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
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 $f(x_1) &gt; f(x_0)$ 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 $f(x_\star) &lt; f(x_0)$ 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.
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 $f(x_0) &gt; f(x_1)$. WLOG, we may assume that $x_0 &lt; x_1$. Since $x_0$ is a local minimum, there exists $\varepsilon&gt;0$ 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 $f(x_0) &gt; f(x_1)$, $x_0$ 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.