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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?

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  • $\begingroup$ If the domain is open (and of course convex), that theorem is still valid. $\endgroup$ – user251257 May 26 '17 at 17:47
  • $\begingroup$ The definition is usually the simplest way. It is in this case. What did you have trouble with, in applying the definition? $\endgroup$ – vadim123 May 26 '17 at 17:47
  • $\begingroup$ Essentially, how I would show that the inequality holds for $|y|<1$. $\endgroup$ – Chaerephon May 26 '17 at 17:50
  • $\begingroup$ Isn't $g(y) =\sqrt{1-y^2}$ strictly concave? Have you forget a minus somewhere? $\endgroup$ – user251257 May 26 '17 at 17:56
  • $\begingroup$ Gosh, my bad...yes...! I'll edit my post (thank you) $\endgroup$ – Chaerephon May 26 '17 at 17:59
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hint

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

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