# Tag Info

1

Following the hint (the definition of gradient): Remember that $\nabla f(x+t h) - \nabla f(x) = Hf (x) t h + o(|t|)$ -- user251257 Take the inner product of both sides with $th$: $$\langle \nabla f(x+t h) - \nabla f(x), th\rangle = t^2 h^THf (x)h + o(|t|^2)$$ If $h^THf (x)h<0$, then the right hand side is negative for sufficiently small $t$, a ...

0

Check that \begin{align}\frac{\partial\operatorname{Var}(x)}{\partial p_k} &=\sum_i\frac{\partial}{\partial p_k}(p_ia_i)^2-\frac{\partial}{\partial p_k}\left(\sum_ip_ia_i\right)^2 \\ &=\frac{\partial}{\partial p_k}(p_ka_k)^2-\frac{\partial}{\partial p_k}\left(\sum_ip_ia_i\right)^2 \\ &= 2a_k^2-2a_k\sum_ip_ia_i\end{align}

2

The definition of convex is slightly off, you mean that any line segment connecting two points in the subset is contained within it. Also, I'm not quite sure what you mean by tangent in this case, perhaps you mean that the extended line is tangent or that there is some point in the interior of the segment that is a boundary point. Let $S$ be a line segment ...

0

We have to prove that for any $x,y\in [0,2]$ the inequality $$(tx+(1-t)y)^2\leq tx^2+(1-t)y^2$$ is hold. We have $(tx+(1-t)y)^2=t^2x^2+2t(1-t)xy+(1-t)^2y^2$ but since $t\in[0,1]$ we have $t^2\leq t$ and $(1-t)^2\leq (1-t)$ also $2t(1-t)xy$ is a nonnegative term so we have$$(tx+(1-t)y)^2\leq tx^2+(1-t)y^2$$

1

As usual, let $\mathbb{R}^{m \times n}$ be the set of all $m$-by-$n$ real matrices and $S^n_{++} \subset \mathbb{R}^{n \times n}$ denote the set of positive-definite $n$-by-$n$ real matrices. Now, using Proposition 2.1(2) of this paper with $C:=Y \in S^n_{++}$, and $B := x^T \in \mathbb{R}^{1 \times n}$, and $A = t \in \mathbb{R}^{1 \times 1}$, we have ...

