[Math] Prove $f$ is a convex function.

calculusconvex optimizationconvex-analysisnumerical methods

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~

Best Answer

Here's the argument:

  1. The function $g(x) = \exp{(-1/x)}$ is convex and nondecreasing on negative numbers.
  2. If $f$ is convex and $g$ is convex and nondecreasing, then $g\circ f$ is convex.
  3. Hence if $f$ is convex and its output is negative, then $\exp{[-1/f(x)]}$ is convex.
  4. Convex functions are closed under sums.
  5. 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:

$$ \begin{align*} (g\circ f)(tx + (1-t)y) &= g( f(tx+(1-t)y)\\ &\leq g(tf(x) + (1-t)f(y)) & f\text{ convex}, g\text{ nondecreasing (!)}\\ &\leq tg(f(x)) + (1-t)g(f(y))& g\text{ convex}\\ &= t(g\circ f)(x) + (1-t)(g\circ f)(y) \end{align*}$$

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