Let \((f_j)\) be an orthonormal sequence in \(L^2([0,1])\). Prove that the Cesaro means \[ S_n=\frac1n\sum_{j=1}^n f_j \] converge to \(0\) almost everywhere.
Solution
Because the \(f_j\) are orthonormal, \[ \|S_n\|_2^2 =\frac1{n^2}\sum_{j=1}^n\|f_j\|_2^2 =\frac1n. \]
1. The square subsequence converges to \(0\) almost everywhere.
For \(n=k^2\), \[ \sum_{k=1}^\infty \|S_{k^2}\|_2^2 =\sum_{k=1}^\infty\frac1{k^2}<\infty. \] Hence, for every \(\varepsilon>0\), Chebyshev gives \[ \sum_{k=1}^\infty m\{|S_{k^2}|>\varepsilon\} \le \frac1{\varepsilon^2}\sum_{k=1}^\infty\|S_{k^2}\|_2^2<\infty. \] Apply Borel–Cantelli for each of the countably many thresholds \(\varepsilon=1/j\), \(j\geq1\) [@Fol13]. Outside the union of their null exceptional sets, each threshold is exceeded only finitely often. Therefore \(S_{k^2}(x)\to0\) for almost every \(x\).
2. The oscillation between consecutive squares tends to \(0\) almost everywhere.
Fix \(k\) and put \(m=k^2\). For \(m<n\le(k+1)^2\), orthogonality gives \[ S_n-S_m =\left(\frac1n-\frac1m\right)\sum_{j=1}^m f_j +\frac1n\sum_{j=m+1}^n f_j, \] with the two sums orthogonal. Therefore \[ \|S_n-S_m\|_2^2 =m\left(\frac1n-\frac1m\right)^2+\frac{n-m}{n^2} =\frac{n-m}{nm}. \] Since \(n-m\leq2k+1\leq3k\), \(m=k^2\), and \(n\geq k^2\), it follows that \[ \|S_n-S_{k^2}\|_2^2\leq\frac{3}{k^3} \] for every \(k\geq1\) and \(k^2<n\leq(k+1)^2\).
For \(\varepsilon>0\), a union bound and Chebyshev give \[ \begin{aligned} m\left\{\max_{k^2<n\le(k+1)^2}|S_n-S_{k^2}|>\varepsilon\right\} &\le\sum_{n=k^2+1}^{(k+1)^2}m\{|S_n-S_{k^2}|>\varepsilon\}\\ &\le\frac1{\varepsilon^2}\sum_{n=k^2+1}^{(k+1)^2}\|S_n-S_{k^2}\|_2^2\\ &\le\frac{9}{\varepsilon^2k^2}, \end{aligned} \] because the block contains \(2k+1\leq3k\) indices. The right-hand side is summable in \(k\). Applying Borel–Cantelli for each \(\varepsilon=1/j\), \(j\geq1\), and taking the countable union of the resulting null exceptional sets gives \[ \max_{k^2<n\le(k+1)^2}|S_n(x)-S_{k^2}(x)|\longrightarrow0 \] for almost every \(x\).
3. Conclude for the full sequence.
Outside the union of the two null exceptional sets from 1 and 2, if \(k^2<n\le(k+1)^2\), then \[ |S_n(x)| \le |S_{k^2}(x)|+|S_n(x)-S_{k^2}(x)|. \] Both terms tend to \(0\) as \(k\to\infty\). Hence \[ S_n(x)\longrightarrow0 \] for almost every \(x\in[0,1]\).