Second derivative test without continuity of the second derivative

Question

Let $S\subset\mathbb{R}$ be an open subset and $f:S\to\mathbb{R}$ such that $f^{''}$ is continuous. Suppose $x_0\in S$ such that $f^{'}(x_0)=0$ and $f^{''}(x_0)>0$, then $x_0$ is a point of local minimum of $f$ by the second derivative test.

I wonder if we remove the continuity assumption of the second derivative function $f^{''}$, can we get $x_0$ is a point of local minimum again or it violates. My guess is that it should violate, since in the proof we use Taylor's theorem, where the continuity of $f^{''}$ is used in the argument.

If it violates, can someone please help with a counter-example of the same?

Thanks in advance.

Answer 1

In the proof of the sufficient condition for a minimum, all you need is the local Taylor's formula (with remainder in the form of Peano) to be able to say that the quadratic term is the leading one near the point. Taylor's formula in that form only requires the derivatives exist at the given point. Of course, the first derivative has to be continuous in a neighborhood for the second to exist at the point, but only the existence of the second is required at the point in question.

Answer 2

Say $f'(x_0) = 0$ and $f''(x_0) = a > 0$. No assumption that $f''$ is continuous. Indeed, no assumption that $f''(x)$ exists for any $x \ne x_0$.

Then $$ \lim_{h\to 0}\frac{f'(x)}{x-x_0} = \lim_{h\to 0}\frac{f'(x)-f'(x_0)}{x-x_0} = a $$ Taking $\varepsilon = a/2$ we get $\delta > 0$ so that $$ |x-x_0| < \delta \quad\Rightarrow\quad\frac{f'(x)}{x-x_0} > \frac{a}{2} $$ thus, for $x$ with $x_0 < x < x_0+\delta$ $$ f'(x) > (x-x_0)\frac{a}{2} > 0 $$ so $f$ is incrreasing on $(x_0,x_0+\delta)$. For $x$ with $x_0-\delta < x < x_0$, $$ f'(x) < (x-x_0)\frac{a}{2} < 0 $$ so $f$ is decreasing on $(x_0-\delta,x_0)$. Thus $f$ has a local minimum at $x_0$.