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Let $f$ be twice differentiable and strictly convex on $[a,b]$. Assume also that at a point $x_0 \in (a,b)$ the derivative $f'(x_0) = 0$. Show that $x_0$ must be a strict local minimum.

I find that when $x<x_0, f'<0$, and when $x>x_0, f'>0$. So if I can conclude that for all $x$ in the interval, $f(x) > f(x_0)$ and therefore $x_0$ must be a strict local minimum?

Rajat
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ash
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  • I think it depends on your definition of convexity: If you use the definition $f$ convex $\Longleftrightarrow f''>0$, then the result is immediate. What is your definition of convexity? – frog Dec 01 '15 at 10:33
  • if f is convex, then f''> 0. – ash Dec 01 '15 at 10:36
  • I hope you are allowed to use that if $x_0$ is a minimum of a differentiable function, then $f'(x_0)=0$. Now if $x_0$ is not a strict local minimum, you find a sequence $x_n\to x_0$ (as $n\to\infty$) of minima of $f$. Now the mean value theorem asserts that $$ f'(x_n)-f(x_0)=f''(\xi)(x_n-x_0) $$ for some $\xi$ between $x_n$ and $x_0$. Can you obtain a contradiction from here? – frog Dec 01 '15 at 12:10
  • I missed a prime, $f(x_0)$ should be $f'(x_0)$ in the last formula, but I cannot edit it anymore, sorry. – frog Dec 02 '15 at 08:44

1 Answers1

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Given Info: $f'(x_0)=0$, here $S=[a,b]$.

Prove: $x_0$ is a strict local minima $\implies f(x_0) < f(x) \forall \ x \in \mathcal{B}_{\delta}(x_0)$ for some $\delta > 0$ (if $x \neq x_0$).

Prop: If $f$ is a strictly convex function defined over the convex set $S$, then $f(x) > f(y) + f'(y)(x-y) \text{ if $x \neq y$ } \forall x \in S$ and $y$ is an interior point.

Sol: From, the given data it is clear that $x_0$ is an interior point. Let say there is a $x \in \mathcal{B}_{\delta}(x_0)$, but $x \neq x_0$. So, from the properties of convex function, $$f(x) > f(x_0)+f'(x_0)(x-x_0) \implies f(x) > f(x_0)$$.

Rajat
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