0
$\begingroup$

In Stephen Boyd & Lieven Vandenberghe's Convex Optimization (2004), section 3.2.1, the following is given without a proof

If $f(x,y)$ is convex in $x$ for each $y\in I$, and $w(y)\ge 0$ for each $y\in I$, then the function $g$ defined as $$g(x)=\int_Iw(y)f(x,y)\, dy$$ is convex in $x$ if the integral exists.

Here $I$ is understood to be an index set, or a more or less arbitrary measure space. One thing that I know may have a connection here is the fact that the pointwise supremum of a family of convex functions is again convex, and thus I intend to express this integral as a pointwise supremum of convex functions in some way. The biggest problem is the sign of $f(x,y)$. If it is constantly positive then I think I can just approximate the integral by a sequence of nonnegative, increasing simple functions which are easily shown to be convex since they can be expressed as finite positive weighted sums of convex functions. If $f(x,y)$ is uniformly bounded below then the problem can still be solved by an offsetting constant. Otherwise I don't know how to proceed. I don't think splitting $f$ as positive and negative parts can help because we are unable to handle $-f^-$ which is not convex.

Can you help? Thanks in advance.

  • 3
    1. Write down definition that $f(x,y)$ is convex. 2. Multiply by $w$. 3. Integrate. Steps 2 and 3 respect the inequality. – 2017-01-13
  • 0
    @A.G. Thank you this is so elegant! I never expected it to be this easy. – 2017-01-13

0 Answers 0