Let \(F:\mathbb R^2\to\mathbb R\) be continuously differentiable, with \(F(0,0)=0\) and \[ \|\nabla F(0,0)\|<1. \] Prove that there is \(r>0\) such that \(|F(x,y)|<r\) whenever \(\|(x,y)\|<r\).
Solution
Put \[ \alpha=\norm{\nabla F(0,0)}<1 \] and choose \[ c=\frac{1+\alpha}{2}. \] Then \[ \alpha<c<1. \]
1. There exists \(r>0\) such that\[ \norm{\nabla F(z)}<c \]whenever\[ \norm z<r. \]
Since \(F\) is continuously differentiable, its gradient is continuous at the origin. Because \[ c-\alpha>0, \] there is \(r>0\) such that \[ \norm{\nabla F(z)-\nabla F(0,0)} < c-\alpha \] whenever \(\norm z<r\). The triangle inequality then gives \[ \norm{\nabla F(z)} \leq \norm{\nabla F(z)-\nabla F(0,0)} + \norm{\nabla F(0,0)} < c. \]
2. For every \(z=(x,y)\in\RR^2\) with \(\norm z<r\),\[ \abs{F(z)}\leq c\norm z. \]
Fix such a \(z\) and define \[ g(t)=F(tz), \qquad 0\leq t\leq1. \] Then \[ g'(t)=\nabla F(tz)\cdot z. \] Since \[ \norm{tz}\leq\norm z<r, \] step 1 and Cauchy–Schwarz give \[ \abs{g'(t)} \leq \norm{\nabla F(tz)}\,\norm z < c\norm z. \] Using \(F(0,0)=0\) and the fundamental theorem of calculus, \[ \begin{aligned} \abs{F(z)} &= \abs{g(1)-g(0)}\\ &= \abs{\int_0^1 g'(t)\,dt}\\ &\leq \int_0^1\abs{g'(t)}\,dt\\ &\leq c\norm z. \end{aligned} \]
3. Whenever \(\norm{(x,y)}<r\),\[ \boxed{\abs{F(x,y)}<r}. \]
By step 2, \[ \abs{F(x,y)} \leq c\norm{(x,y)} < cr. \] Since \(c<1\), \[ cr<r. \] Combining these inequalities gives the claim.
4. Q.E.D.
Step 3 gives the required radius and estimate.