Let \(f:(0,1)\to\mathbb C\) be \(C^1\). Suppose \[ f(t)\to0, \qquad f'(t)\to C\ne0 \qquad(t\to0^+). \] Show that \(g(t)=|f(t)|\) is \(C^1\) for all sufficiently small \(t>0\), and that \[ \lim_{t\to0^+}g'(t) \] exists. Evaluate this limit.
Solution
1. There is a number \(\delta_0>0\) such that \(f'\) is bounded on\((0,\delta_0]\).
The hypothesis \[ f'(t)\longrightarrow C \qquad (t\to0^+) \] implies that there is \(\delta_0>0\) such that \[ \abs{f'(t)-C}<1 \] whenever \(0<t\leq\delta_0\). Thus \[ \abs{f'(t)} \leq \abs{C}+1 \] on this interval.
2. For every \(0<t\leq\delta_0\),\[ f(t) = \int_0^t f'(s)\,ds, \]where the integral is taken componentwise in \(\CC\cong\RR^2\).
Fix \(t\leq\delta_0\). For \(0<a<t\), the fundamental theorem of calculus gives \[ f(t)-f(a) = \int_a^t f'(s)\,ds. \] By hypothesis, \[ f(a)\longrightarrow0 \] as \(a\to0^+\). By step 1, the derivative is bounded near \(0\), so the improper integral \[ \int_0^t f'(s)\,ds \] exists. Letting \(a\to0^+\) in the displayed identity gives the formula.
3. One has\[ \boxed{ \frac{f(t)}{t}\longrightarrow C } \]as \(t\to0^+\).
By step 2, \[ \frac{f(t)}t-C = \frac1t \int_0^t \bigl(f'(s)-C\bigr)\,ds. \] Therefore \[ \abs{ \frac{f(t)}t-C } \leq \sup_{0<s\leq t}\abs{f'(s)-C}. \] The right-hand side tends to zero because \(f'(s)\to C\) as \(s\to0^+\).
4. There is \(\delta\in(0,\delta_0]\) such that\[ f(t)\neq0 \]for every \(0<t<\delta\).
Since \(C\neq0\), step 3 gives a \(\delta\in(0,\delta_0]\) such that \[ \abs{ \frac{f(t)}t-C } < \frac{\abs{C}}{2} \] for \(0<t<\delta\). Hence \[ \abs{ \frac{f(t)}t } \geq \abs{C} - \abs{ \frac{f(t)}t-C } > \frac{\abs{C}}{2} > 0. \] Thus \(f(t)\neq0\) on that interval.
5. The function\[ g(t)=\abs{f(t)} \]is \(C^1\) on \((0,\delta)\), and\[ g'(t) = \operatorname{Re} \left( f'(t) \frac{\overline{f(t)}}{\abs{f(t)}} \right). \]
The modulus map \[ \CC\sm\{0\}\longrightarrow\RR, \qquad z\longmapsto\abs{z}, \] is \(C^1\). By step 4, the image of \((0,\delta)\) under \(f\) avoids \(0\), so the composition \(g=\abs{\cdot}\circ f\) is \(C^1\).
Writing \(f=u+iv\) with real-valued \(C^1\) functions \(u,v\), \[ g(t) = \sqrt{u(t)^2+v(t)^2}. \] Since \(g(t)>0\), \[ \begin{aligned} g'(t) &= \frac{u(t)u'(t)+v(t)v'(t)}{g(t)}\\ &= \operatorname{Re} \left( (u'(t)+iv'(t)) \frac{u(t)-iv(t)}{\abs{f(t)}} \right), \end{aligned} \] which is the displayed formula.
6. One has\[ \frac{f(t)}{\abs{f(t)}} \longrightarrow \frac{C}{\abs{C}} \]as \(t\to0^+\).
Because \(t>0\), \[ \frac{f(t)}{\abs{f(t)}} = \frac{f(t)/t}{\abs{f(t)/t}}. \] Step 3 gives \[ \frac{f(t)}t\longrightarrow C\neq0. \] The map \[ z\longmapsto\frac z{\abs{z}} \] is continuous on \(\CC\sm\{0\}\), so the stated limit follows.
7. The derivative of \(g\) has the limit\[ \boxed{ \lim_{t\to0^+}g'(t)=\abs{C}. } \]
By step 5, \[ g'(t) = \operatorname{Re} \left( f'(t) \frac{\overline{f(t)}}{\abs{f(t)}} \right). \] The hypotheses give \[ f'(t)\longrightarrow C, \] while step 6 gives \[ \frac{\overline{f(t)}}{\abs{f(t)}} \longrightarrow \frac{\overline C}{\abs{C}}. \] Therefore \[ \begin{aligned} \lim_{t\to0^+}g'(t) &= \operatorname{Re} \left( C\frac{\overline C}{\abs{C}} \right)\\ &= \operatorname{Re}(\abs{C})\\ &= \abs{C}. \end{aligned} \]
8. Q.E.D.
Step 5 proves that \(g\) is \(C^1\) for all sufficiently small positive \(t\), and step 7 evaluates the required derivative limit.