1
$\begingroup$

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?

$\endgroup$
  • 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
2
$\begingroup$

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 $f$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.