Let φ: R → R.
Show that φ is convex if and only if: for any y ∈ R, there exists $a$ constant a and there exists a function L: R → R defined by:
$L(x) = a(x−y) +φ(y),x ∈ R$,
such that L(y) = φ(y) and such that L(x) ≤ φ(x) for all x ∈ R. (In the case that φ is differentiable, the latter condition says that φ lies above all of its tangent lines.)
I have a hint provided by the question Suppose φ is convex. If x is fixed and y varies, show that $\frac{φ(y)−φ(x)}{y−x}$ increases as y increases. Draw a picture. What slope a should L have at x?
I have no idea how to use the hint and it's also hard for me to prove backward.
Thank you very much!
Fix the point $y$.
If $z < y < x$ the hint leads to $$\frac{\phi(z) - \phi(y)}{z-y} \le \frac{\phi(x) - \phi(y)}{x-y}.$$ Thus there is a number $a$ with the property that $$\sup_{z < y}\frac{\phi(z) - \phi(y)}{z-y} \le a \le \inf_{x > y}\frac{\phi(x) - \phi(y)}{x-y}.$$
Work with this inequality to show that $a(x-y) + \phi(y)$ has the desired property.