Let \(\{g_n\}\) be a sequence of Riemann-integrable functions from \([0,1]\) to \(\mathbb R\) such that \[ |g_n(x)|\le 1 \] for every \(n\) and \(x\). Define \[ G_n(x)=\int_0^x g_n(t)\,dt. \] Prove that a subsequence of \(\{G_n\}\) converges uniformly.
Solution
1. For every \(n\) and every \(x\in[0,1]\),\[ \abs{G_n(x)}\le1. \]
Using \(\abs{g_n(t)}\le1\), \[ \abs{G_n(x)} = \abs{\int_0^x g_n(t)\,dt} \le \int_0^x\abs{g_n(t)}\,dt \le x \le 1. \]
2. For every \(n\) and every \(x,y\in[0,1]\),\[ \abs{G_n(x)-G_n(y)} \le \abs{x-y}. \]
Assume first that \(x\ge y\). Then \[ G_n(x)-G_n(y) = \int_y^x g_n(t)\,dt, \] so \[ \abs{G_n(x)-G_n(y)} \le \int_y^x\abs{g_n(t)}\,dt \le x-y. \] The case \(y\ge x\) is identical after interchanging \(x\) and \(y\).
3. The family \(\{G_n\}\) is uniformly bounded and equicontinuousin \(C([0,1])\).
1. Step 2 shows that every \(G_n\)is \(1\)-Lipschitz, so for every \(\varepsilon>0\), choosing\(\delta=\varepsilon\) gives\[ \abs{x-y}<\delta \quad\Longrightarrow\quad \abs{G_n(x)-G_n(y)}<\varepsilon \]simultaneously for all \(n\). Thus the family is equicontinuous.The same Lipschitz estimate also shows that every \(G_n\) iscontinuous.
4. Some subsequence of \(\{G_n\}\) converges uniformly on\([0,1]\).
The interval \([0,1]\) is compact. By step 3, the sequence lies in a uniformly bounded equicontinuous family of continuous functions. The Arzelà–Ascoli theorem therefore supplies a subsequence \(\{G_{n_j}\}\) that converges uniformly on \([0,1]\).
5. Q.E.D.
Step 4 is the required conclusion.