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So, my typical approach to showing that a function is strictly convex would be to make use of the rule that if $f''_{11} \cdot f''_{22}-(f''_{12})^{2}>0$ and $f''_{11}>0$, then $f(x,y)$ is strictly convex.

Unfortunately, I lack the mathematical toolkit to show this (or rather, strict convexity generally) for a certain range of values. Are there any suggestions as to how I might prove that this function is strictly convex in a "simple" way?

Chaerephon
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  • If the domain is open (and of course convex), that theorem is still valid. – user251257 May 26 '17 at 17:47
  • The definition is usually the simplest way. It is in this case. What did you have trouble with, in applying the definition? – vadim123 May 26 '17 at 17:47
  • Essentially, how I would show that the inequality holds for $|y|<1$. – Chaerephon May 26 '17 at 17:50
  • Isn't $g(y) =\sqrt{1-y^2}$ strictly concave? Have you forget a minus somewhere? – user251257 May 26 '17 at 17:56
  • Gosh, my bad...yes...! I'll edit my post (thank you) – Chaerephon May 26 '17 at 17:59
  • I am not sure that the edit helps, as $e^{-x}>0$ and $f(x, \cdot)$ is strictly concave for any fixed $x$. – user251257 May 26 '17 at 18:02
  • I'm really sorry, that's two mistakes on my part...it's $1+y^{2}$ – Chaerephon May 26 '17 at 18:03
  • OK, so inspecting the two equations by eye, I can tell that if $|y|<1$ then the function is concave--but this seems a quite unsatisfying way to show that the inequality DOES hold for these two equations: too hand-wavy. Or would a verbal explanation be sufficient? This is for a maths for economists course. – Chaerephon May 26 '17 at 18:05
  • No no, after your edit, $f$ is indeed strictly convex. You can check that using the second derivative or notice that $x\mapsto e^{-x}$ and $y\mapsto \sqrt{1+y^2}$ are positive and strictly convex. – user251257 May 26 '17 at 19:05
  • I really am on a roll--I meant convex. Your explanation makes sense, I think I was over-complicating things. – Chaerephon May 26 '17 at 19:13

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hint

observe that $$f'_1=f''_{11}=f $$

hamam_Abdallah
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