Prove $f$ is a convex function. Prove $f$ is a convex function where 
$$f(x)=\sum^m_{i=1}e^{-\frac{1}{f_i(x)}}$$ dom$f=\{x|f_i(x)<0, i=1, \dots, m\}$, where the functions $f_i$ are convex and $x \in \mathbb R^n$.
I was trying to prove it by the definition of a convex function:
$f$ is called convex if:
$\forall x_1, x_2 \in X, \forall t \in [0, 1]:$ 
$$f(tx_1+(1-t)x_2)\leq t f(x_1)+(1-t)f(x_2)$$ But when I plug in the vectors the while formula is kind of messy. Any ideas? Thanks a lot~
 A: Here's the argument:


*

*The function $g(x) = \exp{(-1/x)}$ is convex and nondecreasing on negative numbers.

*If $f$ is convex and $g$ is convex and nondecreasing, then $g\circ f$ is convex.

*Hence if $f$ is convex and its output is negative, then $\exp{[-1/f(x)]}$ is convex.

*Convex functions are closed under sums. 

*Hence if $f_1,\ldots, f_n$ are functions as described in the question, then $f \equiv \sum_{i} \exp{[-1/f_i]}$ is convex.


Proof of 2: If $f$ is convex, and $g$ is both convex and nondecreasing, then $g\circ f$ is convex.
Proof: Fix $x, y$ in the domain of $f$, and $t\in [0,1]$. Then:
$$
\begin{align*}
(g\circ f)(tx + (1-t)y) &= g( f(tx+(1-t)y)\\
&\leq g(tf(x) + (1-t)f(y)) & f\text{ convex}, g\text{ nondecreasing (!)}\\
&\leq tg(f(x)) + (1-t)g(f(y))& g\text{ convex}\\
&= t(g\circ f)(x) + (1-t)(g\circ f)(y)
\end{align*}$$
A: Hints:
The projection functions $x \rightarrow x_i$ are convex.
The $1/x$ function is convex.
The $e^{-x}$ function is convex. 
The composition of a convex function with an increasing convex function is convex.
The sum of convex functions is convex. 
You are done.
You can prove each of the above (proofs can also be easily found by Googling).
