Function $\mathbb{R} \rightarrow \mathbb{R}$ local minimum is global Could you tell me how to prove or disprove that if a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ has only one critical point and it is a local minimum, then it is a global minimum?
I know that if $f$ is convex, then its local minimum is global. Is it true in general for real functions with one variable?
I would really appreciate all your help.
Thank you.
 A: Without loss of generality, assume the critical point (or critical number) is $0$, and $f(0)=0$.  You can do this by sliding the graph of the function around in the plane.  
Argue by contradiction and assume $f$ does not have a global minimum at $x=0$.  Than $f(b) < 0$ for some $b \neq 0$. For convenience, assume $b>0$.  Since $f$ has a local minimum at $x=0$, either $f\equiv 0$ on $[0,b/2]$ or $f(c)>0$ for some $c\in (0,b/2]$.  The first alternative is impossible since it implies $f'(b/4)=0$ (and $0$ is the only critical point of $f$).  So the second alternative follows.  By the Intermediate Value Theorem, there exists $d \in (c, b)$ with $f(d)=0$.
To conclude the proof you essentially need Rolle's Theorem.  I looked up Rolle's Theorem and the hypotheses are that $f$ is continuous on $[0,d]$ and "differentiable" on $(0,d)$.  I'm not sure if the term "differentiable" requires $f'$ to be continuous.  You actually don't need $f'$ to be continuous, you just need $f'(x)$ to exist for all $x \in (0,d)$.   So I'll go ahead and finish the proof of this problem using basically the standard proof of Rolle's Theorem.
Since $f$ is continuous, $f$ attains maximum and minimum values on the interval $[0,d]$.  If $f$ is constant on $[0,d]$, then $f'(d/2)=0$, which gives you a contradiction.  If $f$ has a positive maximum on $[0,d]$, which occurs at some $w \in (0,d)$, then Fermat's Theorem gives you $f'(w)=0$, also a contradiction.     If $f$ has a negative minimum on $[0,b]$, which occurs at some $w \in (0,b)$, then Fermat's Theorem gives you $f'(w)=0$, also a contradiction.  
Therefore $f$ has a global minimum at $x=0$.
Note that I never assumed that $f'$ was continuous, only that $f'(x)$ exists for all $x\neq 0$ and $f$ is continuous at $x=0$.
If you use the standard Calc I definition of critical point (number): "$c$ is a critical number of $f$ if $f'(c)=0$ or doesn't exist", the conclusion holds if you replace "$f$ is differentiable" with "$f$ is continuous": if $f$ has a nonzero critical point, you're done.  Otherwise, $f'(x)$ exists for all $x \neq 0$, and if you also assume  $f$ is continuous at $x=0$, just repeat the above proof.
Without assuming $f$ is continuous, the conclusion seems to be false: consider $f:\mathbb{R}\to \mathbb{R}$ defined by $f(0)=0$ and $f(x)=1-|x|$ for $x \neq 0$ (sorry, I forgot how to do \cases).  $f$ has a local minimum at $x=0$, but no nonzero critical points.  I'm not sure whether $0$ is officially considered a critical point here ($f$ is discontinuous at $x=0$).
