Let \(f:\mathbb R\to\mathbb R\) be \(C^\infty\) and suppose \(f\) has a local minimum at \(0\). Prove that there is a disk centered on the \(y\)-axis which lies above the graph of \(f\) and is tangent to the graph at \[ (0,f(0)). \]
Solution
Set \[ c\coloneqq f(0). \]
1. One has\[ f'(0)=0. \]
A differentiable function at an interior local minimum has derivative zero.
2. There are constants \(\delta>0\) and \(C\geq0\) such that\[ f(x)\leq c+Cx^2 \]whenever\[ \abs{x}\leq\delta. \]
Since \(f''\) is continuous, it is bounded on some interval \([-\delta,\delta]\). Let \[ M\coloneqq \max_{\abs{t}\leq\delta}\abs{f''(t)}. \] Taylor’s theorem, together with step 1, gives for each \(\abs{x}\leq\delta\) a point \(\xi\) between \(0\) and \(x\) such that \[ f(x) = c+\frac12f''(\xi)x^2. \] Hence \[ f(x) \leq c+\frac M2x^2. \] Take \(C=M/2\).
3. Choose \(R>0\) such that\[ R\leq\delta \]and\[ 2CR\leq1. \]Let\[ \mathcal D \coloneqq \left\{ (x,y)\in\RR^2: x^2+(y-c-R)^2\leq R^2 \right\}. \]Then \(\mathcal D\) is a disk centered on the \(y\)-axis and\((0,c)\in\partial\mathcal D\).
If \(C=0\), choose any \(0<R\leq\delta\). If \(C>0\), choose \[ 0<R\leq\min\left\{\delta,\frac1{2C}\right\}. \] The displayed set is the closed disk of radius \(R\) centered at \[ (0,c+R), \] which lies on the \(y\)-axis. The distance from this center to \((0,c)\) is \(R\), so \((0,c)\) lies on its boundary.
4. For every \(\abs{x}\leq R\), the lower boundary of \(\mathcal D\) is\[ \ell(x) \coloneqq c+R-\sqrt{R^2-x^2}, \]and\[ \ell(x) \geq c+\frac{x^2}{2R}. \]
Rationalizing gives \[ \begin{aligned} R-\sqrt{R^2-x^2} &= \frac{x^2}{R+\sqrt{R^2-x^2}}\\ &\geq \frac{x^2}{2R}, \end{aligned} \] because \[ \sqrt{R^2-x^2}\leq R. \] Adding \(c\) gives the claim.
5. The disk \(\mathcal D\) lies above the graph of \(f\).
If \(\abs{x}\leq R\), then step 2 applies because \(R\leq\delta\), and step 3 gives \[ C\leq\frac1{2R}. \] Therefore \[ f(x) \leq c+Cx^2 \leq c+\frac{x^2}{2R} \leq \ell(x) \] by step 4.
Every point \((x,y)\in\mathcal D\) has \(\abs{x}\leq R\) and \[ y\geq\ell(x). \] Hence \[ y\geq f(x) \] for every point of the disk, which is precisely that the disk lies above the graph.
6. The disk \(\mathcal D\) is tangent to the graph at \((0,c)\).
The lower boundary function satisfies \[ \ell(0)=c \] and \[ \ell'(0)=0. \] By step 1, \[ f(0)=c \qquad\text{and}\qquad f'(0)=0. \] Thus the graph and the circle boundary meet at \((0,c)\) with the same horizontal tangent line. Step 5 shows that the disk stays on the upper side of the graph there.
7. Q.E.D.
Steps 3, 5, and 6 give a disk centered on the \(y\)-axis, lying above the graph, and tangent to it at \((0,f(0))\).