$x_1, x_2 &gt; 0$  $f(x_1,x_2) = \frac{1}{x_1^3x_2}$  Is this function  convex?  I think it is because Hessian matrix is $$\begin{pmatrix} \frac{12}{x_1^5x_2} &amp; \frac{3}{x_1^4x_2^2} \\ \frac{3}{x_1^4x_2^2} &amp; \frac{2}{x_1^3x_2^3}\end{pmatrix}$$  How do i show that the matrix is positive? multiply it by vector a^T and a isn't working..  Thanks!  can i just say its positive cause every element is positive?
123793
For $x_1, x_2 &gt; 0$, let $$f(x_1,x_2) := \frac{1}{x_1^3x_2}$$ Is function $f$ convex?  I think it is because Hessian matrix is $$\begin{pmatrix} \frac{12}{x_1^5x_2} &amp; \frac{3}{x_1^4x_2^2} \\ \frac{3}{x_1^4x_2^2} &amp; \frac{2}{x_1^3x_2^3}\end{pmatrix}$$ And I think the matrix is positive because each value is positive (is it true?) Thanks!
48948
I'm trying to solve the following problem:  Let $f:K\rightarrow \mathbb{R} $, $f$ convex and $K \subseteq \mathbb{R}^n$ convex. Then $f$ is continuous on $K$.   I have proved the only case $n=1$, but for an arbitrary $n$??
106188
Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function such that for all $a,b\in [0,1]$ with $a&lt;b$ $$f\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_a^b f(x)\,dx.$$ How to prove that $f$ is a convex function? Thank you for your help.
63901
I was reading this method related to proximal newton methods  . I came across this page    I didn't get what this part means  $ \frac{1}{2}||y-x||^2_H $ what does H signify here and how do I calculate $ \frac{1}{2}||y-x||^2_H $ given x and y.
71449
Let $f:\mathbb{R}^n\to \mathbb{R}$ be a convex function. It is well-known that (see e.g.~) that there exists a set $A$ such that  $$ f(x)=\sup_{a\in A}A(a)x+b(a), \quad \forall x\in \mathbb{R}^n, $$ where $A(a)\in\mathbb{R}^{n\times n}$ and $b(a)\in \mathbb{R}^n$.  I was wondering, is there any reference saying we can choose $A$ as a countable set?
50249
$$f(x_1,x_2,...x_n):\mathbb{R}^n \rightarrow \mathbb{R}$$ The definition of the gradient is  $$ \frac{\partial f}{\partial x_1}\hat{e}_1 +\  ... +\frac{\partial f}{\partial x_n}\hat{e}_n$$  which is a vector.  Reading this definition makes me consider that each component of the gradient corresponds to the rate of change with respect to my objective function if I go along with the direction $\hat{e}_i$.  But I can't see why this vector (defined by the definition of the gradient) has anything to do with the steepest descent.  Why do I get maximal value again if I move along with the direction of gradient?
87424
I am minimizing a convex function $f(x,y)$ using the steepest descent method: $$\mathbf{x}_{n+1}=\mathbf{x}_n-\gamma \nabla F(\mathbf{x}_n),\ n \ge 0$$  My function is defined over a specific domain $D = \{(x, y) \in R^2 : 2x^2+y^2 &lt; 10\}$, if my $x_{n+1}$ goes out of bound, my method diverge. I was told to choose $\gamma$ using armijo rule to stay inbound.   Can someone explain to me the method and also describe a pseudo code for implementing it?
833
If $(X,\|\cdot\|)$ is a normed linear space, then how to show any ball $B(x,r)$ is convex? I know that if $x,y\in A\subset V$ then $[x,y]\subset A$, where $A$ is a convex subset of vector space $V$ and $[x,y]=\{(1-t)x+ty\mid 0\leq t \leq 1\}$. Please give me some hint.  I tried the following:  Claim. $[a,b] \subset B(x,r)$ Let $a,b \in B(x,r).$ Then I get $\|x-a\|&lt;r$ and $\|x-b\|&lt;r$.  Is it right?
94046
I have two functions $f_1(x)$ and $f_2(x)$ which are strictly quasiconcave (since they have positive derivative before a certain point and negative derivative after that point. The points where derivative become zero can be different for each of the functions). I want to ask whether $f(x)=f_1(x)f_2(x)$ will be a quasiconcave function or not? Any help in this regard will be highly appreciated. Thanks in advance.
88705
According to ,  It describes the local curvature of a function.   AFAIK, for one-variable function $f(x)$, its local curvature is  $$\kappa = \frac{|f''|}{(1 + f'^2)^{3/2}},$$ and its Hessian matrix is  $$\mathcal{Hess}(f) = [f''],$$ right? And here is my problem, I think the local curvature is not just described by its Hessian matrix, because $f'$ also has its role in it, doesn't it?  And furthermore, for 2-variable function $f(x,y)$, its Hessian matrix is  $$\mathcal{Hess}(f) = \left[ \begin{array}{cc} f_{xx}'' &amp; f_{xy}'' \\ f_{xy}'' &amp; f_{yy}'' \end{array} \right].$$ How does it relate to the local curvature of $f(x,y)$?
