Let \((f_n)\) be a sequence of real-valued \(C^1\) functions on \([0,1]\) such that for every \(n\), \[ |f_n'(x)|\le\frac1{\sqrt{x}} \qquad(0<x\le1), \] and \[ \int_0^1 f_n(x)\,dx=0. \] Prove that \((f_n)\) has a subsequence converging uniformly on \([0,1]\).
Solution
1. For every \(n\) and every \(0\le x<y\le1\),\[ \abs{f_n(y)-f_n(x)} \le 2(\sqrt y-\sqrt x) \le 2\sqrt{y-x}. \]
First suppose \(x>0\). By the fundamental theorem of calculus and the derivative bound, \[ \begin{aligned} \abs{f_n(y)-f_n(x)} &\le\int_x^y\abs{f_n'(t)}\,dt\\ &\le\int_x^y t^{-1/2}\,dt\\ &=2(\sqrt y-\sqrt x). \end{aligned} \] For \(x=0\), apply this estimate with \(x=\varepsilon>0\) and let \(\varepsilon\downarrow0\), using continuity of \(f_n\) at \(0\).
Finally, \[ \sqrt y = \sqrt{x+(y-x)} \le \sqrt x+\sqrt{y-x}, \] which gives the second inequality.
2. The family \((f_n)\) is equicontinuous on \([0,1]\).
By step 1, \[ \abs{f_n(y)-f_n(x)} \le 2\sqrt{\abs{y-x}} \] for every \(n\) and every \(x,y\in[0,1]\). Given \(\varepsilon>0\), choosing \[ \delta=\left(\frac{\varepsilon}{2}\right)^2 \] makes the right-hand side less than \(\varepsilon\) whenever \(\abs{x-y}<\delta\), uniformly in \(n\).
3. For every \(n\), there exists \(c_n\in[0,1]\) such that\[ f_n(c_n)=0. \]
If \(f_n\) never vanished, continuity on the connected interval \([0,1]\) would force it to be everywhere strictly positive or everywhere strictly negative. Its integral would then have the same strict sign, contradicting \[ \int_0^1f_n(x)\,dx=0. \]
4. The family \((f_n)\) is uniformly bounded:\[ \abs{f_n(x)}\le2 \]for all \(n\) and all \(x\in[0,1]\).
3. Then step 1 gives\[ \abs{f_n(x)} = \abs{f_n(x)-f_n(c_n)} \le 2\sqrt{\abs{x-c_n}} \le2. \]
5. The sequence \((f_n)\) has a subsequence converging uniformly on \([0,1]\).
By steps 2 and 4, the sequence is equicontinuous and uniformly bounded on the compact interval \([0,1]\). The Arzelà–Ascoli theorem therefore gives a uniformly convergent subsequence.
6. Q.E.D.
Step 5 is the required conclusion.