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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\setA}{\mathcal{A}}$If the points $(x_{j})_{j=1}^{N}$ are linearly independent as vectors in $\Reals^{n}$, the set $\setA$ is the parallelipiped[1] centered at the origin, with a vertex at $-\sum_{j} x_{j}$ and edges $(2x_{j})_{j=1}^{N}$.[2] If the points are linearly dependent, $\setA$ is a projection of a ...


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My favorite example is $$ f(x) =\sqrt{1+|x|^2}. $$ The Hessian is always positive definite, but degenerates for $|x|\to\infty$.


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Zonotope. (I have nothing more to say, but say more to satisfy the computer.)


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$f(x)=e^x$ is strictly convex but not strongly or uniformly convex. The scalar definition of uniform convexity reduces to $f''(x)>\theta$. For any fixed $\theta>0$, this inequality fails for any $x\le\log(\theta)$. Indeed, consider any smooth, positive function that approaches zero asymptotically; integrate twice (from, say, the origin) and you have ...


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No, Evans means that a supporting hyperplane at $ (x, f(x)) $ is the graph of the mapping $ y \mapsto f(x) + r\cdot (y - x) $.


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The idea can be illustrated on the real line. If a function $f:\mathbb{R} \to \mathbb{R}$ has a discontinuity at a point $a$, but the one-sided limits $f(a-)$ and $f(a+)$ exist, then we can define $$ g(x) = \begin{cases}f(x),\quad & x<a \\ f(a-)+\frac{ x-a}{\epsilon}( f(a+)-f(a-)),\quad & a\le x\le a+\epsilon \\ f(x-\epsilon),\quad &x> ...


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The easiest way to go is something like this: Prove that $f_1(x)=\log(1+\exp(x))$ is a convex function of $x$. Note that $f_2(x,y)=\alpha f_1(x) + (1-\alpha) f_2(y)$ is a convex function of $x,y$ for fixed $\alpha\in(0,1)$. The product of a convex function and a positive constant is convex, and the sum of convex functions is convex. Note that ...


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I do not think that this result is true, you need that $K$ is closed w.r.t. the weak-* topology. The following is a counterexample: Take $X = [0,1]$ and $m$ the Lebesgue measure. $K$ is the closed convex hull of the Dirac measures. Then, $m \not\in K$ (see below) but you cannot separate $m$ from $K$ with a continuous function. It remains to show $m \not\in ...


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Short answer: $y_n=x_n$ is a convex combination of $x_1,\dots,x_n$. Clearly, this need not converge strongly. Longer answer: for any particular choice of coefficients one can cook up a sequence $(x_n)$ where the strong convergence of convex combinations fails. Here's an idea. For any sequence $a_n\to \infty$ we have $\sin a_n t\to 0$ weakly in $L^2[0,1]$. ...


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Yes, $\phi(x)$ is convex if and only if $ \nabla_x^2\phi(x) >0$, which is means that $\nabla_x\phi(x)$ be non decreasing.


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Let $f_i(x) = \sum_{j=1}^n x_{(j)}$. This is a concave function of $x$. To prove it, define $$S_{(i)}\triangleq \{s\in\{0,1\}^n\,|\,\textstyle\sum_j s_j = i\}$$ In other words, $S_{(i)}$ is the set of all $\{0,1\}$ vectors with exactly $i$ nonzeros. Then $$f_i(x) = \inf_{s\in S_{(i)}} s^Tx.$$ Because $f_i(x)$ is the pointwise infimum of affine functions, it ...


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We have $f(0)=f(1)=0$ and $f(x)$ is positive for $0\lt x\lt 1$. Thus there is a local (and global) maximum in the interval $(0,1)$. To show there cannot be two or more local maxima, note that (1) $f'(x)=0$ at local maxima, and (2) if we have a local maximum at $a$ and $b$, then $f'(x)=0$ somewhere between $a$ and $b$. But calculation shows that the first ...


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The proposed method will not work, as $f''(x)$ is not always negative; for example try $p=3$. However, all is not lost. Taking just one derivative: $$f'(x)=\frac{x^{p-1}}{(2-x)^2}(px^2+(-3p-1)x+2p)$$ Note that $\frac{x^{p-1}}{(2-x)^2}>0$ for all $p>1$, $x\in(0,1)$. The second bit is an upward-facing parabola. It is positive for $x=0$ and negative ...