1

Your condtion $(*)$ is certainly not sufficient for the global minimum: There could be several local extrema of the length, and all of them satisfy $(*)$. In the following I look at the problem in my pedestrian way. It then becomes clear that the global minimum has to be found numerically. Let the curve $\gamma$ be given by $$\gamma:\quad ... 2 Lemma: K is convex \Leftrightarrow \forall line L\colon K\cap L is convex. Proof: \Rightarrow Intersections of convex sets are convex. \Leftarrow By definition: \forall z_1,z_2\in K take the line L through z_1,z_2, so z_1,z_2\in K\cap L, hence, the segment [z_1,z_2]\subset K\cap L\subset K. \blacksquare Now for a compact set K ... 1 Let us assume that a subset K \subseteq V (where V is a normed vector space) satisfies \lambda x + (1 - \lambda) y \in K for all x,y \in \partial K and all \lambda \in [0,1] (in particular, K is closed). We want to show that K is convex. For this, let x,y \in K be given and assume that t_0x + (1 - t_0)y is not contained in K for some t_0 ... 0 If you calculate the Hessian matrix, you should get something like$$ \frac{1}{\left(e^x+e^y+e^z\right)^2} \left( \begin{array}{ccc} e^x \left(e^y+e^z\right) & -e^{x+y} & -e^{x+z} \\ -e^{x+y} & e^y \left(e^x+e^z\right) & -e^{y+z} \\ -e^{x+z} & -e^{y+z} & e^z \left(e^x+e^y\right) \\ \end{array} \right) Now, in order ... 0 I guess this is pretty clear - or I'm missing something. The measure of a convex combinaton \alpha x+(1-\alpha)y is the corresponding convex combination of measures, and this relation survives the application of f, hence f' is indeed affine. It is also invertible as (f')^{-1}=(f^{-1})'. Thus we have indeed a map H(\partial K_1,\partial K_2)\to ... 3 It is unclear where you could fail: We have f'(x)=\ln x , f''(x)=\frac1x, so f''(x)>0 for all x. 1 Hint: Try f\colon\mathbb R\to\mathbb R, x\mapsto x^2 with I=[-1,1] and with I=[1,4]. 2 You need to show that f(x) \leq \alpha, where x is chosen as convex combination of the points x_1 and x_2, i.e. x=(1-\lambda)x_1 + \lambda x_2. Now, \begin{align*} f(x)& =f((1-\lambda)x_1 + \lambda x_2) \\ &\leq (1-\lambda)f(x_1) + \lambda f(x_2) \;\; \text{(using convexity of f(\cdot))}\\ & \leq(1-\lambda)\alpha + \lambda \alpha ... 1 No, this is not true. You could take the rotated Lorentz coneC = \{(x,y,z) \in \mathbb{R}^3: y^2 \le x z; x \ge 0; z \ge 0\}$$and project it onto the y-z-plane. The result is an open half-plane together with the origin. Some conditions ensuring that the image is closed, can be found in this paper. 1 We need only show f(x,h)=\frac{1}{2}(|(x+h)^3|+|(x-h)^3|) is strictly greater than |x^3| for all h>0,x\in\mathbb{R}. For x=0 we have f(x,h)=\frac{1}{2}(|h^3|+|-h^3|)=h^3>0 For x\ne0:$$\begin{align} f(x,h)&=\frac{1}{2}(|(x+h)^3|+|(x-h)^3|) \\ &=\frac{1}{2}(|x^3+3hx^2+3h^2x+h^3|+|x^3-3hx^2+3h^2x-h^3|) \\ &\ge ...

1

Proximal operator of the indicator function $\iota_C$ is given by \begin{equation*} prox_{\iota_C}(z)= \begin{aligned} & \underset{x}{\text{argmin}} & & \frac{1}{2}\|x-z\|^2_2 + \iota_C(x) \\ \end{aligned} \end{equation*} which can be re-written as: \begin{equation*} prox_{\iota_C}(z)= \begin{aligned} & \underset{x \in C}{\text{argmin}} ...

0

It seems that without loss of generality we may suppose that $K$ contains the zero of the linear space and then the set $\mbox{rint}(K)$ is open as an interior of a convex subset of a linear space $\mbox{aff}(K)$.

3

Hint: the (non-empty) intersection of convex sets is convex, and the sets $\{(x,y):x>1\}$ and $\{(x,y):y>1\}$ are quite trivially convex (and not disjoint).

