Let $I$ be an interval and $f\colon I \to \mathbb{R}$ a differentiable function. Suppose the following definitions:

For $x_0 \in I$ the point $(x_0,f(x_0))$ is called saddle point if $f'(x_0) = 0$ but $x_0$ is not a local extremum of $f$.

For $x_W \in I$ the point $(x_W,f(x_W))$ is called point of inflection if there is a neighborhood $U$ from $x_W$ in $I$ such that $f'$ is strictly monotonic increasing (resp. decreasing) for $x < x_W$ on $U$ and strictly monotinic decreasing (resp. increasing) for $x > x_W$ on $U$.

What is the logical relation between saddle points and points of inflection?

My first intuitive guess was that a point $(x,f(x))$ is a saddle point iff it is a point of inflection and $f'(x) = 0$. However the implication "$\implies$" seems to be wrong. Consider the following counterexample: $$ f(x) = \begin{cases} x^4 \cdot \sin\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases} $$

Then $(0,0)$ is a saddle point but not a point of inflection because the derivativative oscillates on every neighborhood of $0$.

Is this correct so far? Is the other implication true? If so, how to prove it?

  • $\begingroup$ Is it OK to assume we have strictly mono. increasing on some $[x_W-\delta,x_W]$ or is it important to only assume this for some $[x_W-\delta,x_W)?$ $\endgroup$ – zhw. Dec 18 '15 at 22:25
  • $\begingroup$ Since the function is differentiable, strictly mono. increasing on the half open intervall implies the same property on the closed intervall. $\endgroup$ – Julia Dec 19 '15 at 13:01
  • $\begingroup$ I was referring to your third paragraph and the strictly increasing/decreasing nature of $f'.$ $\endgroup$ – zhw. Dec 20 '15 at 1:42

Your example does indeed show that a saddle point need not be an inflection point. (The function $x^2\sin(1/x)$ also works, but your example has the virtue of being continuously differentiable.)

In the other direction, if $(a,f(a))$ is a point of inflection and $f'(a) = 0,$ then $(a,f(a))$ is a saddle point. To see this, suppose WLOG that for some small $\delta > 0$ that $f'$ strictly increases in $[a-\delta,a]$ and strictly decreases in $[a,a+\delta].$ In $[a-\delta,a)$ we have $f'(x)<0,$ because these values must be less than $f'(0)=0.$ The same reasoning shows that $f'(x) < 0$ for $x\in (a,a+\delta].$ The mean value theorem then shows $f$ strictly decreases on both $[a-\delta,a]$ and $[a,a+\delta].$ Hence $f$ strictly decreases on $[a-\delta,a+\delta].$ It follows that $f(a)$ is neither a local max. nor min. for $f$ at $a.$

  • $\begingroup$ Why did you delete the worked out mean value argument. Was there something wrong with it? $\endgroup$ – Julia Dec 21 '15 at 20:58
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    $\begingroup$ No, I just thought, reading it over, that it was better to give the main idea. Thinks like $f'>0\implies f$ is strictly increasing are pretty well known at this level so I thought it would be better to not repeat it. $\endgroup$ – zhw. Dec 21 '15 at 21:14

OK for the first part.

Let's take a look at the second part (the $\Leftarrow$ implication). It is indeed true.

Let us suppose $(x,f(x))$ is a point of inflexion and $f'(x) = 0$. Then there exists an interval $J = [a,b]$, $x \in J$ and $J \subset I$, where we can suppose by symmetry, that $f'$ is increasing on $[a,x]$ and decreasing on $[x,b]$ (otherwise, we can study $-f$).

If $f' < 0$ on $[a,x[$ and $f' < 0$ on $]x,b]$, we have: $\forall u \in [a,x[$, $f(u) > f(x)$ and $\forall v \in ]x,b]$, $f(v) < f(x) $. Consequently, $(x,f(x))$ is not an extremum and is a saddle point.

Then, let us look at the left interval, with a proof by contradiction. If we suppose that there exists $c$ in $[a,x[$ such as $f'(c) >= 0$, as $f'$ is strictly increasing on $[c,x]$, we also have a $d \in [c,x[$ such as $f'(d) > 0$ and $\forall u \in [d,x[$, $f'(u) > f'(d) > 0$.

By integrating on $[x-h,x]$ with $h \in [0,x-d[$, we have $\forall h \in ]0,x-d[$, $f(x)-f(x-h) > f'(d)h$, then $\frac{f(x-h)-f(x)}{h} <-f'(d) < 0$. Consequently, by taking the limit, $f'(x) \neq 0$, and we have our contradiction.

We can have a similar reasoning for the right interval, and conclude that $f' < 0$ on $[a,x[$, $f'>0$ on $]x,b]$, which means that $(x,f(x))$ is a saddle point.


Can only be brief, sorry you can fill in the gaps. Information available in Wiki.. For z = f(x,y) ; second derivative test. <0 for max, >0 for min, test fails but at saddle points the both signs prevail. E.g., monkey saddle.$ f(x,y) = x^3 - 3 x y^2 $. when considering inflection points along certain directions ( 3 of 6 directions). Like a Col point in mountainous range, one direction upward, one downward, one is neither, topography negative Gauss curvature, inflection along asymptotic direction, normal curvature vanishes.. See also Minimax, Nash equilibrium.

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    $\begingroup$ The problem posed is clearly a one-variable problem. $\endgroup$ – zhw. Dec 20 '15 at 1:37

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