Convex function has unique zero

Question

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?

Remarking that $f'$ is increasing over $[a,b]$, we get that this post and that post are related. — EditPiAf, Jan 22 '18 at 15:30

Robert Z · Accepted Answer · 2018-01-22T16:07:36.380

4

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 $f$.

edited Jan 22 '18 at 16:07

answered Jan 22 '18 at 15:37

Robert Z

145,942

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? – Oria Gruber Jan 22 '18 at 15:38
I feel as if that should be proven as well, as that wasn't "given" in the question. Otherwise good answer! – Oria Gruber Jan 22 '18 at 15:41
@OriaGruber Actually $t=(b-x_2)/(b-x_1)$. – Robert Z Jan 22 '18 at 15:43

score 1 · Answer 2 · answered Mar 15 '23 at 05:57

1

Let $c \in (a,b)$ a zero. Since $f$ is convex, the graph of $f$ on $(c,b)$ is below the chord determined by $c$, $b$ , so below the $x$ axis. Conclusion: there are no zeroes to the right of any zero in $[a,b]$ ; thus, the zero is unique.

answered Mar 15 '23 at 05:57

orangeskid

53,909

Convex function has unique zero

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