I want to show that if $y = f(x) > 0$ is a concave function on $\mathbb{R}$, then $z = \frac{1}{f(x)}$ is a convex function.

Since $f(x) > 0$ then if we applied the second derivative test, wouldnt $z = \frac{1}{f(x)}$ always come out to be greater then or equal to zero which implies convexity? That is as much as I can come up with when it comes to showing why its true.

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    $\begingroup$ Well, did you actually compute the second derivative of $1/f(x)$ and see what you could get from it? (On a side note, the only functions positive and concave on all of $\mathbb{R}$ are constants.) $\endgroup$ – Harald Hanche-Olsen Jan 16 '14 at 21:30
  • $\begingroup$ It's not generally true that the reciprocal of a convex function is concave or vice versa. Think e.g. of $f(x)=x^2+1$, which is convex but whose reciprocal is $1/(x^2+1)$, which is clearly not concave. $\endgroup$ – JPi Jan 17 '14 at 0:42

The only concave functions on $\mathbb{R}$ satisfying $f(x) >0$ everywhere are the constant functions, hence $x \mapsto {1 \over f(x)}$ is trivially convex.

Suppose $a<b<c$, then using concavity we have $f(b) \ge f(a)+{b-a \over c-a} (f(c)-f(a))$, and rearranging gives $f(c) \le f(a) + { c -a \over b-a} ((f(b))-f(a))$. Hence if $f(b) <f(a)$, we see that $\lim_{c \to \infty} f(c) = -\infty$. Similarly, if $f(b)>f(a)$, a similar calculation shows that $\lim_{c \to -\infty} f(c) = -\infty$. Hence $f(a)=f(b)$ for all $a,b$.

(Note: The conclusion from the above is that any concave function that is bounded below must be constant.)

  • $\begingroup$ That makes sense. Thanks! $\endgroup$ – user121895 Jan 16 '14 at 23:53
  • $\begingroup$ Just to refresh my memory real quick, how did you come up with the fact that $f(b) \geq f(a) + \frac{b-a}{c-a}(f(c) - f(a))$? $\endgroup$ – user121895 Jan 17 '14 at 0:27
  • $\begingroup$ @user121895: From the definition of a concave function, $f(\lambda x + (1-\lambda)y) \ge \lambda f(x) + (1-\lambda) f(y)$ for $\lambda \in [0,1]$. $\endgroup$ – copper.hat Jan 17 '14 at 0:38
  • $\begingroup$ Ah yes. I learned it with different symbols but at the end of the day, it is all the same thing :) $\endgroup$ – user121895 Jan 17 '14 at 0:42
  • $\begingroup$ I am sorry to keep asking questions. Maybe I am not good at translating from the definition but where did you get $\frac{b-a}{c-a}$? I understand the rest of it though. $\endgroup$ – user121895 Jan 17 '14 at 0:52

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