Strongly convex, bounded from below by a quadratic function.

Question

A strongly convex function $V: \mathbb{R}^d \rightarrow \mathbb{R}$ with negative parameter is given, i.e.

$$ V(tx + (1-t)y) \leq tV(x) + (1-t) V(y) - \lambda t(1-t) | x -y |^2 , $$ with $\lambda<0$. Why is then $V$ bounded from below by a quadratic function?

I considered the cases $x=y$ and $x=-y$, but I do not know how to go on.

Are you sure that the question is right? What mean $\ge$ in $\mathbb R^d$?I think your mind is: $||V(tx+(1−t)y)||\ge t||V(x)||+(1-t)||V(y)||-λt(1−t)||x−y||^2$ — hamid kamali, May 19 '15 at 14:43
Oh, I am sorry! My first question, and immediately my first mistake! The function is of course real valued. — Jean Valjean, May 19 '15 at 14:45
No. the question is right. I'm sorry. I made a mistake that $V:\mathbb R^d \to \mathbb R^d$ — hamid kamali, May 19 '15 at 14:49
Then the quadratic term vanishes, I think. I also tried to set x=-y. But I do not know how to go on. But thank you, Agha kamali :) — Jean Valjean, May 19 '15 at 15:19
The sum $V(x)+c|x|^2$ will be convex for suitable $c>0$, and therefore bounded below by a linear function. — , May 19 '15 at 15:42
@JeanValjean: Is the norm in your question, Euclidean norm on $\mathbb R^d$? — hamid kamali, May 19 '15 at 16:21
I think you can solve the question similar this question: prove this is a strongly convex function — hamid kamali, May 19 '15 at 16:26

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

Proof for completeness (compiling the comments). Let $f(x)=-2\lambda\|x\|^2$. Then $f$ is strongly convex with parameter $-\lambda>0$. Therefore, $V+f$ is convex. It follows that the graph of $V+f$ lies above any of its tangent planes. So, $V+f$ is bounded from below by a linear function $L$, and thus $V\ge L-f$.

edited Apr 13 '17 at 12:21

Community

1

answered May 19 '15 at 21:32

Strongly convex, bounded from below by a quadratic function.

1 Answers1