2

Assume that $\phi$ is continuous real function on $(a,b)$ such that $$\phi\Big(\frac{x+y}{2}\Big)\le \frac{1}{2}\phi (x) +\frac{1}{2} \phi (y)$$ for all $x,y \in (a,b)$. Prove that $\phi$ is convex.

We need to show that for $\lambda \in [0,1]$ we have $\phi(\lambda x+ (1-\lambda)y) \le \lambda \cdot\phi(x)+(1-\lambda)\cdot \phi (y) \quad\forall x,y \in (a,b)$. I had drawn some figures and it was coming out to be convex, for rigorous proof I was thinking of proving this for rationals $x,y$ then using continuity argument we can prove it for all $x,y$ but I'm not able to do so.

Mathronaut
  • 5,120
  • This looks like a good candidate for the contrapositive: Assume $\phi$ is continuous and not convex (i.e., the graph lies strictly above the secant line over some closed, bounded interval) and find $x$ and $y$ violating the inequality in the hypothesis. – Andrew D. Hwang May 08 '15 at 12:02
  • yes that will be a good geometrical argument but I'm looking for some rigorous proof. – Mathronaut May 08 '15 at 12:05
  • Jensen's inequality? – Alex May 08 '15 at 12:31
  • But this is the definition of convex function. Unless you are using other definition. – Tolaso May 08 '15 at 12:36
  • @Tolaso I'm using the definition I mentioned in my sketch of proof – Mathronaut May 08 '15 at 12:38
  • Then replace $\lambda$ with $1/2$ and you are done. You've proved it. Actually this is known as Jensen's inequality. – Tolaso May 08 '15 at 12:40
  • @Tolaso, I think you're not getting my point , I've to show that $\phi(\lambda x+ (1-\lambda)y) \le \lambda \cdot\phi(x)+(1-\lambda)\cdot \phi (y) \quad\forall x,y \in (a,b)$ given that $\phi\Big(\frac{x+y}{2}\Big)\le \frac{1}{2}\phi (x) +\frac{1}{2} \phi (y)$ – Mathronaut May 08 '15 at 12:42
  • @Neeraj: At this stage my comment is moot (since the question is a duplicate), but it's a perfectly rigorous analytic suggestion. :) (Naturally, you have to translate the statement about secant lines into inequalities: "There exist $x < y$ such that $\phi(\lambda x + (1 - \lambda)y) > \lambda \phi(x) + (1 - \lambda) \phi(y)$ for $0 < \lambda < 1$" and establish the truth of this assertion.) – Andrew D. Hwang May 08 '15 at 13:26
  • You can prove that $\phi(\lambda x + (1-\lambda) y) \le \lambda \phi(x) + (1-\lambda)\phi(y)$ whenever $\lambda = p/2^n$ and $p$ is odd and $0 < p < 2^n$, by induction on $n$. (For $n=1$, this is exactly your assumption.) Consider the cases $p < 2^{n-1}$ and $p>2^{n-1}$ separately. Then use continuity to prove the result for all $\lambda\in[0,1]$. – mjqxxxx May 11 '15 at 20:18

0 Answers0