In S. Boyd textbook p.105 (button): (cvx = convex)
Let F = log f & G = log g are convex (i.e. Let f & g are log-cvx)
(This guarantees f & g are cvx, since log-cvx is included in cvx)
Now, the book says:
From the composition rules for cvx functions: log(exp(F) + exp(G)) = log(f + g) is cvx.
I do not understand the last step?
What I think is:
exp(F) & exp(G) are cvx .... (OK)
exp(F) + exp(G) are cvx .... (OK)
However, logarithm is a concave function in R++
Composition rules:
Thanks!!
I was puzzled by the statement too. Here is my proof:
$F$ and $G$ are convex. With $e(*) = \exp(*)$, $$Z=\log[e(F) + e(G)]$$ $$Z' = \frac{[e(F)F' + e(G)G']}{[e(F) + e(G)]}$$ $$Z" = -\frac{[e(F)F' + e(G)G']^2}{[e(F) + e(G)]^2}+\frac{[e(F)(F')^2+ e(F)F''+ e(G)(G')^2+e(G)G'']}{[e(F) + e(G)]}$$ Check: $$T = [e(F)(F')^2+ e(F)F''+ e(G)(G')²+e(G)G'']\cdot [e(F) + e(G)] - [e(F)F' + e(G)G']^2$$ $$T = e(F)(F')^2(G) + e(G)(G')^2e(F) - 2e(G)e(F)F'G' + [e(F)F''+e(G)G'']\cdot [e(F)+e(G)]$$ $$= e(F)e(G)[F'-G']^2+[e(F)F''+e(G)G'']\cdot [e(F)+e(G)]$$ $$e(f)>0, \ e(G)>0, \ F''\ge 0, \ G''\ge 0.$$
So: $$T\ge0 \implies Z''\ge0 \implies Z \ \text{is convex}$$ Also: $$Z=\log(f+g)$$ $$\implies f+g \ \text{is log-convex.}$$