# 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 answer seems to be yes. Can you show some work you've done on the problem so far, or your thoughts about it? – Jeff Snider Jan 28 '14 at 18:21

The definition of a local extremum uses "soft" ineqaulities. For example for all $x$ in a neigborhood of your $x_0$ we have $f(x) \ge f(x_0)$.

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.

• Global minimums also can be allowed to be "soft" (i.e. $f(x_\star)\le f(x_0)$) – Squirtle Jan 28 '14 at 18:15
• @Squirtle Not sure I understand. If $f(x_\star)=f(x_0)$, then $x_\star$ is no longer a "witness" that the original $x_0$ wasn't global, because as I said in the beginning, extrema are defined with non-strict inequality signs. – Jeppe Stig Nielsen Jan 28 '14 at 18:21
• @Squirtle As far as I can see, this proof works no matter if you define extrema using strict or non-strict inequalities? Did you downvote? – Jeppe Stig Nielsen Jan 28 '14 at 18:42
• I did downvote, but I have removed it because I see honest effort here and I downvoted based on the hard vs soft and wasn't sure what "witness" was referring to. I like it, but only have one question..... I don't like that you are saying, "$f$ restricted to some compact set $[x_\star,x_0]$ obtains a maximum, hence contradiction." That isn't a contradiction at all, in fact when we say "$f$ obtains only one local extremum", it is implicit that we mean "on all of $(a,b)$". For example, $f$ restricted to a single point obtains an extremum, but that's obviously no contradiction. Agreed? – Squirtle Jan 28 '14 at 18:50
• @Squirtle Yes. The global maximum of the restriction of $f$ to $[x_\star,x_0]$ will, because it falls in the interior of this interval, also be a local maximum for the non-restricted $f$. That is one additional local extremum of our original $f$, and the desired contradiction is reached. – Jeppe Stig Nielsen Jan 28 '14 at 19:12

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.

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.

• You have a subtle but important fallacy... namely, you are assuming a minimum $x_0$ cannot be a maximum. – Squirtle Jan 28 '14 at 18:14
• @Squirtle No, that can be dealt with, can't it? See my answer (which is really the same as this one, only mine is more clumpsy). – Jeppe Stig Nielsen Jan 28 '14 at 18:16
• If it were, the function would haver to be locally constant and $x_0$ would not be the only local extremum any more. – ChrisBln Jan 28 '14 at 18:16
• If you are going to answer something... don't have any holes in it, rather than racing to be the first to give an answer with holes in it... consider all possible mistakes. For example, include "local extremum" argument. – Squirtle Jan 28 '14 at 18:18

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.

• No. $f(x_1)<f(x_0)$ does not imply that $f$ has a local minimum at $x_1$. Indeed, $f$ needn't have any other local minima. – ChrisBln Jan 28 '14 at 18:27
• If $f(x_1) \to \infty$ we don't consider this a minimum because we can't identity a single point that obtains this minimum especially because $(a,b)$ is an open set. – Squirtle Jan 28 '14 at 18:28
• @ChrisBln, I believe this is incorrect: You have to consider continuitiy – Bento Jan 28 '14 at 18:35
• Even a continuous function on an open interval could tend to infinity at the borders. – ChrisBln Jan 28 '14 at 18:37
• To give an explicit example of what @ChrisBln mentioned in his first comment, with $f(x)=x^2-x^4$ and $(a,b)=(-2,2)$ (see graph), the function has only one local minimum, at zero. There are local maxima, however. – Jeppe Stig Nielsen Jan 28 '14 at 18:50

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.

• I don't understand. Your function has the global minimum at 2. – twinky Jan 28 '14 at 17:58
• I agree with @twinky see: Plot in WA – Thomas Produit Jan 28 '14 at 18:00
• My mistake. I got myself confused by looking for a function that had no global min on $\mathbb{R}$ and then restricting to meet interval conditions and forgot that changed the function. – John Habert Jan 28 '14 at 18:01
• Your answer.... isn't an answer at all, " EDIT: Previous counterexample was badly formed. Reconsidering problem now." – Squirtle Jan 28 '14 at 18:06
• I know. I was removing the poorly formed example and attempting to fix it on the spot. Though I seem to have run into the same problem you call ChrisBln out for. Will continue looking at this. Analysis was never my strongest subject. – John Habert Jan 28 '14 at 18:16