3

It is straightforward to show that if $C_1, C_2$ are convex sets, then so is $C_1 \cap C_2$ (this holds in general for any collection of convex sets). Also, if $S_1,S_2$ are convex, then so is $S_1 \times S_2$. Let $C_1 = \{ ((x,y) | x>1 \} = (1,\infty) \times \mathbb{R}$, $C_2 = \{ ((x,y) | y>1 \} = \mathbb{R} \times (1,\infty)$, and note that $S = ... 0 Hint: Let$v,w$be two such vectors and$tv+(1-t)w$be a convex combination. Can you show that the$x$and$y$components of this convex combination satisfy$x>1$and$y>1$? 0 This is false as stated. Quasiconvexity definition says that the lower level sets$\{f\le \lambda\}$are convex. Unimodularity says that$f$increases along the rays emanating from its global minimum. The first certainly implies the second. But the converse is false. Example:$f(x_1,x_2) = \sqrt{|x_1|}+\sqrt{|x_2|}$. Clearly,$-f$is unimodal with global ... 2 As frank000 points out, any cyclic group of odd order is$1/2$-convex. So if you have a non-cyclic$1/2$-convex group and take an element of odd order, the subgroup generated will be a proper$1/2$-convex subgroup. For an example, let$n$be an odd positive integer and consider$\mathbb G=Z_n^2$. This is a$1/2$-convex group, and the subgroup$H$generated ... 1 If you look along the line$x_1 = x_2$you will see the function is actually convex along this line for$x_1 \geq 0$. As far as 'what's going on', your Hessian is negative-semidefinite, not negative definite. When you have semidefinite instead of definite, you can't draw any global conclusions about convexity or concavity. 1 You have $$u( t(0,1)+(1-t)(1,0) ) = u(1-t,t) = t(1-t)$$ $$tu(0,1)+(1-t)u(1,0) = 0$$ So clearly, there exists$t=\frac{1}{2}$such that $$u( t(0,1)+(1-t)(1,0) ) < tu(0,1)+(1-t)u(1,0)$$ So$u$is not concave (by the definition of concavity). 3 This is false in general. There exists a convex function$f$on the open upper halfdisk$U=\{(x,y): x^2+y^2<1,\ y>0\}$such that$0\le f\le 1$everywhere,$f(x,y)\to 0$as$(x,y)$converges to the diameter (except endpoints), and$f(x,y)\to 1$as$(x,y)$converges to semicircle (except endpoints). There is no limit as$(x,y)\to (\pm 1,0)$. Example: ... 0 Okay, this is convoluted so bear with me. For any point x in the plane you can construct a line, L, from the origin of the plane through x. For each real number r (positive or negative) you can construct a line perpendicular to L that is r distance from the origin. Precisely one of these lines at a specific distance r from the origin will cut A is half. ... 0 This is only a rough sketch of an idea, but I think it should work. Put some kind of Cartesian coordinates down on the plane. For any$\theta \in [0, \pi]$,consider also a coordinate system obtained by rotation through this angle about the origin. Let$f_A$and$f_B$be functions defined by the property: the line defined by$x=f_A(\theta)$in the ... 1 Hint: Let$x, y \in \{x \in \Bbb R^n \mid Ax+b \in C\}$. You want to check that given$0 < t < 1$, you also have$tx+(1-t)y$in that set. Meaning, you want to prove that$A(tx+(1-t)y)+b \in C$. But$A$is linear and$C$is convex, so? (You might want to write$b= tb + (1-t)b$, by the way) 0 More generally, the polyhedron$\Delta = \{x \in \mathbb{R}^n| Ax \le b\}$is convex. Your triangle can be written in this form with the choice$A := \begin{bmatrix}-1\hspace{.5em}0\hspace{.5em}1\\0\hspace{.5em}-1\hspace{.5em}1\end{bmatrix}^T$and$b := (0, 0, 1)^T$. On the Convexity of$\Delta$. Now, if$(x, y, t) \in \Delta^2 \times [0, 1]$, then ... 0 With respect to the last "why?": Let$\lambda^{-1}<\delta^{-1}$and let$(x,\mu) \in \lambda^{-1}\text{epi h}, i.e. (\lambda x,\lambda \mu) \in \text{epi h}$. Note that$\kappa:=\frac{\lambda^{-1}}{\delta^{-1}}<1$. As$\text{epi h}$is convex and contains the origin$(0,0)$and the point$(\lambda x,\lambda \mu)$it also contains the point ... 1 Hint: If$(x_1,x_2)$and$(y_1,y_2)$are both points with the described properties, can you show that$(tx_1+(1-t)y_1,tx_2+(1-t)y_2)$for$0\leq y\leq 1$also has those properties? If$x_1,y_1\geq 0$, and$0\leq t\leq 1$, then$tx_1+(1-t)y_1$is the sum of nonnegative numbers, and therefore nonnegative. The same goes for the second component. If ... 2$\newcommand{\Co}{\operatorname{Co}}$If you want to prove that $$\Co\left( \bigcup_{i=1}^k X_i \right) \subseteq \left\{\sum_{i=1}^k t_i x_i : x_i \in X_i,t_i\geq 0,\sum_{i=1}^k t_i =1\right\}, \tag 1$$ you should just assume$x$is an element of the left side and prove that it follows that it's an element of the right side. For now I'll take$\Co\left( ...

