$e^{|t|} - |t| -1 , t \in \mathbb{R}$ is convex Consider the function $e^{|t|} - |t| - 1, t \in \mathbb{R}$. I want to show that such function is convex. I am trying to use that $e^x - x - 1, x>0$ is convex (this follows by the test of the second derivative) but I am getting anywhere . Someone could help me?
Thanks in advance

If $f(x)$ is a convex function on $\mathbb{R}^+$,
$g(x)=f(\left|x\right|)$ is a convex function on $\mathbb{R}$, since it is the composition of two convex functions. As you noticed, the convexity of $f(x)=e^x-x-1$ on $\mathbb{R}^+$ is trivial since
$$ e^x-1-x = \frac{x^2}{2}+\frac{x^3}{6}+\ldots $$
obviously has a positive second derivative on $\mathbb{R}^+$.
A small variation: since $g$ is a continuous function (as a composition of two continuous functions) in order to prove its convexity it is enough to prove its midpoint-convexity. On the other hand,
$$g\left(\frac{x+y}{2}\right)=f\left(\frac{|x+y|}{2}\right)\stackrel{(1)}{\leq}f\left(\frac{|x|+|y|}{2}\right)\stackrel{(2)}{\leq} \frac{f(|x|)+f(|y|)}{2}=\frac{g(x)+g(y)}{2}$$
holds, since $(1)$ follows from the fact that $|\cdot|$ is convex on $\mathbb{R}$ and $f$ is increasing on $\mathbb{R}^+$ and $(2)$ follows from the fact that $f$ is (midpoint-)convex on $\mathbb{R}^+$.
