Let \(f:(-1,1)\to\mathbb R^2\) be \(C^1\), with \[ f(0)=0, \qquad f'(0)\ne0. \] Prove that there exists \(\varepsilon\in(0,1)\) such that \(\|f(t)\|\) is increasing on \((0,\varepsilon)\).
Solution
1. One has\[ \frac{f(t)}{t}\longrightarrow f'(0) \qquad\text{as }t\to0. \]
Since \(f(0)=0\) and \(f\) is differentiable at \(0\), \[ f'(0) = \lim_{t\to0}\frac{f(t)-f(0)}{t} = \lim_{t\to0}\frac{f(t)}{t}. \]
2. One has\[ \left\langle\frac{f(t)}{t},f'(t)\right\rangle \longrightarrow \norm{f'(0)}^2>0 \]as \(t\to0\) through nonzero values.
By step 1, \[ \frac{f(t)}{t}\to f'(0), \] while continuity of \(f'\) gives \(f'(t)\to f'(0)\). Continuity of the inner product gives the stated limit. The limit is positive because \(f'(0)\ne0\).
3. There exists \(\varepsilon\in(0,1)\) such that\[ \left\langle\frac{f(t)}{t},f'(t)\right\rangle>0 \]for every \(t\in(0,\varepsilon)\).
This follows from the positive limit in step 2.
4. The function\[ h(t)\coloneqq\norm{f(t)}^2 \]is strictly increasing on \((0,\varepsilon)\).
3. Hence \(h\) is strictly increasing there.
5. The function \(t\mapsto\norm{f(t)}\) is strictly increasing on \((0,\varepsilon)\).
The square-root function is strictly increasing on \([0,\infty)\). Since \[ \norm{f(t)}=\sqrt{h(t)}, \] the conclusion follows from step 4.
6. Q.E.D.
Step 5 proves the assertion.