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Hint: Suppose $f(a) = c$ and $f(b) = d$, and $0 < t < 1$. Thus there exist $s_1$ and $s_2$ so $g(s_1) + h(a - s_1)$ is arbitrarily close to $c$, and $g(s_2) + h(b - s_2)$ is arbitrarily close to $d$. Now taking $s = t s_1 + (1-t) s_2$, $$g(s) = g(t s_1 + (1-t) s_2) \le t g(s_1) + (1-t) g(s_2)$$ $$h(t a + (1-t) b - s) = h(t(a - s_1) + ...


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No, it is not enough. Consider this example. Partition $(0,1]$ into $\cup_{n=0}^\infty (2^{-n-1},2^{-n}]$. For even $n$, let $f$ be the line segment connecting $(2^{-n-1},2^{-n-2})$ and $(2^{-n},2^{-n-1})$ on the interval $(2^{-n-1},2^{-n}]$. For odd $n$, let $f$ be the line segment connecting $(2^{-n-1},2^{-n-1})$ and $(2^{-n},2^{-n-1})$ on the interval ...


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Let's suppose $a\neq 0$. Then $f''(x) \le 0 $ for all $x$ implies $4a^2x^2+4abx+b2+2a$ has a constant sign, which can happen only if $\Delta'=-8a^3\le 0\iff a\ge 0$. In such a case, this sign is the sign of $4a^2$, so $f$ can't be concave unless $a=0$ If $a=0$, similarly we must have $b=0$. Thus $f$ is concave if and only if $a=b=0$, i. e. if and only ...


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Here we need that $f''=(2ax+b)^2e^{ax^2+b}+2ae^{ax^2+b}\le 0$, since the exponential is positive we need $$(2ax+b)^2+2a\le 0.$$


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Suppose S satisfies the first definition, and let $u\in S$. $\;\;$If we define $V=\{s-u:s\in S\}$, then 1) $0\in V$ since $u\in S$. 2) If $v\in V$, then $tv\in V$ since $\;\;\;\;v\in V\implies v+u\in S\implies t(v+u)+(1-t)u\in S\implies tv+u\in S\implies tv\in V.$ 3) If $v\in V$ and $w\in V$, then $v+w\in V$ since $\;\;\;v, w\in V\Rightarrow v+u\in S ...


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Well, according to your lemma, your idea seems trivial, but is totally fine! The affine space $M$ generated by $S$ is \begin{align*} M=\{a(1,1,1)+b(2,3,4)+c(1,2,3)+d(2,1,0)\ |\ a,b,c,d\in\mathbb R,a+b+c+d=1\}. \end{align*} Taking $a=1-b-c-d$ we obtain \begin{align*} M=\left\{\begin{pmatrix}1+b+d\\1+2b+c\\1+3b+2c-d\end{pmatrix}\ |\ b,c,d\in\mathbb ...


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You have $\{v_1,v_2,v_3,v_4\}$. An affine space containing those points is $v_1+\text{ span }\{v_2-v_1,v_3-v_1,v_4-v_1\}$ which is $v_1$ plus an ordinary subspace. Find a basis for that subspace.


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Reference : 355p in the book ${\it mathematical\ analysis}$- Apostol Since $U$ is convex then there exists a line between $u$ and $v$ when $u,\ v\in U$. Fix $u,\ v$: MVT : Let ${\bf f} : \mathbb{R}^n\rightarrow \mathbb{R}^m$. For every ${\bf a}\in \mathbb{R}^m$, there exists ${\bf z}\in \overline{{\bf uv}} $ : $$ \langle {\bf a}, {\bf f}({\bf ...


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Here is one possibility: $$\begin{cases} \alpha_1x^2+\alpha_2x+\alpha_3 & x\le a\\f(x) & a<x<b \\ \beta_1x^2+\beta_2x+\beta_3 & x\ge b\end{cases}$$ You choose $\alpha_1, \alpha_2, \alpha_3$ so that the function and its first two derivatives agree at $a$ (on both sides). Similarly, you choose $\beta_1, \beta_2, \beta_3$ so that the ...


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Here's a second-derivative proof for $\mathbb{R}^n$. The gradient and Hessian do not exist at the origin, but everywhere else they are given by $$\nabla f(x) = \|x\|^{-1} x, \quad \nabla^2 f(x) = \|x\|^{-1} I - \|x\|^{-3} xx^T$$ Strict convexity requires that the Hessian be positive definite; that is, $v^T(\nabla^2 f(x))v>0$ for all $v\neq 0$. But suppose ...


