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My question is the function $(x,y)\mapsto f(x,y)$ convex where

$$f(x,y)=x\log\left(1+\frac{1}{1+y}\right).$$

I have found the partial derivatives as follow:

$$\dfrac{\partial^2 f(x,y)}{\partial x^2}=0.$$ $$\dfrac{\partial^2 f(x,y)}{\partial y^2}=\dfrac{x(3+2y)}{(1+y)^2(2+y)^2}.$$ $$\dfrac{\partial^2 f(x,y)}{\partial x\partial y}=-\dfrac{1}{(1+y)(2+y)}.$$

But I cannot continue.

Ribz
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1 Answers1

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The function $y \mapsto \ln(1+ {1 \over 1+y} )$ is not convex for $y<2$, hence $f$ is not convex on $\{1\} \times (-\infty, -2)$.

copper.hat
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  • Why do you have the singleton ${1}$? $f(x, \cdot)$ is convex on $\mathbb{R}$ no? – Ribz Oct 31 '16 at 18:17
  • If $f$ is not convex on some convex subset, then it is not convex. I just picked the half line above as an example. $y \mapsto f(x,y)$ is not convex on $\mathbb{R}$. Just plot it. – copper.hat Oct 31 '16 at 18:19