Prove that $f(x)=-\exp(-g(x))$ is convex if $g(x)$ is convex...

Question

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.

Answer 1

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.

Answer 2

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 $H - (\nabla g) (\nabla g)^T$.

Answer 3

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.$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \Box$

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Now apply the lemma with $h(a) := -\exp(-a)$.