Let $\phi:\mathbb [0,\infty) \to [0,\infty)$ be convex, and strictly convex on $[a,\infty)$, for some $a \in (0,\infty)$.

Let $x>a\ge y$, and suppose that $$\phi(t x + (1-t) y) = t\phi(x) + (1-t)\phi(y).$$

How to prove that $t\in \{0,1\}$?

I am quite sure this should not be complicated, but I am not sure how to approach this.


I think that this should follow from the following claim: If a convex function agrees with a chord at some interior point, then it must coincide with this chord all the way-i.e. be affine on the relevant segment.

  • 1
    $\begingroup$ @KaviRamaMurthy $\phi$ is strictly convex only on $[a,\infty).$ $\endgroup$
    – zhw.
    May 19, 2020 at 7:42

1 Answer 1


Hint: If $f$ is convex on $[0,\infty)$ and $0\le x < y <z,$ then

$$\frac{f(y)-f(x)}{y-x} \le \frac{f(z)-f(y)}{z-y}.$$

When does equality hold?


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