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Prove that $u_b(x)=1-\frac12(e^{-2x}+e^{-bx^2})$ is concave for $\frac12<b<\frac32$. What about b=$\frac1{20}$? $b=2$?

By disregarding the 1 and then the $\frac12$ we can turn it into the equivalent problem to show that $e^{-2x}+e^{-bx^2}$ is convex.

the second derivative is $2be^{-b x^2} (2 b x^2-1)+4 e^{-2 x}$, but showing its nonnegativity requires somehow to manipulate the different exponents, and neglecting terms has led to nothing.

Emolga
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1 Answers1

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HINT:

Note that $$\begin{array}{rcl} 2b\mathrm{e}^{-bx^2}>0&&\forall x\in\mathbb{R}\\ 4\mathrm{e}^{-2x}>0&&\forall x\in\mathbb{R} \end{array}$$

So you only need to verify the sign of $(2bx^2-1)$.

cjferes
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  • Actually I've already done that and I can't see how does it help. This seems too 'brutal' since now for $b=\frac12$ we no longer have that it is $\geq 0$ for all x, e.g. not for $x=0$. – Emolga Aug 19 '14 at 18:51
  • but the condition says $\frac{1}{2}<b<\frac{3}{2}$. What's the problema there? – cjferes Aug 19 '14 at 18:56
  • Exactly the same argument works for $b=1, x=0$. – Emolga Aug 19 '14 at 19:00
  • Are there assumptions on $x$? I can't seem to find a suitable answer... although the hint of discarding everything with a known sign stands. – cjferes Aug 19 '14 at 19:11