Let \(I \subseteq \mathbb{R}\) be an open interval, and let \(f : I \to \mathbb{R}\) have continuous \(k\)-th derivatives \(f^{[k]}\) on \(I\) for \(k \leq n - 1\). Let \(a \in I\) be a point such that \(f^{[k]}(a) = 0\) for all \(1 \leq k \leq n - 1\), \(f^{[n]}(a)\) exists and \(f^{[n]}(a) > 0\). Prove that \(f\) has a local minimum at \(a\) if \(n\) is even, and has no local extremum at \(a\) if \(n\) is odd.
Solution
1. If \(n=1\), then \(f\) has no local extremum at \(a\).
Here \(f'(a)>0\). By the definition of the derivative, there is \(\delta>0\) such that whenever \(0<\abs{x-a}<\delta\), \[ \frac{f(x)-f(a)}{x-a}>0. \] Thus \[ f(x)<f(a) \quad\text{for }a-\delta<x<a, \] while \[ f(x)>f(a) \quad\text{for }a<x<a+\delta. \] Hence \(a\) is neither a local minimum nor a local maximum.
2. Suppose \(n\geq2\). Then for all \(t\) sufficiently close to \(a\) with\(t\neq a\),\[ \frac{f^{[n-1]}(t)}{t-a}>0. \]
Since \(n\geq2\), the hypothesis includes \[ f^{[n-1]}(a)=0. \] The existence and positivity of \(f^{[n]}(a)\) give \[ \lim_{t\to a} \frac{f^{[n-1]}(t)-f^{[n-1]}(a)}{t-a} = f^{[n]}(a) >0. \] Therefore the displayed quotient is positive on some punctured neighborhood of \(a\).
3. Under the hypotheses of step 2, for every \(x\) sufficiently closeto \(a\) with \(x\neq a\), there exists a point \(c\) strictly between \(a\) and\(x\) such that\[ f(x)-f(a) = \frac{f^{[n-1]}(c)}{(n-1)!}(x-a)^{n-1}. \]
Apply Taylor’s theorem with Lagrange remainder to order \(n-2\) on the interval with endpoints \(a\) and \(x\). It gives \[ f(x) = f(a) + \sum_{k=1}^{n-2} \frac{f^{[k]}(a)}{k!}(x-a)^k + \frac{f^{[n-1]}(c)}{(n-1)!}(x-a)^{n-1} \] for some \(c\) strictly between \(a\) and \(x\). Since \(f^{[k]}(a)=0\) for \(1\leq k\leq n-2\), the sum vanishes.
4. If \(n\) is even, then \(f\) has a strict local minimum at \(a\).
Take \(x\) sufficiently close to \(a\), as in steps 2 and 3.
If \(x>a\), then \(c>a\), so step 2 gives \(f^{[n-1]}(c)>0\). Also \(x-a>0\), hence step 3 gives \(f(x)-f(a)>0\).
If \(x<a\), then \(c<a\), so step 2 gives \(f^{[n-1]}(c)<0\). Since \(n\) is even, \(n-1\) is odd and therefore \((x-a)^{n-1}<0\). The product in step 3 is again positive.
Thus \(f(x)>f(a)\) for every sufficiently close \(x\neq a\).
5. If \(n\geq3\) is odd, then \(f\) has no local extremum at \(a\).
For \(x>a\) sufficiently close to \(a\), the argument in step 4 gives \(f^{[n-1]}(c)>0\) and \((x-a)^{n-1}>0\), hence \[ f(x)>f(a). \]
For \(x<a\) sufficiently close to \(a\), one has \(f^{[n-1]}(c)<0\). Now \(n-1\) is even, so \((x-a)^{n-1}>0\), and step 3 gives \[ f(x)<f(a). \] Therefore every neighborhood of \(a\) contains values of \(f\) both below and above \(f(a)\).
6. The required parity conclusion holds for every positive integer \(n\).
If \(n\) is even, then \(n\geq2\) and step 4 gives a local minimum. If \(n\) is odd, either \(n=1\) and step 1 applies, or \(n\geq3\) and step 5 applies.
7. Q.E.D.
Step 6 is exactly the required conclusion.