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    Uniform decay of the derivative of $f/(1+cf)$ for a nonnegative periodic function

    P-BERK83SU-20

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    1983
    Status
    draft

    Let \(f:\mathbb R\to\mathbb R\) be continuously differentiable, nonnegative, and periodic of period \(1\). Prove that \[ \frac{d}{dx}\left(\frac{f(x)}{1+cf(x)}\right)\longrightarrow0 \qquad(c\to\infty) \] uniformly in \(x\).

    Solution

    For \(c\ge0\), set \[ g_c(x)=\frac{f(x)}{1+cf(x)}. \]

    1. One has\[ g_c'(x)=\frac{f'(x)}{(1+cf(x))^2}. \]

    Differentiating by the quotient rule gives \[ g_c'(x) = \frac{f'(x)(1+cf(x))-cf(x)f'(x)} {(1+cf(x))^2} = \frac{f'(x)}{(1+cf(x))^2}. \]

    2. If \(f(x_0)=0\), then \(f'(x_0)=0\).

    Since \(f\ge0\), any point \(x_0\) with \(f(x_0)=0\) is a local minimum of the differentiable function \(f\). Hence Fermat’s theorem gives \(f'(x_0)=0\).

    3. Fix \(\varepsilon>0\), and let\[ K = \left\{ t\in[0,1]: \abs{f'(t)}\ge\varepsilon \right\}. \]If \(K\) is nonempty, then\[ m\coloneqq\min_{t\in K}f(t)>0. \]

    The set \(K\) is closed in the compact interval \([0,1]\), hence compact. By step 2, no point of \(K\) can be a zero of \(f\). Thus \(f>0\) on \(K\). If \(K\) is nonempty, continuity of \(f\) and compactness of \(K\) imply that \(f\) attains there a strictly positive minimum \(m\).

    4. There exists \(C>0\) such that for every \(c\ge C\) and every\(x\in\RR\),\[ \abs{g_c'(x)}<\varepsilon. \]

    Let \[ M=\max_{t\in[0,1]}\abs{f'(t)}. \] If the set \(K\) from step 3 is empty, then \(\abs{f'(x)}<\varepsilon\) for every \(x\) by periodicity, and step 1 gives \[ \abs{g_c'(x)} \le \abs{f'(x)} < \varepsilon \] for every \(c\ge0\).

    3. Choose \(C>0\) so large that\[ \frac{M}{(1+Cm)^2}<\varepsilon. \]For any \(x\in\RR\), periodicity lets us choose \(t\in[0,1]\) with\[ f(x)=f(t), \qquad f'(x)=f'(t). \]If \(t\notin K\), then \(\abs{f'(t)}<\varepsilon\), and step 1again gives \(\abs{g_c'(x)}<\varepsilon\).

    If \(t\in K\), then \(f(t)\ge m\), so for \(c\ge C\), \[ \abs{g_c'(x)} \le \frac{M}{(1+cm)^2} \le \frac{M}{(1+Cm)^2} < \varepsilon. \] Thus the same bound holds for every \(x\).

    5. Therefore\[ \boxed{ g_c'\longrightarrow0 \quad\text{uniformly on $\RR$ as $c\to\infty$} }. \]

    Step 4 is exactly the \(\varepsilon\)-definition of uniform convergence of \(g_c'\) to \(0\).

    6. Q.E.D.

    Step 5 is the required conclusion.