Let \((K_n)\) be nonnegative integrable functions on \(\mathbb R/\mathbb Z\) such that \[ \int K_n=1 \] and for every \(0<\varepsilon<1/2\), \[ \int_{\varepsilon\le |t|\le1/2}K_n(t)\,dt\longrightarrow0. \] For continuous \(f\) on \(\mathbb R/\mathbb Z\), prove that \(f*K_n\to f\) uniformly.
Solution
1. The contribution from \(|t|<\delta\) can be made uniformly small.
Fix \(\eta>0\). Since \(f\) is continuous on the compact circle, it is uniformly continuous. Choose \(0<\delta<1/2\) so that \[ |f(x-t)-f(x)|<\frac{\eta}{2} \qquad(|t|<\delta) \] for every \(x\). Since \(K_n\ge0\) and \(\int K_n=1\), \[ \int_{|t|<\delta}|f(x-t)-f(x)|K_n(t)\,dt \le\frac\eta2 \] uniformly in \(x\) and \(n\).
2. The contribution from \(\delta\le|t|\le1/2\) tends uniformly to zero.
For every \(x\) and \(t\), \[ |f(x-t)-f(x)|\le2\|f\|_\infty. \] Therefore \[ \int_{\delta\le|t|\le1/2}|f(x-t)-f(x)|K_n(t)\,dt \le 2\|f\|_\infty \int_{\delta\le|t|\le1/2}K_n(t)\,dt. \] The last integral tends to \(0\) by hypothesis, and the bound is independent of \(x\).
3. The convolutions converge uniformly to \(f\).
Because \(\int K_n=1\), \[ (f*K_n)(x)-f(x) =\int (f(x-t)-f(x))K_n(t)\,dt. \] By steps <1>1–<1>2, for all sufficiently large \(n\) the absolute value of the right-hand side is \(<\eta\) for every \(x\). Hence \[ \|f*K_n-f\|_\infty\longrightarrow0. \]
4. Q.E.D.
Step 3 is exactly the required uniform convergence.