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0

There is no general rule, and you must carefully analyse the function. The first function is $$f(x_1,x_2) = \|(x_1,x_2)\|^2 +2x_1 -3x_2 -2$$ and it is coercive because the term $2x_1 -3x_2 -2$ grows linearly at infinity. The second example is not coercive, since $$f(0,x_2) = -x_2^2 \to -\infty$$ when $|x_2| \to +\infty$

2

If $x=0$ you're done, since $p(y)\le p(y)$. Similarly if $y=0$. So assume $x\ne0$ and $y\ne0$. Let $C=\{x:p(x)\le 1\}$. Assume $C$ is convex. Now $$\frac x{p(x)},\frac y{p(y)}\in C.$$ Let $$t=\frac{p(x)}{p(x)+p(y)}\in[0,1].$$Then $$C\ni t\frac{x}{p(x)}+(1-t)\frac{y}{p(y)}=\frac{x+y}{p(x)+p(y)}.$$So $$p\left(\frac{x+y}{p(x)+p(y)}\right)\le1,$$hence ...

4

Hint: Suppose that $f(c_1x_1+...+c_nx_n)\leq \sum c_if(x_i)$, $\sum_ic_i=1$. Consider $c_1,...,c_{n+1}: \sum_ic_i=1, c_{n+1}\neq 0$. $f(c_1x_1+...+c_{n+1}x_{n+1})=f(c_1x_1+..+c_{n-1}x_{n-1}+(c_n+c_{n+1})({{c_nx_n+c_{n+1}x_{n+1}}\over {c_n+c_{n+1}}}))\leq c_1f(x_1)+..c_{n-1}f(x_{n-1})+(c_n+c_{n+1})f({{c_nx_n+c_{n+1}x_{n+1}}\over {c_n+c_{n+1}}})$. We have ...

0

Take your favourite, convex, smooth function $f \colon \mathbb{R} \to \mathbb{R}$ with $f''(2) = 1$ and $f''(1) = 3$. Then, with $x = 2$ and $y = 1$ we get $$(x \, f''(x) - y \, f''(y)) \, (x - y) = (2 \cdot 1 - 1 \cdot 3) \, (2 - 1) = - 1 < 0.$$ A final note: If you need to determine the sign of something like $x^\top A \, x$, you have to study the ...

