When is a local minimum a global minimum over a closed interval

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be twice differentiable function. Suppose $a<c<b$ such that $f'(c)=0$ and $f''(c)>0$. Then $c$ is a local minimum of $f$. Does it follow that over $[a,b]$, $c$ is also global minimum? It seems no.

By EVT, there exists $y\in[a,b]$ such that $f(y)\leq f(x)$ for all $x\in[a,b]$. Then $f(y)\leq f(c)$. We also know that there exists $\delta>0$ such that for all $x\in(c-\delta,c+\delta)$, $f(c)\leq f(x)$. Since it's not necessary for $y$ to be in the interval $(c-\delta,c+\delta)$, we may not have $f(c)\leq f(y)$; i.e. $f(c)$ is not necessarily a global min over $[a,b]$.

Is there anything we can say about the global minimum of $f$ over $[a,b]$, once we find a local minimum in $(a,b)$?

Answer 1

You generally cannot expect local information to translate to global information. Consider $f(x)=x^3-x^2$ with $a = -100$, $b=1$ and $c=2/3$.

The issue here is that the curve was allowed to turn around again to create lower points. If you assume that your function is differentiable* and the critical point is unique, then yes: the local min is a global min.

*You can either take this assumption to mean differentiable on an open interval containing $[a,b]$, or you can assume differentiability on $(a,b)$ and continuity on $[a,b]$.

Answer 2

It's indeed possible to say something about a global minimum in your setting: you can at least give a lower bound:

Scince $f'$ is continuous on the Interval $[a,b]$, it is bounded, say $| f'(x)| \le M$ for some $M > 0$ and all $x \in [a,b]$. The main Theorem of Calculus states:

$$ f(x) = f(c) + \int_c^x f'(x) dx$$

were $c$ is the local minimum. Then:

$$ f(x) \ge f(c) - M |x - c|,$$

giving a lower bound for the global minimum over the Interval. Note that this bound is sharpest if $c$ is not a local minimum but a global maximum.