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I need to prove that every homogeneous function on the domain of all positive real numbers is either concave, or convex.

UPD

Let f(x) — homogeneous of degree k, then f(ax)=a^kf(x). Then, for x1,x2>0 and a within the interval [0,1], f(ax1)=a^k*f(x1) and f((1-a)x2)=(1-a)^kf(x2).

For a convex function it should hold that f(a*x1+(1-a)x2) is less or equal than af(x1)+(1-a)*f(x2). (For a concave function the logic should be similar, so I don’t consider it).

Then I’m stuck with that fact that:

• f(a*x1+(1-a)x2) can’t be separated into a a sum f(ax1)+f((1-a)*x2), so I have to stick to using inequalities for the proof

• there is no specification about k, so f(ax1) can be less, bigger or equal to af(x1), depending on the value (or different values of k just correspond to different convexity/concavity?)

• I don’t get where the condition of an all-positive domain should be used

Ksenia
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  • Okay. Where are you stuck? – Jacky Chong Feb 06 '23 at 17:12
  • Let me also give you a hint: it suffices to consider $f(x) = x^k$. – Jacky Chong Feb 06 '23 at 17:14
  • I don’t think that it is suitable to use some function in this kind of proof.

    If the function of homogeneous of degree k, it should hold that f(ax)=a^kf(x). If it is convex (concave), f(ax+(1-a)y) is less or equal (more or equal) that af(x)+(1-a)*f(y). a belongs to [0,1] and x and y should be positive. I can’t get, where I should use the fact that the function is on the domain of all positive real numbers.

     – Ksenia Feb 06 '23 at 18:06
  • Well, I kinda got the idea of the proof while I was typing all of this, but I still can’t guess where the property of an all-positive domain is used. – Ksenia Feb 06 '23 at 18:11

1 Answers1

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Hint: By definition, a function $f(x)$ is homogeneous of order $k$ provided \begin{align} f(\lambda x) = \lambda^k f(x) \end{align} for all $\lambda>0$.

By definition a function is convex (concave) provided \begin{align} f(tx+(1-t)y) \le t f(x)+(1-t) f(y) \quad (\ge \text{ if concave}) \end{align} for all $t \in [0, 1]$.

Case $1$, $k=1$: Here we see that \begin{align} f(t x+(1-t) y) =&\, f(1) (t x+(1-t) y) \\ =&\, tx f(1)+(1-t)y f(1)\\ =&\, t f(x)+(1-t) f(y). \end{align} Do you notice where I have used $x, y>0$?

Case $2$, $k>1$: Think about $f(x) = x^2$ and how $(tx+(1-t)y)^2\le tx^2+(1-t)y^2$. Then look at case 1 for inspiration.

Case $3$, $0<k<1$: Consider $f(x) = \sqrt{x}$ and repeat the procedure in the previous case.

Case $4$, $k<0$: Same type of argument as above.

Jacky Chong
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