Consider a vector x ∈ R + + N . Also consider two functions, g ( x ) : R N → R , and a ( x ) : R N → R , representing the geometric and arithmetic means respectively.We also know that 0 ≤ g ( x ) a ( x ) ≤ 1 is always true.Is the geometric-to-arithmetic function convex or concave?

Question

Consider a vector       x    ∈            R              +      +        N  . Also consider two functions,   g  (      x    )  :            R        N    →      R  , and   a  (      x    )  :            R        N    →      R  , representing the geometric and arithmetic means respectively.We also know that   0  ≤            g      (              x            )              a      (              x            )        ≤  1 is always true.Is the geometric-to-arithmetic function convex or concave?

blogpolisft · Accepted Answer

g  (  x  )      /    a  (  x  )  ≥  α    ⟺    g  (  x  )  ≥  α  a  (  x  )  .(This depends upon your claim that   x  ∈            R              +      +        n  , so   a  (  x  )  &amp;gt;  0). You can even maximize the ratio by solving a sequence of convex optimization problems.The arithmetic mean, incidentally, is both convex and concave; that is, it is affine (indeed, linear). So unlike the more general answer I gave on the other page, the constraint conversion works even if   α  &amp;lt;  0. Of course, if   α  &amp;lt;  0, it is trivially satisfied.

Consider a vector x in R_++^N. Also consider two functions, g(x):R^N->R, and a(x):R^N->R, representing the geometric and arithmetic means respectively. We also know that 0<=g(x)a(x)<=1 is always true. Is the geometric-to-arithmetic function convex or concave?

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