0

After understanding the logic of creating mr-convex hull of a set of points in a plane from A.G's answer, I tried making a sketch that explains the idea clearly (hopefully). You can access the sketch here.

1

I assume your definition of convexity is as follows: $K$ is said to be convex if whenever $x,y$ are in $K$, so too is $tx+(1-t)y$ for all $0\leq t\leq 1$. If so, all you need to do is use induction. Edit: Here's the answer: Let $\sum_{i=1}^{n+1}t_{i}x_{i}$ be a convex combination. Then $\sum_{i=1}^{n+1}t_{i}x_{i}=\sum_{i=1}^{n}t_{i}x_{i}+t_{n+1}x_{n+1}$ ...

1

The maximal orthogonal convex hull is what is left of the plane when you take away all set-free quadrants, that is basically the intersection of the sets 1,2,3 below plus cutting off the rest outside the rectangle (I have shown only three the most important quadrants on the pictures as the others cannot penetrate in the rectangle, so they make the obvious ...

0

Let us suppose that $\mathcal{F}$ is defined by $\mathcal{P} \cap \{ \mathbf{x}: f_1(\mathbf{x}) := \mathbf{a}\cdot \mathbf{x} - c = 0 \}$, where $f_1(\mathbf{x}) \geq 0$ is a valid inequality for $\mathcal{P}$. Let us take any face $\mathcal{F}' \subset \mathcal{F},$ given by $\mathcal{F} \cap \{ \mathbf{x}: f_2(\mathbf{x}):=\mathbf{b}\cdot \mathbf{x} - d = ... 0 Idea If you consider$C^2$-functions, you can use that:$f(\cdot,y):\mathbb{R}\to\mathbb{R} \text{ convex }\Longleftrightarrow f''(\cdot,y)\geq 0 \text{ for arbitrary }y\in\mathbb{R}\Longleftrightarrow \partial_x^2 f\geq 0f(x,\cdot):\mathbb{R}\to\mathbb{R} \text{ convex }\Longleftrightarrow f''(x,\cdot)\geq 0 \text{ for arbitrary ...

0

Take $f(x,y)=x^2+y^2-100xy$. For fixed $y_0$, the function $x\to f(x,y_0)$ is convex because it's second derivative is $2$. Similarly for fixed $x_0$, the function $y\to f(x_0,y)$ is convex. However, $f$ is not convex in $\mathbb{R}^2$, because $f(0,0)=0$, $f(1,1)=-98$, but $f({1\over 2},{1\over 2})$ is ${1\over 2}-25$, which is not smaller than ...

0

I agree with Tintarn about this inequality being a straightforward consequence of Karamata's inequality, but it is also quite easy to prove from scratch. Given a convex and differentiable function $f$, if we define $\Delta_f$ as: $$\Delta_f(a,b)=\left\{\begin{array}{rcl}\frac{f(a)-f(b)}{a-b}&\text{if}&a\neq b,\\ ... 3 You can use the definition twice in a row; the trick is to choose values of t that will produce x+y and x+z as arguments on the left-hand side. First take t=\dfrac{z}{y+z}, whence 1-t=\dfrac{y}{y+z}:$$ f\left(tx + (1-t)(x+y+z)\right) \le tf(x) + (1-t)f(x+y+z)\\ f\left(\frac{z}{y+z}x + \frac{y}{y+z}(x+y+z)\right) \le \frac{z}{y+z}f(x) + ...

