Suppose $f(x_1, \dots, x_n): \mathbb{R}^n \to \mathbb{R}$. Is there an example of such $f$ that is convex in every argument and not convex? Are there any additional conditions, under which 'marginal' convexity would be sufficient to call $f$ convex?
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?
2383
Suppose $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ and for all bounded measurable $f$, $$ \varphi\Big(\int_0^1fd\lambda\Big) \le \int_0^1\varphi(f)d\lambda $$ I'm asked to prove that $\varphi$ is a convex function. I have no idea how to even begin, only idea I've had is to try to suppose that $\varphi''(x)&lt;0$ for some $x\in(0,1)$ but then I haven't got a clue.
22494
Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?  Edit:  The proof mentioned above is from : Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is "as otherwise $C\cap T$ would be contained in a hyperplane".  Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of . To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.
123458
Let $f\colon X \to [ -\infty, +\infty]$ be proper and strictly convex. Show that $f$ has at most one minimizer. Give examples where $f$ is strictly convex, and either (i) $f$ has one minimizer; or (ii) $f$ has no minimizer.  For the strict convexity part, you may use the fact that f is strictly convex whenever its Hessian is positive definite.   I am not sure how to show that f has at most one minimizer, could someone help get me started with the question?
72223
Let $X$ be compact Hausdorff, and $C(X)$ the space of continuous functions over $X$. Denote the closed unit ball in $C(X)$ by $(C(X))_1$, then it can be shown $f$ is an extreme point of $(C(X))_1$ if and only if $|f(x)|=1$ for all $x\in X$.  In , we are asked to show that the convex hull of extreme points of $(C(X))_1$ is dense in $(C(X))_1$. By using a theorem of Fejer we can show this is true when $X=[0,1]$. But I do not know how to do it for general $X$.   Since it is only the 7th problem in the first chapter, I am guessing the solution should be elementary.   Thanks!
59814
$\newcommand{\prox}{\operatorname{prox}}$ $\newcommand{\argmin}{\operatorname{argmin}}$ $\newcommand{\dom}{\operatorname{dom}}$  Recall again that the proximal operator for vectors $\prox_{f}: R^n \rightarrow R^n$ of $f$ is defined as:  $\prox_f(v) := \argmin_{x} \left(f(x) +(1/2)\|x-v\|_2^2 \right) $,  where $f: R^n \rightarrow R \cup \infty$ is a closed proper convex function and $\|\cdot \|_2$ is the Euclidean norm. $\prox$ is strongly convex and not everywhere infinite, so there it will have a unique minimizer for every $v \in R^n$  The crucial property of proximal operator is that $x^*$ minimizes $f(x)$ iff $x^* = \prox_f(x^*)$, i.e. $x^*$ is a fixed point of $\prox_f$.  Let us consider the proof of this property. The first part, namely, that if $x^*$ minimizes $f$ then $\prox_f(x^*)=x^*$, is trivial:  $f(x) +(1/2)\|x-x^*\|_2^2 \le f(x^*) = f(x^*) +(1/2)\|x^*-x^*\|_2^2$  The second part uses the notion of subdifferential. In the proof authors say that $\tilde{x}$ minimizes  $f(x) +(1/2)\|x-v\|_2^2$  i.e. $\tilde{x} = \prox_f(v)$ iff $0 \in \partial f(\tilde{x}) + (\tilde{x}-v)$,   where the sum is of a set and a point. Recall the definition of the subdifferential:  $\partial f(x) = \{ y$ |  $ f(z)\le f(x) + y^T(z-x)$ $\forall z \in \dom{f}\}$  Take $\tilde{x}=v=x^*$, it follows that $0 \in \partial f(x^*)$ so $x^*$ minimizes $f$.  Question 1:  Consider proximal operator defined for matrices now:  $\prox_f(Y) := \argmin_{X} \left(f(X) +(1/2)\|X-Y\|^2 \right)$,  where $X$ is some real $m$ by $n$ matrix. What norm function should be taken in the definition of the $\prox$ in this case? Frobenius? Or spectral norm (induced 2nd norm) which is the largest eigenvalue (is spectral norm even differentiable?)? Or something else?  Question 2:   Can the norm for the definition of $\prox$ in case of matrices be chosen in different ways? What conditions does the proof above impose on the norm function?
99044
I was wondering if linear function is convex or concave? For example f(x)=x, is function whose second derivate is 0 so we cant tell anything using this criteria. Can someone help?
116901
Let  $f:X\times Y\to \mathbb{R}$. I could not come up with an example that satisfies $$\sup_{y\in Y}\inf_{x\in X}f(x,y)&lt;\inf_{x\in X}\sup_{y\in Y}f(x,y).$$ Any help would be appreciated. Thanks!
42968
I have seen that if a set $K$ on an Hilbert space $H$ is convex and strongly sequentially-closed, it is weakly closed. The teacher said that if you take a convex and weakly lower semicontinuous functional $F$, using the fact that the sets $F^{-1}(-\infty, \lambda]$ are convex and that closure implies weak closure, it is easy to conclude that convexity and strong lower semicontinuity imply weak lower semicontinuity. I do not see how to do that though. I would like to see a proof not involving weak topologies etc. The way he said it menat it was supposed to be done using only the definitions, or little more.
56436
I'm trying to prove that the following set ($B$) defined by an ellipse is convex, but I am getting stuck.  $$B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} $$  My sad attempt:  $$ \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1]$$  $$ \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2  )$$ $$3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)$$  and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form $(\lambda x_1 -3)^2$, $((1-\lambda) y_1 -3)^2$, $(\lambda x_2 -3)^2$, and $((1-\lambda) x_2 -3)^2$, but I don't think that helps the situation any.
85235
Today I was going through my text book exercise and got hold of the following question.     Let $X$ be a normed linear space with norm $\lVert\cdot\rVert$ and $A$ is a non empty convex subset of $X$ then prove that $\operatorname{Closure}(A)$ is also a convex subset of $X$.   I gave it a try but could not succeed.
416
If $C\subset\mathbb{R}^m$ is a convex set, $A$ is an $m\times n$-matrix and $b\in\mathbb{R}^m$, how do I prove that the set $S=\{x\in\mathbb{R}^m|Ax+b\in C\}$ is convex?  I know that the definition of convexity is that all combinations $\lambda x+(1-\lambda)y$ (with $\lambda\in[0,1]$) is in $S$ for pairs $x,y\in S$, but I don't see how to prove this for my $S$.
60696
Consider a real-valued convex function $f$ defined on an open interval $(a,b) \subset \mathbb{R}$. Let $x,y \in (a,b)$.  I want to prove that \begin{equation} f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y) , \lambda \in [0,1]\implies f(y) + f'(y)(x-y) \leq f(x) \end{equation} No other assumptions (regarding the second derivatives) is allowed.  While both facts are really obvious and clear, I am unable to prove this without making additional assumptions.
61671
How to prove that the intersection of any finite number of convex sets is a convex set?  I have no idea.
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 $n$??
85253
I am working through a bank of previous exams and couldn't figure a problem out to my satisfaction.     Let $f(x) : \mathbb{R} \to \mathbb{R}\,$ be a continuous function.         Show that $f$ can have at most countably many strict local maxima.   Assume that $f$ is not monotone on any interval. Then show that the local   maxima of $f$ are dense in   $\mathbb{R}$.
19467
How can we show that a bounded and convex function on $\mathbb R$ is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all.  Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.