Attempt at Proving A Lemma (critical point, 2nd derivative, global maximum).

Asked Mar 11 '13 at 03:53

Active Mar 11 '13 at 05:12

Viewed 105 times

could you please check my attempt at proving the following lemma.

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ has two continuous derivatives, has only one critical point $x_{0}$ and $f''(x_{0})< 0$. Then $f$ achieves its global maximum at $x_{0}$, that is, $f(x)\leq f(x_{0})$ $\forall x \in R$.

Attempt:
By Cauchy's Mean Value Theorem, there is a point $c \in (x,x_{0})$ such that $$f''(c)=\frac{f'(x)-f'(x_{0})}{x-x_{0}}<0,$$ hence $$f'(x)<f'(x_{0}).\tag1$$

Integrating both sides of $(1)$ with respect to $x$, we have

$$\int f'(x) dx<\int f'(x_{0})dx,$$ Therefore $f(x)<f(x_{0})$ for all $x \in \mathbb{R}.$
Q.E.D.

Thank you for your time :)

edited Mar 11 '13 at 05:12

Gerry Myerson

179,216

asked Mar 11 '13 at 03:53

user4167

1,453

2

$f'(x_0)$ is a constant, so integrating does not give $f(x_0)$ – L. F. Mar 11 '13 at 04:01
1

Also, if $x<x_0$ then $$f''(c)=\frac{f'(x)-f'(x_{0})}{x-x_{0}}<0$$ doesn't imply that $f'(x)<f'(x_0)$ - indeed, when $x<x_0$ the opposite is true. – Thomas Andrews Mar 11 '13 at 04:05
Thank you for pointing that out. Is there any other way that I can approach this question? I'll give it a try. – user4167 Mar 11 '13 at 04:06
1

Also, you only know that $f''(x_0)<0$ - you don't have as a condition that $f''(c)<0$ for all $c\in\mathbb R$ – Thomas Andrews Mar 11 '13 at 04:07
1

@euler Have a look here for inspiration. – L. F. Mar 11 '13 at 04:13
Thank you L.F. I'll look into it. – user4167 Mar 11 '13 at 04:17

Attempt at Proving A Lemma (critical point, 2nd derivative, global maximum).

0 Answers0