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Theorem:

If $f(x)$ , a function of one variable , is continuous on an interval I and has only one critical point in I (where a critical point is a point c at which $f'(c)$ does not exist or $f'(c)=0$),and that critical point is a local minimum in I , then it is the global minimum in I.

(Same applies for maximum)


This theorem does not hold in higher dimensions
eg.$f(x,y)=e^{3x}+y^3-3ye^x$(You can show using second derivative test that it fails the theorem)

But it does hold in one dimension .
I want to know how to prove this theorem in one dimension(Whatever the proof is fails in higher dimensions)
To prove it i think you have to use some property of functions from analysis

Edit: if f is differentiable on (c,x] and the derivative is zero nowhere and has the same signed derivative on (c,x) then can you make any conclusion about the sign derivative of x or not?

Edit: The theroem that is critcial for the proof of this theorem is that If derivative has different signs at $a, b\in I$ then by applying Darboux on $[a, b] $ the derivative vanishes somewhere on $(a, b) $. So to apply to our case if the sign is different with any two points within (c,x) then the derivative must vanish in (c,x) which contracdicts the hypothesis hence the sign must be same for all (c,x]

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  • $\begingroup$ If $f: \mathbb{R}^2\rightarrow \mathbb{R}$, then what is $f'$? It is not like the usual derivative in one variable. $\endgroup$
    – Eminem
    Jun 22 '20 at 18:22
  • $\begingroup$ About your edit: if $f$ is differentiable on some interval $I$ and $f'$ does not vanish on $I$ then $f'$ maintains same sign on $I$. Now $I$ can be any interval (closed, open, whatever). $\endgroup$
    – Paramanand Singh
    Jun 24 '20 at 12:11
  • $\begingroup$ @ParamanandSingh you have said that is due to Darboux but the proof you gave me only worked for open intervals.... $\endgroup$
    – Logic
    Jun 24 '20 at 12:13
  • $\begingroup$ No I talked about any interval, not just open intervals. You can read my comments again. $\endgroup$
    – Paramanand Singh
    Jun 24 '20 at 12:14
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By your assumption every point of $I$ except $c$ is not a critical point. By definition of critical point in your question, it means that if $x\in I, x\neq c$ then $f'(x) \neq 0$.

Let us deal with the case when $c$ is an interior point of $I$. It is also given that $c$ is a local minimum. Consider any $x\in I$ and $x>c$. The derivative $f'$ does not vanish in interval $(c, x) $ and hence by Darboux theorem maintains a constant sign. Now if this derivative $f'$ were negative in $(c, x) $ then we would have $f(y) <f(c) $ for all $y\in(x, c) $ and this contradicts that $c$ is a local minimum. It follows that $f'$ is positive in $(c, x) $ and hence $f(x) >f(c) $.

Similarly for $x\in I$ with $x<c$ we can prove that $f(x) > f(c) $ so that $c$ is indeed a global minimum on $I$.

The proof is simpler and similar if $c$ is an end point of $I$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Jun 24 '20 at 10:26
  • $\begingroup$ @VivaanDaga: there is no need to know sign of derivative at $x$ to answer the current problem. But yes derivative maintains a constant sign on $(c, x) $ as well as $(c, x] $. I still don't know what is the real confusion with you. $\endgroup$
    – Paramanand Singh
    Jun 24 '20 at 12:14
  • $\begingroup$ @ParamanandSingh can you please check my second edit and tell if i am right or wrong $\endgroup$
    – Logic
    Jun 24 '20 at 12:28
  • $\begingroup$ @ParamanandSingh in my second edit am i correct i have tried to take your theorem and apply it to this case . Also this does fail in higher dimensions becasue of things like going up in one dimension and going down in another and alll...am i correct? $\endgroup$
    – Logic
    Jun 24 '20 at 13:39
  • $\begingroup$ @ParamanandSingh what part of this proof fails in higher dimensions $\endgroup$
    – Logic
    Jun 24 '20 at 14:23
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Suppose there is $c_1$ in $I$ with $f(c_1) \le f(c)$. Because $f$ is continuous, there is $c_2$ in $I$ with $f(c)=f(c_2)$. If. $f$ is differentiable in $(c,c_2)$. Then there is $c'$ in $(c,c_2)$ with $f'(c')=0$, by Rolle's Theorem. If $f$ is not differentiable in $(c,c_2)$, let's say in $c'$, again $c'$ is another critcal point of $f$. Either way, $c'$ is another critical point of $f$.

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