convexity of inverse function

I have a question on the reverse of a convex function.

Let $f(x)$ be a convex function.

Is the reverse function, say $g(x)=f(x)^{-1}$, is necessarily a concave function ? Considering that such function $g$ does exist ($f$ is invertible):

-Is necessarily a concave function?

-What about the opposite $-f(x)$, is concave?

Any hints?

Solution 1:

I'm not sure whether you mean the inverse of the function, i.e. $f^{-1}$ such that $f(f^{-1}(x))=f^{-1}(f(x))=x$ or of the reciprocal, i.e. $f(x)^{-1}=1/f(x)$.

The reciprocal of a convex function need not be concave, for example look at $f(x)=e^x$. $f$ is convex. Its reciprocal $f(x)^{-1}=1/f(x)=e^{-x}$ is also convex.

On the other hand, if $f$ is convex, then $-f$ is concave and vice versa. This is because you just multiply the defining inequality of convexity by $-1$ and it turns the inequality sign around.

For the inverse function there are two cases: Since $f$ is continuous and bijective, it must be either strictly increasing or strictly decreasing. If it is strictly increasing, then its inverse is concave. If it is strictly decreasing, its inverse is convex. An example is again $f(x)=e^{-x}$. This is convex and its inverse function is $f^{-1}(x)=-\log x$, also convex.

Solution 2:

I just stumbled upon this question, wondering the same thing as the OP. YourAdHere's answer was what I was looking for, but I also needed a formal proof, so I thought I might try and add that to complement his/her answer.

Consider a continuous, convex, bijective function $f: A \rightarrow B$ and its inverse $f^{-1}: B \rightarrow A$. By definition of convexity we have for $\lambda \in [0, 1]$ and $x, y \in A$: \begin{align} f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y). \end{align} For $u, v \in B$ define \begin{align}%\label{} a :=& f^{-1}(\lambda u + (1-\lambda) v) \\ b :=& \lambda f^{-1}(u) + (1-\lambda)f^{-1}(v) \,. \end{align} If $a \leq b$ then $f^{-1}$ is convex, else it is concave. We have \begin{align}%\label{} f(a) =& f(f^{-1}(\lambda u + (1-\lambda) v) ) \\ =& \lambda u + (1-\lambda) v = \lambda f(f^{-1}(u)) + (1-\lambda)f(f^{-1}(v)) \\ \geq& f(\lambda f^{-1}(u) + (1-\lambda) f^{-1}(v)) = f(b) \,, \end{align} where the inequality is due to the convexity of $f$. So if $f' \geq 0$ we have $f(a) \geq f(b) \Leftrightarrow a \geq b$ and hence $f^{-1}$ is concave. If conversely $f' \leq 0$, then $f(a) \geq f(b) \Leftrightarrow a \leq b$ and $f^{-1}$ is convex.

Solution 3:

Because you didn't specify the domain and range, assume f is a real function. Also, so the inverse is defined, assume f is one-to-one. Thus f is either increasing or decreasing. Now the inverse of f reflects f about the line y=x. Can you visualize what this reflection does to convex or concave functions?