# Minima of a continuous function

A function $f:\mathbb{R}\mapsto \mathbb{R}$ is continuous and smooth in $[a,b]$. How can I prove that the minimum of the function will be either at the end points or where the derivative goes to zero.

The extreme value theorem states that there is at least one minimum and one maximum. However, I was looking for a proof that the global minimum is either at the end points or the derivative.

EDIT

As pointed out in the comments, it would be more appropriate to say if $f(c)$ is a minimum, then how do we prove that $c$ is either the end points or $f'(c)=0$ ?

• One problem is in the use of "the minimum"; it's true that there's a smallest element in the image of $f$ on $[a, b]$; but this smallest value might be attained at many points: consider the function given by $f(x) = 3$, which attains its minimum at every point. The right statement is "if $f(c)$ is a minimum on the interval, then either $f'(c) = 0$ or $c$ is one of the endpoints." – John Hughes Dec 22 '16 at 17:53

## 3 Answers

Proof by contradiction to show that the global minimum is either at the end points or at an interior point where the derivative is zero.

Suppose it is not at the end points (say, at $x_0 \in(a,b)$) and the derivative is non-zero. Then for all $\epsilon > 0$, there exist $\delta > 0$ such that $$\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\epsilon$$ for $|x-x_0|<\delta$. In other words $$f'(x)-\epsilon <\frac{f(x)-f(x_0)}{x-x_0}<f'(x_0)+\epsilon$$ If $f'(x)<0$, we can choose small enough $\epsilon$ such that $$\frac{f(x)-f(x_0)}{x-x_0}<0$$ or in other words for $x_0+\delta>x>x_0$ $$f(x)<f(x_0)$$ Similarly, for $f'(x)>0$, we can choose small enough $\epsilon$ such that $$\frac{f(x)-f(x_0)}{x-x_0}>0$$ or in other words for $x_0-\delta<x<x_0$ $$f(x)<f(x_0)$$ Contradiction.

Suppose that $f$ attains its minimum at $x=c\in(a,b)$. then

$$\forall x\in(a,c) \;\;\frac{f(x)-f(c)}{x-c}\leq 0$$

$$\implies \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}=f'(c)\leq 0.$$

You can finish for $x\in (c,b)$.

• How do we incorporate the end points? I understand it intuitively, but not able to formalize it in mathematical terms. – Saurav Dec 22 '16 at 17:59
• @Saurav There are two cases. end point or not. – hamam_Abdallah Dec 22 '16 at 18:01

A minimum occurs at a non-endpoint $c$ if and only if for every $x\in[a,b]$ you have $f(c)\le f(x).$ That is equivalent to $f(x)-f(c)\ge0.$

If $x>c$, then $x-c>0$ so $\dfrac{f(x) - f(c)}{x-c} \ge0.$ Therefore $\displaystyle\lim_{x\,\downarrow\,0} \frac{f(x)-f(c)}{x-c}\ge0.$

If $x<c$, then $x-c<0$ so $\dfrac{f(x) - f(c)}{x-c} \le 0.$ Therefore $\displaystyle\lim_{x\,\uparrow\,0} \frac{f(x)-f(c)}{x-c}\le0.$

Therefore $\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c} \ge0$ and $\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c} \le0.$

Therefore $\displaystyle0\le\lim_{x\,\to\,0} \frac{f(x)-f(c)}{x-c}=0.$