Define equicontinuity and state the Arzela-Ascoli theorem.
Let \(\mathcal{F}\) be the family of real valued functions on \([0,1]\) satisfying \(f(0)=0\) and \(\int_0^1 f'(x)^2\,dx\le 1\). Show that any sequence in \(\mathcal{F}\) has a subsequence that converges uniformly.
Solution
The functions in \(\mathcal F\) are taken absolutely continuous, so that \(f(x)=\int_0^xf'(t)\,dt\) for \(f\in\mathcal F\).
1. (i) A family \(\mathcal F\subseteq C([0,1])\) is equicontinuous if for every \(\eps>0\) there is \(\delta>0\) with \(\abs{f(x)-f(y)}<\eps\) whenever \(\abs{x-y}<\delta\) and \(f\in\mathcal F\). The Arzelà–Ascoli theorem: for a compact metric space \(K\), a family in \(C(K)\) that is pointwise bounded and equicontinuous has the property that every sequence in it has a uniformly convergent subsequence.
These are the requested definition and statement; on the compact interval \([0,1]\) equicontinuity at every point is equivalent to the uniform version stated.
2. \(\abs{f(x)-f(y)}\le\sqrt{\abs{x-y}}\) for \(f\in\mathcal F\) and \(x,y\in[0,1]\).
For \(y\le x\), the Cauchy–Schwarz inequality gives \(\abs{\int_y^xf'}\le(x-y)^{1/2}\bigl(\int_0^1f'^2\bigr)^{1/2}\le\sqrt{x-y}\).
3. Q.E.D.
By step 2 with \(y=0\), \(\abs f\le1\) on \([0,1]\) for all \(f\in\mathcal F\), and with \(\delta=\eps^2\) the family is equicontinuous. By the Arzelà–Ascoli theorem of step 1, every sequence in \(\mathcal F\) has a uniformly convergent subsequence.