$f'$ changes strict monotonicity but $f''$ isn't of strictly opposite signs

Question

Let $I \subset \mathbb{R}$ be an open interval and $f \in C^2(I,\mathbb{R})$.

I am looking for an (simple) example of $f$ with the following properties ($x_0 \in I$)

$f'$ is strictly monotonic increasing for $x < x_0$ and strictly monotonic decreasing for $x > x_0$.

However the following property must be violated on every neighborhood of $x_0$: $f''(x) > 0$ for $x < x_0$ and $f''(x) < 0$ for $x > x_0$.

The background for this question is studying points of inflection i.e. $(x_0,f(x_0))$ satisfying the first property above (or increasing/decreasing swapped) on a neighborhood of $x_0$. Now if there is a neighborhood $U \subset I$ of $x_0$ where $f''(x) < 0$ for $x < x_0$ and $f''(x) > 0$ for $x > x_0$ (or the other way around), the point $(x_0,f(x_0))$ is a point of inflection. The intention of my question was to find an (simple) example which shows that the reversed implication is wrong.

Answer 1

Let $$ g(x)=-\int_{x_0}^xt\sin^2\frac1t\,dt. $$ Then $g$ is strictly decreasing on $x>x_0$ and strictly increasing on $x<x_0$, but $g'$ has zeroes on any neighborhood of $x_0$. Now define $$ f(x)=\int_{x_0}^xg(t)\,dt. $$

Answer 2

How about $$ f(x) = \int_0^x (1-t^2)^3\,dt $$ This gives $f'(x) = (1-x^2)^3$ which satisfies the monotonicity requirement.

But $f''(-1)=f''(1)=0$.

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If you want $f''(x)=0$ arbitrarily close to $x_0$, that can be arranged too: Let $$ g(x) = \begin{cases} 0 & \text{when }x=0 \\ \dfrac{1}{1/x+\sin(1/x)} & \text{otherwise} \end{cases} $$ and then $$ f(x) = \int_0^x g(-t^2)\,dt $$

The point here is that $g$ is strictly increasing and differentiable everywhere with $g'(0)=1$, but $g'(x)=0$ whenever $x$ is the reciprocal of an odd multiple of $\pi$ -- that is, arbitrarily close to $0$.

Answer 3

Edit: The Question has in effect been modified to require $f''(x)$ to have isolated zeros in any neighborhood of $x_0$, both above and below $x_0$, thus invalidating my answer (and those of other responders). Since analytic functions cannot have zeros that "accumulate", a "natural" example becomes less tractable. Probably the easiest approach would amount to defining $f$ "piecewise" on subintervals $[(n+1)^{-1},n^{-1}]$ and $[-n^{-1},-(n+1)^{-1}]$ for each $n\in \mathbb{N}$ in a way that makes the second derivative continuous with zeros at each $\pm 1/n$.

If a "natural" example is wanted, we can start with the nonnegative even function:

$$ f'(x) = | x - \sin x | $$

which is strictly decreasing when $x \lt 0$ and strictly increasing when $x \gt 0$. Moreover the function $f'(x)$ is differentiable (even at $x=0$ by symmetry), and its derivative is the odd function:

$$ f''(x) = (1 - \cos x)\cdot \text{signum}(x) $$

An indefinite integral of $f'$ is thus the odd function:

$$ f(x) = \left(\frac{x^2}{2} - 1 + \cos x \right)\cdot \text{signum}(x) $$

We avoid $f''(x) \gt 0$ for all $x \gt 0$ (resp. avoid $f''(x) \lt 0$ for all $x \lt 0$) because $f''(2k\pi) = 0$ for all integers $k\in \mathbb{Z}$.