Let \(f\in L^1([0,1])\). For \(k\in\mathbb N\), define the step function \(f_k\) by \[ f_k(x)=k\int_{j/k}^{(j+1)/k}f(t)\,dt, \qquad \frac jk\le x<\frac{j+1}{k}. \] Show that \(f_k\to f\) in \(L^1([0,1])\) as \(k\to\infty\).
Solution
Let \(A_k:L^1([0,1])\to L^1([0,1])\) denote the averaging operator \(A_k f=f_k\).
1. \(A_k\) is an \(L^1\) contraction.
On each interval \[ I_{j,k}=\left[\frac jk,\frac{j+1}{k}\right), \] one has \[ A_k f=k\int_{I_{j,k}}f. \] Therefore \[ \int_{I_{j,k}}|A_k f(x)|\,dx = \left|\int_{I_{j,k}}f(t)\,dt\right| \le \int_{I_{j,k}}|f(t)|\,dt. \] Summing over \(j\) gives \[ \|A_k f\|_1\le\|f\|_1. \]
2. \(A_k g\to g\) uniformly for every continuous \(g\).
Let \(g\in C([0,1])\). Since \(g\) is uniformly continuous, its modulus of continuity \[ \omega_g(\delta)=\sup_{|x-y|\le\delta}|g(x)-g(y)| \] satisfies \(\omega_g(\delta)\to0\) as \(\delta\downarrow0\). If \(x\in I_{j,k}\), then \[ \begin{aligned} |A_k g(x)-g(x)| &=\left|k\int_{I_{j,k}}(g(t)-g(x))\,dt\right|\\ &\le k\int_{I_{j,k}}|g(t)-g(x)|\,dt\\ &\le \omega_g(1/k). \end{aligned} \] Hence \[ \|A_k g-g\|_\infty\le\omega_g(1/k)\longrightarrow0, \] and therefore also \(\|A_k g-g\|_1\to0\).
3. Approximate an arbitrary \(L^1\) function by continuous functions.
Fix \(\varepsilon>0\). Choose \(g\in C([0,1])\) with \[ \|f-g\|_1<\frac\varepsilon3, \] using density of \(C([0,1])\) in \(L^1([0,1])\). By 2, for all sufficiently large \(k\), \[ \|A_k g-g\|_1<\frac\varepsilon3. \] Then the contraction estimate from 1 gives \[ \begin{aligned} \|A_k f-f\|_1 &\le\|A_k(f-g)\|_1+\|A_k g-g\|_1+\|g-f\|_1\\ &\le2\|f-g\|_1+\|A_k g-g\|_1\\ &<\varepsilon. \end{aligned} \] Thus \(f_k=A_kf\to f\) in \(L^1\).