# Prove this function is convex

we have $g: R^n \rightarrow R$ is a concave function and $S$={$x :g(x)> 0$} and $f:S \rightarrow R$ and $f(x)$=$1/g(x)$ so we must show that $f$ is a convex function

• Are you getting this from Boyd & Vandenberghe's book? Because if so you can use their composition rules, if you carefully define what the outer function is. – Michael Grant Dec 9 '14 at 3:58
• sorry,my master get me this exercise .I don't know which book has this – user157745 Dec 9 '14 at 4:15
• That's fine. I just wanted to know what principles you are permitted to use. – Michael Grant Dec 9 '14 at 12:33
• thanks .but the question didn't say any thing about domain.so we must assume the domain is true – user157745 Dec 9 '14 at 17:50
• Ah, but it did: it's $S$. What you do is create a function that equals $1/x$ on $S$, but has the domain you need to make the proof work. – Michael Grant Dec 9 '14 at 17:51

Since $g$ is concave and positive on $S$, for all $x,y \in S$ and $\lambda \in [0,1]$,

$$g[\lambda x + (1- \lambda)y] \geqslant \lambda g(x) + (1-\lambda)g(y)> 0.$$

For every $x,y \in S$, we have $f(x),f(y) > 0$ and

$$f[\lambda x + (1- \lambda)y] = \frac{1}{g[\lambda x + (1- \lambda)y]} \\\leqslant \frac{1}{\lambda g(x) + (1-\lambda)g(y)}= \frac{1}{\lambda [f(x)]^{-1} + (1-\lambda)[f(y)]^{-1}} \\\leqslant \frac{1}{[f(x)]^{-\lambda}[f(y)]^{-(1-\lambda)}}= [f(x)]^{\lambda}[f(y)]^{(1-\lambda)} \\ \leq \lambda f(x) + (1-\lambda)f(y).$$

Therefore, $f$ is convex.

The following inequality -- for nonnegative $a,b$ and $0 \leqslant \lambda \leqslant 1$ -- is used above:

$$a^\lambda b^{1-\lambda} \leqslant \lambda a + ( 1 - \lambda) b.\tag{*}$$

Proof of (*).

Note that $h(t) = \lambda t - t^\lambda$ has a minimum at $t = 1$. Hence, $t^\lambda \leqslant \lambda t + 1 - \lambda$. Let $t = a/b$. Then $(a/b)^\lambda\leqslant \lambda(a/b) + 1 - \lambda.$ Multiplying both sides by $b$, we get $a^\lambda b^{1-\lambda}\leqslant \lambda a + (1 - \lambda)b.$

• I can't understand this part in the proof of * $t^\lambda \leqslant \lambda t + 1 - \lambda$ . – user157745 Dec 13 '14 at 9:03
• @user157745: $h(t)$ is minimized at $t=1$. So $h(t) \geqslant h(1) \implies \lambda t - t^{\lambda} \geqslant \lambda -1 \implies t^{\lambda} \leqslant \lambda t +1 - \lambda$. – RRL Dec 13 '14 at 9:09
• yes ,thank you for all things – user157745 Dec 16 '14 at 16:24

Read Convex Optimization by Boyd & Vandenberghe---Chapter 3 specifically, but you need to be comfortable with the first two chapters to fully digest that. There you will find a set of composition rules that govern when convexity/concavity is preserved in the composition of nonlinear functions.

You can see a proof of one of those rules here on Math.SE, but that's not the one we want here. The rule we need to use is: the composition of a convex, decreasing (or nonincreasing) function with a concave function is convex.

That rule is satisfied here by $f(x)=h(g(x))$, where $$h(x)=1/x \qquad \mathop{\textrm{dom}} h=(0,+\infty)\supseteq S$$ Note that the restricted domain is important: without it, $1/x$ is neither convex nor concave. But with it, $h$ is a convex, decreasing function. So its composition with a concave function produces a convex result.

EDIT: the OP commented: "but the question didn't say any thing about domain.so we must assume the domain is true". In a sense, this is correct. But note that the function $g(x)$ is known to be strictly positive. Therefore, my function $h(g(x))$ is exactly equivalent to $1/g(x)$, and can serve as the basis for the proof. I would say that it was an error on my part not to explicitly point this out at first. Obviously, if $g(x)$ were not known to be positive, you could not do this.