0

If I have a function $f:\mathbb{R}^n\to\mathbb{R}$ that is convex in ${\bf x} = (x_1,x_2,\ldots,x_n)$ and strictly convex in one of the variables, say $x_1$, then is $f({\bf x})$ strictly convex in ${\bf x}$? If so, how would I prove this?

jonem
  • 383

2 Answers2

1

No, just consider $f(x) = x_1^2$. This is strictly convex in $x_1$ and weakly convex in $x$ but not strictly convex in $x$.

user2566092
  • 26,142
  • Wouldn't $f$ then be a function from $\mathbb{R}$ to $\mathbb{R}$? I would like to place the additional restriction that each $x_i$ needs to show up in the function. – jonem Apr 02 '14 at 19:37
  • isn't $f(\lambda x+(1-\lambda)y)<\lambda f(x) + (1-\lambda) f(y)$ for $x,y\in\mathbb{R}^n$, $\lambda\in(0,1)$? Meaning that $f$ is strictly convex. – jonem Apr 03 '14 at 19:51
1

Take any convex (but not strictly convex) function $c:\mathbb{R} \to \mathbb{R}$ and any strictly convex function $s:\mathbb{R} \to \mathbb{R}$, then the function $f:\mathbb{R}^2 \to \mathbb{R}$ given by $f(x,y) = c(x)+s(y)$ satisfies the criteria in the question, but clearly $f$ is not strictly convex for any fixed $y$.

copper.hat
  • 172,524
  • Thanks, I asked a similar question a while ago, care to explain the difference between this case and the one seen here: http://math.stackexchange.com/questions/720680/is-this-function-strictly-convex I'm a bit confused. – jonem Apr 02 '14 at 19:39
  • 1
    If I get a chance I will have a look at it. – copper.hat Apr 02 '14 at 19:57
  • What do you mean by 'explain the difference'? There are many differences... – copper.hat Apr 02 '14 at 20:08
  • Sorry for the ambiguity. I don't understand what additional property of the function (in the link) is making it strictly convex compared to only convexity of the function in the question posed here. If I were to add an additional assumption to the question above to make $f$ strictly convex, what would it be? Is it because each function in the sum of the three terms (in the link) is convex in the same variable that the function is strictly convex in (i.e. $x_1$)? Relating it back to this post, being convex in ${\bf x}$ and strictly convex in one variable is not enough. Thanks for your attention – jonem Apr 02 '14 at 20:15
  • 1
    I added an answer to the other question. Hopefully that helps answer your question. Note that if the function $c$ above was strictly convex, then $f$ would be too. – copper.hat Apr 02 '14 at 22:57