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

Let $$f:\mathbb{R}\rightarrow\mathbb{R}$$ be twice differentiable function. Suppose $$a such that $$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 $$(a,b)$$?

• In general you cannot deduce something about the global minimum, once you have found a local minimum. However if $f$ is a convex function (en.wikipedia.org/wiki/Convex_function) then any local minimum is also a global minimum. Dec 4, 2021 at 18:28
• I see, thank you. Dec 4, 2021 at 18:30

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 $$[a,b]$$.

• Ok thank you. Does anything change if we, e.g., also assume that $f(a)=f(b)$? Dec 4, 2021 at 18:29
• No, that won't help: draw a curve with two local minima of different heights. The taller one is a local min that is not global (because of the lower one). Dec 4, 2021 at 18:31
• Right, thanks again. Dec 4, 2021 at 18:36

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.