Prove that $f(x)=-\exp(-g(x))$ is convex if $g(x)$ is convex... Show that the following function $f:\Re^{n}\rightarrow \Re$ is convex.
\begin{equation}
f(x)=-\exp(-g(x))
\end{equation}
where $g:\Re^{n}\rightarrow \Re$ is a twice differentiable function with convex domain and satisfies
\begin{eqnarray*}
\left(
\begin{array}{cc}
\nabla^{2} g & \nabla g  \\
\nabla^{T} g & 1 \\
\end{array}
\right)\geq 0 \;(semidefinite\; positive\; matrix)
\end{eqnarray*}
for $x\in Dom$ $g$.\
My idea to prove that is to show that the Hessian of $f$ is a semidefinite positive matrix. So, I computed the Hessian of $f$
\begin{eqnarray*}
\nabla^{2}f(x)&=&-\exp(-g(x))\left(
\begin{array}{cccc}
g_{x_{1}}^{2}-g_{x_{1}x_{1}} & g_{x_{2}}g_{x_{1}}-g_{x_{2}x_{1}} & \ldots & g_{x_{n}}g_{x_{1}}-g_{x_{n}x_{1}}  \\
g_{x_{2}}g_{x_{1}}-g_{x_{2}x_{1}} & g_{x_{2}}^{2}-g_{x_{2}x_{2}} & \ldots & g_{x_{n}}g_{x_{2}}-g_{x_{n}x_{2}}  \\
\vdots & \vdots & \ddots & \vdots \\
g_{x_{n}}g_{x_{1}}-g_{x_{n}x_{1}} & g_{x_{2}}g_{x_{n}}-g_{x_{2}x_{n}} & \ldots & g_{x_{n}}^{2}-g_{x_{n}x_{n}}
\end{array}
\right)\\
&=&\exp(-g(x))\left( \nabla^{2}g-\nabla g \nabla^{T}g\right) .
\end{eqnarray*}
Until now I havent been able to use the hipotesis to prove what I want, just that $\nabla^{2}g$ is a semidefinite positive matrix. Also, I know that $\nabla g \nabla^{T}g$ is a semidefinite positive matrix (but I dont know if this result is useful). Thanks for the help.
 A: In the one-variable case, what you need for a function $-\exp(-g(x))$ (where $g$ is twice differentiable) to be convex is $g''(x) \ge g'(x)^2$.  Thus in the many-variable case, you need $$ \dfrac{d^2}{dt^2} g(x_t) \ge \left(\frac{d}{dt} g(x_t)\right)^2 $$
on every line $x_t = a + b t$ in the domain.  This translates to
$$ b^T H b \ge (b \cdot \nabla g)^2 = b^T (\nabla g) (\nabla g)^T b \ \text{for all } b$$
where $H$ is the Hessian of $g$, and that is equivalent to positive semidefiniteness of $H - (\nabla g) (\nabla g)^T$.
A: You don't need differentiability of $g$...
Lemma: If $g: \mathbb R^n \rightarrow (-\infty, +\infty]$ is convex and $h: \mathbb R \rightarrow \mathbb R$ is convex non-decreasing, then $h \circ g$ is convex.
Proof:
Let $x, y \in \mathbb R^n$ and $t \in [0, 1]$. Then
$$
\begin{split}
(h \circ g)(tx + (1-t)y) &:= h(g(tx + (1-t)y)) \\
&\le h(t g(x) + (1-t)g(y))\text{ ($g$ convex, $h$ non-decreasing)}\\
&\le t h(g(x)) + (1-t)h(g(y))\text{ ($h$ convex)} \\
&=: t(h \circ g)(x) + (1-t)(h \circ g)(y),
\end{split}
$$
showing that $h \circ g$ is convex.$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad  \Box$

Now apply the lemma with $h(a) := -\exp(-a)$.
A: Let $x\in \Re^{n}$ then by the hipotesis
\begin{eqnarray*}
(x^{T},-x^{T}\nabla g)
\left(
\begin{array}{cc}
\nabla^{2} g & \nabla g  \\
(\nabla g)^{T} & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
x \\
-(\nabla g)^{T} x \\
\end{array}
\right)=x^{T}\nabla^{2}gx-x\nabla g(\nabla g)^{T}x\geq 0 
\end{eqnarray*}
since $x\in \Re^{n}$ is arbitrary, then the Hessian of $f$ is positive-semidefinite which implies that $f$ is convex.
