# Showing that $e^{-x} \sqrt{1+y^2}$ is strictly convex for $\lvert y\rvert<1$

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?

• 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

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