Local minimum global Let $f:(a,b)\to\Bbb R$ be continuous. Assume that $f$ has a local minimum at some point $x_0$. Further assume that this is the only point where $f$ has a local extremum. Does it follow that $f$ has a global minimum at $x_0$.
Thanks
 A: The definition of a local extremum uses "soft" ineqaulities. For example for all $x$ in a neigborhood of your $x_0$ we have $f(x) \ge f(x_0)$.
Then the answer to your question would be yes, I think.
We know there is a neighborhood around $x_0$ such that $f(x) \ge f(x_0)$ for $x$ there. Choose a point from in there, $x_1 \ne x_0$, then actually $f(x_1) > f(x_0)$ strictly, because otherwise $f$ would have a local minimum (of the same size) at $x_1$ too.
Now, we want to prove that this minimum at $x_0$ is global. Suppose it was not. Then some $x_\star$ exists with $f(x_\star) < f(x_0)$ strictly. Without loss of generality, $x_\star$ lies on the same side of $x_0$ as does $x_1$. Then look at $f$ restricted to the compact interval between $x_\star$ and $x_0$. The image of this compact interval contains its supremum. In other words $f$ restricted to the compact interval has a maximum. By construction this maximum is attained in the interior of the interval. So there is a local maximum, which contradicts our hypothesis.
A: In one variable, it is true that a differentiable function with only one local extremum must also have a global extremum at that point. It is not true in several variables, however. Intuitively, the reason for this is that a function of one variable can't go from being increasing to being decreasing (or vice versa) without passing through a local extremum; but in several variables, a function can start to decrease in one direction without necessarily decreasing in all other directions. Wikipedia's article on maxima and minima provides this example:
$$f(x,y)=x^2+y^2(1-x)^3$$
The only critical point is a local minimum at the origin, with $f(0,0)=0$. However, since $f(4,1)=-11$, this is clearly not a global minimum.
A: Let's try this again more carefully. (I'll eventually get it right.)
Assume, by way of contradiction, that $x_0$ is not a global minimum. So there exists $x_1 \in (a,b)$ such that $f(x_0) > f(x_1)$. WLOG, we may assume that $x_0 < x_1$. Since $x_0$ is a local minimum, there exists $\varepsilon>0$ such that for all $x\in(x_0-\varepsilon,x_0+\varepsilon)$ we have $f(x) \geq f(x_0)$. Then we can take the interval $[x_0,x_1]$, which is compact, and consider the image under $f$. Since $f$ in continuous, there must be a local maximum and a local minimum in the interval. Since $f(x_0) > f(x_1)$, $x_0$ is not the local minimum. Since there is some $x_2 \in (x_0, x+\varepsilon) \subset [x_0,x_1]$ with $f(x_2) \geq f(x_0)$, $x_0$ is not the local maximum. Hence one of the local extrema occurs in the interval $(x_0,x_1)$. But this gives a contradiction since it implies there is a second local extremum. Therefore $x_0$ is a global minimum.
A: Yes. If it wasn't the global minimum, there would have to be some $x_1\in(a,b)$ s.t. $f(x_1)<f(x_0)$. Now, $[x_1,x_0]$ (or $[x_0,x_1]$) is compact and a continuous function must take a minimum and a maximum on this interval. As the maximum can neither be $x_0$ nor $x_1$, it must be in the interior of the interval and thus would also be a local maximum for f on $(a,b)$ which contradicts your assumption that $f$ didn't have any other local extremal points.
A: If it was not a global minimum, then there exists a $x_1$ such that $f(x_1)<f(x_0)$, so it would be a local minimum against the hypothesis that $x_0$ is the only one.