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The function $$ \lVert \mbox{ }\rVert:\mathbb{R}^n\to \mathbb{R},\, f(x)=\lVert x\rVert $$ is certainly convex. In fact, given $t\in [0,1]$ and $x,y\in \mathbb{R}^n$ we have: $$ \lVert tx+(1-t)y\rVert=\lVert tx+(1-t)y\rVert\le \lVert tx\rVert+\lVert (1-t)y\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert. $$ But is it STRICTLY CONVEX? i.e. if $t\in (0,1)$ and $x,y ...


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It is easier to work with the definition that a function f is convex if for any $\lambda \in [0,1]$ and for all $x,y \in (a,b)$: $$ f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y)$$ Then: $$ ||(1-\lambda x) + \lambda y ||^2 = \sum_{i} |(1-\lambda x_i) + \lambda y_i|^2 \leq \sum_{i} (1-\lambda) |x_i|^2 + \lambda |y_i|^2 = (1-\lambda)||x||^2 ...


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I just stumbled upon this question, wondering the same thing as the OP. YourAdHere's answer was what I was looking for, but I also needed a formal proof, so I thought I might try and add that to complement his/her answer. Consider a continuous, convex, bijective function $f: A \rightarrow B$ and its inverse $f^{-1}: B \rightarrow A$. By definition of ...


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Take $f$ strictly positive, increasing log-convex function with $f'(0)=1$, put also $h=2f$. Now $$g_n(x) = \frac 32f(x) +\frac 12 f(x)\sin (nx)$$ Easy to see that $g_n'(0)=\frac{3+n}{2}$, which can take any value we want.


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If $$ M:=\pmatrix{1&-1\\-1&17}, \quad \Omega:=\pmatrix{3&4&3&5\\3&5&2&2\\2&4&1&4\\4&2&2&4}, \quad F_1:=\pmatrix{1&1\\0&4}, \quad F_2:=\pmatrix{1&4\\0&1}, $$ the function $$ \phi(t):=f(tF_1+(1-t)F_2)-[tf(F_1)+(1-t)f(F_2)] $$ attains both positive and negative values on $[0,1]$. EDIT ...


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Consider two points with $x = a > 0$ and $y > 0$. They lie on a straight line which is both convex and concave. Now consider two points on $x=y,\quad x>0$, here $f=x^2$ which is convex. Now, two points on $x+y=5$ (say) gives you a concave function, so $f$ is clearly neither convex nor concave in your region (or indeed, anywhere).


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You can write it as $(g(x)+\sin(x))-g(x)$ where $g(x)$ is any function with $g''(x)\geq 1$. For instance, $(x^2+\sin(x))-x^2$ as was given in the comments.


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The derivative of $\sin(x)$ can be written as the difference of two continuous positive functions.


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This is not true. Take $X = [0,1] \subset \mathbb{R}^1$ and $f_n(x) = 1-x^n$.


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There is no formal proof, because it is not always true. In fact, it is possible to select $A$ and $b$ such that (2) is feasible but (1) is not. The properties of the objective function are basically irrelevant. You just happen to be lucky for your particular instance. I am sure there are particular cases where the two problems are equivalent, mind you. But ...


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Let $x$ and $y$ be real numbers. Then $f(y) + f'(y)(x-y) = y^2 + 2y(x-y) = 2xy -y^2$. Note that $2xy - y^2 - x^2 = -(x-y)^2 \le 0$. Particularly this says that $2xy - y^2 \le x^2$. Translating this into terms involving $f$, we have: $f(y) + f'(y)(x-y)\le f(x)$. Hence $f$ is convex. I suspect you started with the expression $x^2 \ge y^2 +2xy(x-y)$ and wanted ...


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Let $z\in \mathbb{R}^n$,by assumption, $\varphi(z)\ge \varphi(x_k)+\langle y_k,z-x_k\rangle$, let $k\to \infty$ and use lsc of $\varphi$ at $x$ and joint continuity of inner product, you can get $$\varphi(z)\ge \varphi(x)+\langle y,z-x\rangle$$ which holds for all $z\in \mathbb{R}^n$. Hence $y\in\partial \varphi(x)$.


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The statement is not true. For example, let $S$ be the union of the open interval from $(0,0)$ to $(1,0)$ and the open interval from $(1,0)$ to $(2,0)$. Then $S$ is not convex but its closure, the closed interval from $(0,0)$ to $(2,0)$, is convex. A similar example works in any $\Bbb R^n$, even for $n=1$.


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Let's focus on that first inequality. We note four things: First: all three terms are concave functions of $x$, $y$, and $z$. The second and third terms are affine, of course, but it is their concavity we need here. Second: the first two terms are positive when the other inequalities are satisfied. For the first term, it helps to note that ...