59173
Let $f\colon\mathbb R \to \mathbb R$ be convex and $X_1,\ldots,X_n$ be i.i.d. random variables (e.g., uniform). Furthermore, suppose $g_1,\ldots,g_n\colon\mathbb R\to\mathbb R$ are increasing. Is it true that $$\mathbb E f(g_1(X_1) + g_2(X_1) + \cdots + g_n(X_1)) \ge \mathbb E f(g_1(X_1) + g_2(X_2) + \cdots + g_n(X_n))? $$ Like the previous version of this question (), which is the special case where the $g_i$ are all the identity, it seems like one approach is to show that $g_1(X_1) + \cdots + g_n(X_1)$ is a mean-preserving spread of $g_1(X_1) + \cdots + g_n(X_n)$. As before, I am also curious if there is a name for this sort of result/inequality.
90814
In  (section 2.3), it is stated that:     A point $x^*$ is a minimizer of a function $f$ (not necessarily convex) if and only if $f$ is subdifferentiable at $x^*$ and $0 \in\partial f(x^*).$   Could anybody provide me with references for a proof of the above statement?   Is there a reference where we can learn more about subgradients of non-convex functions?  In Section 3. Calculus of subgradients of the above notes,many properties of subgradients are presented for convex functions. I would like to know which ones among these properties still hold for non-convex functions.  Thank you in advance!
61631
How to prove the real value function $f$ is convex on $[a,d]$ if $f$ is convex on both $[a,c]$ and $[b,d]$? ($a&lt;b&lt;c&lt;d$).    I try to show that: for arbitrary $x\in(a,b), z\in(c,d)$, and for all $y\in (x,z)$ the inequality   $\dfrac{f(y)-f(x)}{y-x}\leqslant \dfrac{f(z)-f(y)}{z-y}$  holds. Then $f$ is convex on $[a,d]$.   But it seems to need some inequality tricks, I can't finish it.
16309
In Milnor's Morse Theory, the Hessian of a smooth function $f : M \to \mathbb R$ defined on a manifold $M$ at a critical point $p$ is the bilinear functional on $T_p M$ defined as follows:  $$f_{**}(v, w) = \tilde v_p(\tilde w(f))$$  where $\tilde v, \tilde w$ are vector fields that extend $v$ and $w$. It is written that $\tilde w(f)$ denotes the directional derivative of $f$ in the direction $\tilde w$ and that $\tilde v_p$ is, of course, $v$. I am assuming that this means $\tilde w(f) = T_p(f)(w)$, but   I do not know what it means to write $v(T_p(f)(w))$ it is claimed that $f_{**}$ is symmetric because $$\tilde v_p(\tilde w(f)) - \tilde w_p(\tilde v(f)) = [v, w]_p(f) = 0$$ since $p$ is critical. If I am not mistaken, the argument used here is that since $p$ is critical, any directional derivative of $f$ at $p$ (e.g. $[v, w]_p(f)$) is zero, but wouldn't that also mean that $\tilde w(f) = \tilde v(f) = 0$?   Hopefully, I am only mixing up notations. I would appreciate any comment.
23509
Suppose a differentiable, convex function $F(x)$ exists. Then $b = a - \gamma\nabla F(a)$ implies that $F(b) \leq F(a)$ given $\gamma$ is chosen properly. The goal is to find the optimal $\gamma$ at each step. In my book, in order to do this, one should minimize  $G(\gamma)=$ $F(x-\gamma\nabla F(x))$ for $\gamma$. It also says that it should be minimized via a line search. Why can't this function be minimized by simple calculus?  I am unsure what it means to perform a line search on this function. I am also unsure why we want to minimize this function to find the optimal step size.
121069
I feel like this should be straightforward, but does anyone have a proof of the following?  Let $f: \mathbb{R}^n \to \mathbb{R}$ satisfy the following. For each coordinate $i$, for an arbitrary vector $x_{i}$, define $ f_i(y)$ to be $f$ restricted to the $i^{th}$ parameter, fixing the others $ x_{i}$; then $ f_i(y)$ is convex. Then $f$ is a convex function.  Also, if there is a more standard math notation I should use to describe this problem, what would that be?  (Edit: or a counterexample...)  Thanks!    I should add what I've tried.  $g(\lambda \vec{x} + (1-\lambda)\vec{y}) = g(\lambda x_1 + (1-\lambda)y_1,\dots,\lambda x_n + (1-\lambda)y_n)$  $\leq \lambda g(x_1,\lambda x_2 + (1-\lambda)y_2,\dots) + (1-\lambda) g(y_1,\lambda x_2 + (1-\lambda)y_2,\dots)$  I don't see this going anywhere good, we end up with something nasty like  $\lambda^n g(\vec{x}) + \lambda^{n-1}(1-\lambda) \sum_i g(y_i,\vec{x}_{i}) + \dots + (1-\lambda)^n g(\vec{y})$  I think. Maybe we can get it from there?
31224
In convex optimization, the feasible region is convex if equality constraints $h(x)$ are linear or affine, and inequality constraints $g(x) \leq 0$ are convex. Does this mean that if $h(x)$ is nonlinear, then the optimization problem is non-convex? Is there any special case where $h(x)$ is nonlinear and the feasible region is still convex?