# How to prove a generalised function is quasiconcave?

I have a question I have been asked to solve:

Given that $$(a_1, a_2,...,a_n)\in R_{++}^n$$ and $$(x_1, x_2,...,x_n)\in R_{++}^n$$, and $$A>0, \mu >0, p \neq 0$$, if there is a function $$f(x)=A(a_1x_1^p+a_2x_2^p+...+a_nx_n^p)^\frac{\mu}{p} \space \space \forall x \in R_{++}^n$$, show that function $$f$$ is a quasiconcave function over $$R_{++}^n$$ when $$p \in (0,1]$$.

I understand that to show a function is quasiconcave, it needs to satisfy this definition:

$$f:𝑅_+^𝑁→𝑅$$ is quasiconcave on $$𝑅_+^𝑁$$ if and only if $$∀𝒙,𝒚∈𝑅_+^𝑁$$ and for all $$𝑡\in(0,1)$$

$$f(𝑡𝒙+(1−𝑡)𝒚)≥𝑚𝑖𝑛\{f(𝒙),f(𝒚)\}$$

I am wondering how I would use this definition to show that a generalised function $$f$$ is quasiconcave, or if I should be approaching this from another concept/definition.

• What do you mean by "generalized" function? This is an ordinary function. – Michael Greinecker May 8 at 13:33
• Sorry, I omitted CES when typing it out. It should read generalised CES function $f$ – DoubleRainbowZ May 8 at 13:39

Note first that the function $$r\mapsto A(r)^{\mu/p}$$ is strictly increasing on $$\mathbb{R}_+$$. When you look at the minimum in the definition of quasi-concavity, you can therefore ignore this part and it suffices to show that the function $$(x_1,x_2,\ldots,x_n)\mapsto a_1x_1^p+a_2x_2^p+...+a_nx_n^p$$ is quasi-concave. Actually, the function is even concave. The sum of concave functions is concave, so it suffices to show that the function $$(x_1,x_2,\ldots,x_n)\mapsto a_ix_i^p$$ is concacave for each $$i=1,\ldots,n$$.
• Correct me if I'm wrong but your argument seems to be valid only for $p \in (0,1]$? For example if $p < 0$ then $A(r)^{\mu/p}$ is not increasing. Also if $p > 1$ then $x^p$ is convex? – tdm May 8 at 16:55
• Ah ok, I saw the $p \ne 0$ qualification but looked over the $p \in (0,1]$ condition. Sorry. – tdm May 11 at 6:05