(x)\,\mathrm dx = 2 $$ $$\int_{-1}^1x^2q(x)\,\mathrm dx = 4 $$ Those are three equations in the three coefficients of $q$, so you should be fine. If you just want to show existence, note that the linear map $V\to V^*$, $q\mapsto ...


0

Posing: $$f(x)=\sum_{j=0}^{n-1}c_jx^j$$ we have: $$\int_0^1 f(x)e^xdx=\int_0^1 \sum_{j=0}^{n-1}c_jx^je^xdx=\sum_{j=0}^{n-1}c_j\int_0^1 x^je^xdx$$ $$\sum_{i=1}^na_if(i)=\sum_{i=1}^na_i\sum_{j=0}^{n-1}c_ji^j=\sum_{j=0}^{n-1}c_j\sum_{i=1}^na_ii^j$$ So, for $0\leq j\leq n-1$: $$\sum_{i=1}^na_ii^j=\int_0^1 x^je^xdx$$ This is a system of $n$ equations in the $n$ ...


0

A beautifull result establishes that if $F_1,...,F_r, F$ are functionals such that $\bigcap_{i=1}^r\mathrm{Ker}(F_i)\subseteq\mathrm{Ker}(F)$, then there are numbers $a_1,...,a_r$ such that $F=a_1F_1+...+a_rF_r$. The proof can be found here Let $V=\{\mbox{polynomials with degree }\le n-1\}$, and set $F_i:V\to\mathbb{R}$ given by $F_i(f)=f(i)$, ...


0

Really, this isn't about integrals. The mapping $f \mapsto \int_0^1 f(x)e^x \, dx$ is a linear functional on the $n$-dimensional space $V$ of polynomials of degree $< n$. The problem asks you to prove that in the dual space $V^{*}$, this functional is in the span of the functionals $$\phi_1 \colon f \mapsto f(1), \quad \phi_2 \colon f \mapsto f(2),\dots, ...


3

The natural isomorphism is not between $W^0$ and $V/W$, but between $W^{\circ}$ and $(V/W)^*$. Consider the linear map \begin{align*} L : W^{\circ} &\to (V/W)^*\\ \varphi &\mapsto \hat{\varphi} \end{align*} where $\hat{\varphi}(v + W) := \varphi(v)$. Note, the map $\hat{\varphi}$ is well-defined because if $v' + W = v + W$, then $v' = v + w$ for ...



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