Convex functions

Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be a convex function. Use the definition of convexity to show that \(f\) is "turning upwards" in the sense that if \(x_{1},x_{2},x_{3}\) are three scalars such that \(x_{1}<x_{2}<x_{3}\), then

\(\hspace{5cm}\frac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}}\leq \frac{f\left( x_{3}\right) -f\left( x_{2}\right) }{x_{3}-x_{2}}.\)

Let \(x_{1},x_{2},x_{3}\) be three scalars such that \(x_{1}<x_{2}<x_{3}\). Then we can write \(x_{2}\) as a convex combination of \(x_{1}\) and \(x_{3}\) as follows

\(\hspace{5cm}x_{2}=\frac{x_{3}-x_{2}}{x_{3}-x_{1}}x_{1}+\frac{x_{2}-x_{1}}{x_{3}-x_{1}}x_{3}\),

so that by convexity of \(f\), we obtain

\(\hspace{5cm}f\left( x_{2}\right) \leq \frac{x_{3}-x_{2}}{x_{3}-x_{1}}f\left(x_{1}\right) +\frac{x_{2}-x_{1}}{x_{3}-x_{1}}f\left( x_{3}\right) .\)

This relation and the fact

\(\hspace{5cm}f\left( x_{2}\right) =\frac{x_{3}-x_{2}}{x_{3}-x_{1}}f\left( x_{2}\right) +\frac{x_{2}-x_{1}}{x_{3}-x_{1}}f\left( x_{2}\right) \)

imply that

\(\hspace{5cm}\frac{x_{3}-x_{2}}{x_{3}-x_{1}}\left( f\left( x_{2}\right) -f\left(x_{1}\right) \right) \leq \frac{x_{2}-x_{1}}{x_{3}-x_{1}}\left( f\left(x_{3}\right) -f\left( x_{2}\right) \right)\)

By multiplying the preceding relation with \(x_{3}-x_{1}\) and dividing it with \(\left( x_{3}-x_{2}\right) \left( x_{2}-x_{1}\right) ,\) we obtain

\(\hspace{5cm}\frac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}}\leq \frac{f\left( x_{3}\right) -f\left( x_{2}\right) }{x_{3}-x_{2}}.\)