2

Have you checked if the proof to theorem A. uses weak lower-semicontinuity or sequential weak lower-semicontinuity? I would guess it is the latter, especially after reading @user127096 response in the link you provided. As for your questions, (i) Note that for $t\in\mathbb R$, $$I(tu)=\frac{|t|^2}{2}\int|\nabla u|^2-\frac{|t|^q}{q}\int|u|^q\sim_{|t|\to0} ... 0 According to Boyd & Vandenberghe (2009, Convex Optimization, p. 105), a function f(x)>0 defined over \mathbb{R} is log-concave iff$$f(x)f''(x)\leq (f'(x))^2$$As shown here, while retaining log-concavity when subtracting the scalar, adding a scalar is more problematic. It requires that f''(f+k)\leq(f')^2 keeps true. While this might be true for ... 1 This is not true. (0,1) and (1,2) are convex sets in \Bbb R^1. Their intersection (and its closure) is empty, but their closures intersect in \{1\}. 1 Suppose that P is not bounded. There is then a sequence (x_i)_{i\geq0} of points in P such that |x_i|\to\infty as i\to\infty. We may assume that x_i\neq x_0 for all i\geq1, and then the sequence of vectors$$d_i=\frac{x_i-x_0}{|x_i-x_0|}$$takes values in the unit sphere, which is compact, so by replacing our sequence by a subsequence, we may ... 1 The condition is exactly the same. If A is a convex subset of a real vector space V, and f: A \to \Bbb{R} is a function, then f is concave whenever$$f((1- \lambda)x + \lambda y) \ge (1- \lambda)f(x) + \lambda f(y)$$for all x,y \in A, and for all \lambda \in [0,1]. Here x,y are meant as vectors, and this condition does not require that the ... 0 In this particular example, there's no need for tedious computation. Fact: The sum of convex (concave) functions is always convex (concave). Fact: The composition of a convex (concave) function with an affine map is convex (concave). Let$$f_i(p_1, \ldots, p_n) = p_i(1-p_i),this is concave (though not strictly concave) because it's a composition of the ... 1 Take the definition of convex set (e.g. chapter 2.1.4 from Boyd & Vandenberghe). Can you say that p(x) = \theta \, p_1(x) + (1-\theta) \, p_2(x) \in P for any 0 \leq \theta \leq 1 and any p_1(x),\,p_2(x) \in P? Yes, p(x) is a valid pdf, it is non-negative and integrates to 1. BTW, the resulting density is called a mixture density distribution ... 0 I assume you are implying a^tx-b>0 to ensure convexity (see Michael's comment). For simplicity I also assume that Q is Hermitian but the solution can be easily generalized. If b \le 0, then x=\mathbf{0} is trivially a solution. If b > 0, you can follow the approach below (forgive some sloppiness in the maths, for brevity). Setting \nabla ... 0 I would write this as a comment but I don't have enough reputation. In your claim x is still undefined, is it \forall x or \exists x or what? I think your claim should be: \hspace{2cm} f_1(x_2^*) - \epsilon \hspace{0.5cm} \leq \hspace{0.5cm} f_2(x_2^*) \hspace{0.5cm} \leq \hspace{0.5cm} f_1(x_1^*) + \epsilon because when you say "From (1) we ... 2 Try f(x)=|x| on \mathbb R. Then, for all t\in[0,1] and x_1,x_2\in\mathbb R, \begin{align*}f(tx_1+(1-t)x_2)&=|tx_1+(1-t)x_2|\leq |tx_1|+|(1-t))x_2|=t|x_1|+(1-t)|x_2|\\ &=tf(x_1)+(1-t)f(x_2), \end{align*} therefore f is convex. 5 It is not in general convex. Use g(x) = -x, f(y) = y, on [0,1]. Along the line x=y, the curve is -x^2 which is strictly concave. 1 Let g(x)=(W-xV)^T(W-xV)\geq 0. Then g'(x)=-2(W-xV)^TV and g''(x)=2V^TV. Now f(x)=g^{k}(x) implies that f''=kg^{k-2}((k-1)g'^2+gg''); if k>0, then f'' has same signum as (k-1)g'^2+gg''=4(k-1)((W-xV).V)^2+2||W-xV||^2||V||^2. By Cauchy-Schwarz, ((W-xV).V)^2\leq ||W-xV||^2||V||^2; thus, if k\geq 1/2, then f''(x) is always \geq 0 and ... 0 For \mu a Borel positive measure on the sphere S^{n-1}, consider a continuous Minkowski sum of segments K_\mu = \int_{S^{n-1}} [0,\theta] \, \mathrm{d}\mu (\theta) .$$The set K_\mu is convex (it could be defined by its support function, then the integral becomes a usual one). Now observe that (1) K_{\mu+\nu} = K_\mu + K_\nu (2) by rotation ... 1 Being bounded, K is contained in some closed ball. A closed ball is compact in weak* topology (Alaoglu). In any topological space, a closed subset of a compact set is compact. Hence, K is weak* compact. It remains to apply Krein-Milman's theorem to K. 2 Yes: f concave implies -f convex, so concave+odd implies concave+convex, and concave+convex implies linear. Your reasoning is fine. 0 No, this is even fails in dimension 2. Just try C = \{(x,y) \in \mathbb R^2 \mid (x-1)^2 + y^2 \le 2 \}. 0 Yes, this is correct. Let C be the convex hull of P. Since x\in\partial P, there is a supporting hyperplane for C passing through x; that is, a hyperplane M such that C is disjoint from one of the open half-spaces that M determines. Consider a convex combination of elements of P that is equal to x. It cannot involve any elements of P ... 4 Take \psi(t)=(1+t^2)^{\frac 1 2}, then \psi''(t)=(1+t^2)^{-\frac 3 2}\geq 0, \psi is convex on \mathbb{R}. By Taylor-lagrange :$$\psi(t)=\psi(s)+(t-s)\psi'(s) +\frac 1 2 (t-s)^2 \psi''(\xi) \geq \psi(s)+(t-s)\psi'(s)~~ \forall t,s \in \mathbb{R}$$So, for all t \in [0,1] :$$(1+f'(t)^2)^{\frac 1 2} \geq (1+F'(t)^2)^{\frac 1 2} + (f'(t)-F'(t)) ...

