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I was wondering if you know a theorem that states that the function $f(x)/x$ is convex in $x$ if $f(x)$ is concave or convex in $x$.

$f(x)$ is convex and increasing in $x$.

when $\lambda>0, \mu>0,k>=0$,

k is an integer

OR

I know this A function should be convex and decreasing but I can't prove it.

$A=\lambda(1-B)/\mu $

$B=((\lambda /\mu) \int_0^\infty e^{(-\lambda /\mu)z}(1+z)^k dz)^{-1}$

$dB/d\mu=B(1-B-x/(\lambda /\mu))<0$

and B is convex.

I can provide the second derivative of B if necessary as well. For beginners if I could show the decreasing part, it would be great!

user130023
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  • An intuition for this is cost functions. A realistic cost function should be increasing as the number of items manufactured increases. If the cost function is concave, this means that the manufacturing process is getting more efficient (since the marginal cost is decreasing). In this context, the question becomes: If your manufacturing process is getting more efficient, what (if anything) does this say about the average cost to produce $x$ items. – Baby Dragon Feb 19 '14 at 18:56
  • welcome to MATHEMATICS. Accept answers if they are helpful please. – Hoseyn Heydari Feb 19 '14 at 19:01

2 Answers2

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You can also have examples of strictly convex functions $f$ such that $\frac{f(x)}{x}$ is not convex. A very easy example is $\frac{e^x}{x}$, which is not convex on $(-\infty, 0)$ (indeed, there it is concave, as you may prove by computing its second derivative).

Giuseppe Negro
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The function $f(x)\equiv 1$ is both convex and concave, whereas $g(x)=\frac{f(x)}{x}=\frac 1x$ has a convex and a concave branch, hence is, globally, neither.

Simon Markett
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