The global minimum is necessarily the image of a zero of the derivative?

Question

Let $f:I\subset\mathbb{R}\to\mathbb{R}$ be a differentiable function.

If I is bounded, it is possible that the global minimum of $f$ is not the image of a zero of $f'$. For example, $\text{id}:[a,b]\to\mathbb{R}$. Its global minimum is $a$. And the derivative is nowhere $0$.

However, if $I$ is the real line it seems to me that this may be true. Just to be explicit, the "theorem" I am suggesting is:

Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function. Then, if $f$ has a global minimum $M$, there exists a real number $a$ such that $f'(a)=0$ and $f(a)=M$.

Is this really true? It seems to me that maybe one weird function could be a counterexample to this.

Answer 1

That's true, and it's a classical theorem in calculus (and it's the same for maxima). It's sometimes referred to as "Fermat's Theorem" (which is not very illustrative itself since there are so many other theorems in other areas called the same name).

The idea of a proof is that if $f$ has a minimum at $x_0$, then for any $x\neq x_0$ we have $f(x)\ge f(x_0)$ (and even '$>$' if it were a unique minima).

Now, for $x>x_0$ (since $x-x_0>0$) $$\frac{f(x)-f(x_0)}{x-x_0}\ge 0$$ (positive/non-negative over positive is positive/non-negative). You can also see easily that for $x<x_0$ $$\frac{f(x)-f(x_0)}{x-x_0}\le 0.$$

Now convince yourself that if the limit $$\lim_{x->x_0}\frac{f(x)-f(x_0)}{x-x_0}$$ happened to exist then it should be zero (think of lateral limits and both previous inequalities). Well that the limit exists is by definition to say that $f$ is differentiable at $x=x_0$ and that limit is precisely the derivative. So yes, under that hypothesis $f'(x_0)=0$.

Comments:

• The hypothesis that $f$ be differentiable at the point where it has a minimum (or maximum; maybe you can think what you would have to change in the argument in that case). Indeed, $f(x)=|x|$ is differentiable at every $x\in \mathbb R$ but at $x=0$, and there has a global minimum (of course, it could have had none).
• How does the fact that $I$ is not $[a,b]$ but $\mathbb R$ changes the situation here? You thought it might have to do with boundedness, but actually it just has to do with the fact that $\mathbb R$ is an open set. The same would happen for intervals like $(a,\infty)$, $(-\infty,b)$ and even $(a,b)$ (which is bounded). Remember that the key property of an open set is that all its points are interior, that is, you can move at least a little in any direction from the point and still be in the set. For example, if $I=(0,1)$, no matter how close you choose a point to $0$ or $1$ (but you can't pick those), you will always have some other numbers in that interval both at the left and at the right of the number that you chose. Think why this is important in the proof I've sketched above.
• Interestingly, the theorem is also valid for local extrema under the same hypothesis, since by definition a local minimum is a global minimum restricting the domain to some (sufficiently small) open interval around the point at the minimum or maximum.