Let $f: [a,b] \to \mathbb R$ be a differentiable, convex function with $f(a) \gt 0$ and $f(b) \lt 0$. Then $f$ has a unique zero in $[a,b]$.

The existence of the zero follows immediately from the intermediate value theorem. But how am I going to prove that this zero is unique?

  • 2
    $\begingroup$ Remarking that $f'$ is increasing over $[a,b]$, we get that this post and that post are related. $\endgroup$ – Harry49 Jan 22 '18 at 15:30

Assume that $a<x_1<x_2<b$ and $f(x_1)=f(x_2)=0$. Then
$$x_2=tx_1+(1-t)b\quad \text{for $t=\frac{b-x_2}{b-x_1}\in (0,1)$}$$ and, by the convexity of the function $f$, $$0=f(x_2)=f(tx_1+(1-t)b)\leq tf(x_1)+(1-t)f(b)=(1-t)f(b)<0$$ which is a contradiction. There is no need of the derivative of $f$.

  • $\begingroup$ Silly question but $x_2 = tx_1 + (1-t)b$ for some $t\in (0, 1)$ relies on the convexity of $[a, b]$ and not $f$, correct? $\endgroup$ – Oria Gruber Jan 22 '18 at 15:38
  • $\begingroup$ I feel as if that should be proven as well, as that wasn't "given" in the question. Otherwise good answer! $\endgroup$ – Oria Gruber Jan 22 '18 at 15:41
  • $\begingroup$ @OriaGruber Actually $t=(b-x_2)/(b-x_1)$. $\endgroup$ – Robert Z Jan 22 '18 at 15:43

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