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You can construct a countable infinite $D_f$ by letting $f$ be piecewise linear in the intervalls $[1-\frac{1}{n},1-\frac{1}{n+1}]$ with gradient $n$. An explicit example is given by: $$f(x) = \begin{cases} \frac{1}{x}-\frac{1}{2} & 0 < x \le \frac{1}{2} \\ n(x-1+\frac{1}{n}) + H(n) & 1-\frac{1}{n} \le x \le 1-\frac{1}{n+1}\,,\,\, n \ge 2 ...


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If $f$ is general, then $X$ may not be compact. Pick an arbitrary, strictly convex function $g$ with $g(0.5) = 0$, and define $f(x) = g(x)$ for $x \neq 0.5$ and $f(0.5) = 1$. You have $\tilde f = g$ and $X = [0,1]\setminus \{0.5\}$ which is not compact. If $f$ is continuous, then the result is true, because if $\tilde f(x)<f(x)$ then this happens on a ...


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The idea of the convex hull is to form a convex set from the set of starting vectors. So it is okay that the vectors themselves are not convex. One way to define the convex hull of a finite set of points is that it is the set of all points that can be written as a convex combination of your initial starting vectors. Formally, say your vectors are ...


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Here's how I ended up solving it. I made reference to this answer for the general principle, and this answer for the part about proving slopes of lines. Let $m = \frac{a+b}{2}$. Suppose we draw a line (on a two-dimensional Cartesian grid) from $(a,f(a))$ to $(b,f(b))$. Then the point $(m,f(m))$ lies on or below this line because $f$ is midpoint convex: the ...


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You're close. Here is a hint. To prove that $\mathbb{T}$ contains all the extreme measures, you might look for a contradiction: Assume $P$ is an extreme measure not in $\mathbb{T}$. Use the fact that $\mathbb{N}$ is the union of the set of atoms $\mathcal{A}$, and that those atoms are pair-wise disjoint. Then, apply the properties of probability measures to ...


0

You decompose $y$ on $d$ and $d^{\bot}$ : there exist $\lambda \in [0,1]$, $\mu \in \mathbb R$ and $e\in d^{\bot}$ such that $$y = \frac{\lambda (\mu d)+ e}{1-\lambda}$$ ie $$(1-\lambda)y + \lambda (-\mu d) = e$$ Then $$g(e) = g( (1-\lambda)y + \lambda (-\mu d) ) \leq \lambda g( -\mu d) + (1-\lambda) g(y)$$ $$g(e) - \lambda g(-\mu d) \leq (1-\lambda) ...


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Given $f$ is a continuous and using the results from this answer, $f$ can be proven to satisfy: $f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda)f(x_2)\ \forall \ \lambda \in [0,1]$ Now, by using Taylor's expansion, $f''(x)$ can be written as: $$ f''(x) = \lim_{h \rightarrow 0} \frac{f(x+h)+f(x-h)-2f(x)}{h^2} $$ $f(\frac{1}{2}(x+h) + ...


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I would set up a proof by contradiction. Assuming a single point where $f''(x) < 0$, you can use the continuity of $f''(x)$ to find an interval $[a,b]$, where $f''(x) < 0$ throughout. The intuition is then clear, in the sense that if you draw a concave down segment, then any secant line lies below your curve. I will leave it to you to fill in the ...


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Linear Matrix Inequalities should have been called Affine Matrix Inequalities but it got stuck with the first. The second form that is confusing you is the base form. The screenshot is in the more natural block variable form. Assume you have a 2x2 LMI given as $$ \begin{bmatrix} a&b\\b&1 \end{bmatrix} \succ 0 $$ Then from this block form to the ...


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Let $m$ be the midpoint of $[a,b].$ We have $f(m)\le (f(a) + f(b))/2.$ This is the same as saying $(m,f(m))$ lies on or below the line through $(a,f(a)), (b,f(b)).$ Thus the slope from $(a,f(a))$ to $(m,f(m))$ is $\le$ the slope from $(m,f(m))$ to $(b,f(b)).$ Apply the MVT to see this implies $f'(c)\le f'(d)$ for some $c\in (a,m), d \in (m,b).$ Apply the MVT ...


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Assuming midpoint convexity means $$ f(a/2+b/2)\leq f(a)/2+f(b)/2 \text{ for all }a,b, $$ I would try to approximate $f''(y)$ as with a two-sided second difference quotient, i.e., $$ (f(y+h)-2f(y)+f(y-h))/h^2\to f''(y),\text{ as }h\to0 $$ and then look at the implications of midpoint convexity on a relationship between the various summands in the numerator ...



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