0

If $a^T F$ weren't zero, $x \mapsto a^T Fx$ were a linear function bounded below (namely by $b - a^T g \in \mathbf R$) that is not zero. To prove your first sentence, suppose $f \colon \mathbf R^n \to \mathbf R$ is a non zero linear function, say $f(x) \ne 0$. To show it is not bounded below, let $M \in \mathbf R$ be arbitrary. Suppose $f(x) > 0$ ...

1

Definition: A function $g:\ R^n\ \longrightarrow\ R^m$ is affine if it is of the form $$g(x)=Mx+v$$ for some matrix $M\in\operatorname{Mat}(n\times m,R)$ and vector $v\in R^m$. Given a pair of matrices $A,B\in\operatorname{Mat}(n\times n,R)$, the function $$f:\ R^n\ \longrightarrow\ R^{2n}:\ x\ \longmapsto\ (Ax,Bx),$$ satisfies $f(x)=Cx$ for all $x\in ... 0 The standard terminology is "convex vs concave", but there are variations (indeed the names are confusing). For example: convex | concave convex downwards | convex upwards concave up | convex cap The "cap" sign, in convex-$\bigcap$, points surely to the latest - and that terminology is not too rare - and, granted, is the ... 0 There are a couple of identities involving the Hadamard ($\circ$) product that you'll need in order to see how the Diag() operation arises. Let$\,\,\,\,x,y$= arbitrary vectors$\,\,\,\,1$= vector of all ones$\,\,\,\,{\rm Diag}(x)$= function which returns a diagonal matrix whose diagonal equals the vector$xThen \eqalign{ {\rm ... 1 Let z be any one of X eigenvectors with eigenvalue \lambda, z^TXz = \lambda ||z||^2 \geqslant0 implies \lambda \geqslant0. Therefore all the eigenvalues of X are nonnegative, which is a motivation of notion of X \geqslant0 . As for the positive cone part, it is easy to verify that if X \geqslant0 , then cX \geqslant0 for c \geqslant0 ... 0\mathit{S} = \{(x,y,z)|x^2+y^2-z^2 \leqslant 0,\ z \geqslant 0\} \implies \mathit{S} = \{(x,y,z)|\sqrt{x^2+y^2}\leqslant z,\ z \geqslant 0\} \implies \mathit{S} = \{(a,z)| \|a\|_2\leqslant z,\ z \geqslant 0\}$$Lest say, (a_1,z_1) ,(a_2,z_2) \in \mathit{S}. So we have to show the following$$(\theta a_1 + (1-\theta)a_2,\theta z_1 + ... 1 This set is the epigraph of the2$-norm, which is a convex function. Hence, this set is convex. By the way, this set is called the "second-order cone" or the "ice-cream cone". (More precisely, this is the ice-cream cone in$\mathbb R^3$.) 1 It's actually pretty simple. Suppose we have two points$x_1,x_2\in S$, and define $$y_1=Ax_1+b, \quad y_2=Ax_2+b.$$ According to the definition of$S$, it must be the case that$y_1,y_2\in S$. Now consider the convex combination$x_3 = \lambda x_1 + (1-\lambda) x_2$,$\lambda\in[0,1]$, and define$y_3=Ax_3+b$. Then$\begin{aligned} y_3 = Ax_3+b &= ... 1 I had to peek at the PDF you linked, to see that the notationa^T_i$denotes the$i^{th}$row of the matrix$A$, rather than the transpose of the$i^{th}$column of the matrix. You can approach the problem using sequential change-of-variables within differential expressions. This is useful in matrix calculus, because quite often the intermediate ... 1 As achille hui explained, the scaling transformation$x\mapsto \frac{x}{r}$can be used to answer this, and in greater generality. Let$C$be any cone, meaning a set such that$x\in C\implies rx\in C$for all$r>0$. All sets$C$you consider are cones in this sense. The scaling transform$x\mapsto x/r$maps$C\cap B(0,r)$onto$C\cap B(0,1)\$. The ...

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