Show that $x \cdot f(x)$ is convex

Question

Given a function: $$ f: (0,\infty)\to\mathbb{R}$$ I need to show that $$xf(x)$$ is convex if and only if $$f\left(\frac{1}{x}\right)$$ is convex

I'm not sure how to approach this type of problems, convexity isn't my thing.

Any help and hints will be appreciated

Answer 1

I will assume $f$ is twice differentiable since it you're only given convexity if $f''(x)\ge 0$, but as others have mentioned, differentiability is not needed in general.
I'll show one direction, and hopefully you can follow the same sort of idea for the other.

Since $f(1/x)$ is convex, $[f(1/x)]''\ge0$, i.e. $$ [f(1/x)]'' = \frac{2}{x^3}f'(1/x) + \frac{1}{x^4}f''(1/x) \ge 0 $$ (make sure you check this computation). this tells us that $\frac 1 x f''(1/x) \ge - 2f'(1/x)$. If $x>0$, so is $1/x$, so we may replace $x$ with $1/x$ in this computation and see that $$ xf''(x) \ge -2f'(x). $$

Next, we can compute the second derivative of $xf(x)$ which is $2f'(x) + xf''(x)$, and by the above, we see $$ 2f'(x)+xf''(x) \ge 2f'(x)-2f'(x) = 0, $$ which proves convexity of $xf(x)$. The other direction is nearly identical, and you should be able to carry it out on your own if you understood this one.

Answer 2

As a start, you might want to consider the perspective function $f: \mathbb{R}^n \to \mathbb{R}^{n-1}$ where $f(\mathbf{x},t) = \mathbf{x}/t$ for $\mathbf{x} \in \mathbb{R}^n, t>0$. Here's a nice reference.

Answer 3

There is a simple answer to this if you the old fashioned definition of convexity : f is convex if $f(x)=\sup \{ax+b:(a,b) \in E\}$ for some set of pairs E. [A convex function is one whose graph is an upper envelope of straight lines]. Today's definition of convexity is equivalent to this. If you assume this you simply have to replace $x$ by $1/x$ and multiply by $x$ to proves the result. (Both parts follows by the same argument!).

Answer 4

A proof using the basic definition of convex functions. If $0 < x < y$ and $0 < t < 1$ then $$ s = \frac{ty}{(1-t)x + t y} $$ satisfies $0 < s < 1$, $$ 1-s = \frac{(1-t)x}{(1-t)x + t y} $$ and $$ \frac{1-s}{x} + \frac{s}{y} = \frac{1}{(1-t)x + t y} \, . $$

If $x f(x)$ is convex and $x, y, t, s$ are as above then $$ f\left( \frac{1}{(1-t)x + t y}\right) = ((1-t)x + t y) \left( \frac{1-s}{x} + \frac{s}{y} \right)f \left( \frac{1-s}{x} + \frac{s}{y}\right) \\ \le ((1-t)x + t y) \left( \frac{1-s}{x} f\left( \frac 1x \right) + \frac{s}{y} f\left( \frac 1y \right)\right) \\ = (1-t) f\left( \frac 1x \right) + t f\left( \frac 1y \right) \, , $$ so that $f(1/x)$ is convex.

Conversely, if $f(1/x)$ is convex and $x, y, t, s$ are as above then $$ \bigl((1-t)x + ty\bigr) f\bigl((1-t)x + ty\bigr) = \bigl((1-t)x + ty\bigr) f\left( \frac{1}{(1-s)/x + s/y}\right) \\ \le \bigl((1-t)x + ty\bigr) \bigl( (1-s) f\left( \frac{1}{1/x}\right) + s f\left( \frac{1}{1/y}\right)\bigr) \\ = (1-t) x f(x) + t y f(y) $$ so that $x f(x)$ is convex.