0

A convex function must be subdifferentiable on the relative interior of its domain (where it is finite). So the only points where one can have an issue is the boundary of its domain. The function constructed by Kevin Holt is one where all of the domain is boundary--hence Theorem 23.4 in Rockafellar's Convex Analysis cannot be used. As another illustration ...

3

(Since you already know that the case $[a,b]=[0,1]$ was enough, it would have been helpful to phrase the original question that way.) My thought was to try a family of quadratic polynomials and hope that, by continuity, one of them would satisfy the required conditions. So consider the quadratic polynomial $f_c(x) = 3x^2+2cx-c-1$, where the constant term ...

1

Since $f$ is affine you have $$f((1-\lambda-\mu)0+\lambda x+\mu y)=(1-\lambda-\mu)f(0)+\lambda f(x)+\mu f(y).$$ One can rewrite this $$f(\lambda x + \mu y) -f(0) =\lambda (f(x)-f(0)) +\mu (f(y)-f(0)).$$ Hence $g=f-f(0)$ is linear.

2

Here is a proof in the setting of complete simply connected Riemannian manifolds of nonpositive sectional curvature. The case of general CAT(0) spaces should be similar, but there are some technical details which make the discussion less pleasant in that case: One needs to smooth out convex functions of one variable (which can be done via convolution with ...

2

They are equal. Let $C$ be the closed convex hull of $A$, defined as the intersection of all closed convex sets containing $A$. (Closed half-spaces are enough here.) Let $B$ be the convex hull of $A$, defined as the intersection of all convex sets containing $A$. (Half-spaces, either open or closed, are not enough here.) Clearly $B\subset C$. Since $C$ is ...

0

Lemma. for $f(x)=x^k$. if $k\geq 1$ then for $a,b>0$,then $a^k+b^k\leq (a+b)^k$. if $0\leq k\leq 1$, then $a^k+b^k\geq (a+b)^k$. proof.when $k\geq 1$, then we have $(\frac{a}{a+b})^k\leq\frac{a}{a+b}$ and $(\frac{b}{a+b})^k\leq\frac{b}{a+b}$ for $\frac{a}{a+b}\leq 1$, $\frac{b}{a+b}\leq 1.$ Hence:$(\frac{a}{a+b})^k+(\frac{b}{a+b})^k\leq ... 1 The complement of$B$is a countable union of intervals $$I_n=(a_n,b_n),$$with$a_n,b_n\in B$, plus possibly$I_\omega=(a_\omega,\infty)$with$a_\omega\in B$, if$B$is bounded. Let$f$be$e^{-x}$or$1/(x+1)$or whatever strictly convex decreasing function you like. Let$g|_B=f|_B$, and on$\overline I_n$let$g$be the straight line such that ... 0$f(x) = h(x)+\mathbf{1}\{x \in B\}e^{-x},g(x) =e^{-x}$and choose$h(x)$convex decreasing such that$h(x) = e^{-x}$on the boundaries of$B$but nowhere else? 1 Hint:${s\over{s+t}}x+{t\over{s+t}}y\in A$if$x,y\in A$and$A$is convex, thus$(s+t)({s\over{s+t}}x+{t\over{s+t}}y)=sx+ty\in (s+t)A0 I think there is a simple way to look at it without using duality. Our goal is to find the minimizer for the problem \begin{align} \tag{\spadesuit} \text{minimize} & \quad \| x \|_2 + \frac{1}{2t} \| x - \hat x \|_2^2 \\ \text{subject to} & \quad x \geq 0. \end{align} First note that if\hat x_i < 0$, then there is no benefit from taking ... 1 Disappointingly, there is no simpler expression than the definition itself. in Combinatorial Convexity and Algebraic Geometry by Günter Ewald, page 105, the author states: The polar body$(K+L)^*$of a sum of convex bodies has, in general, no plausible interpretation in terms of$K^*, L^*\$. Only in the case of direct sums do we present such an ...

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