Let \(f\in L^1(\RR)\) and let \(\mathcal{U}\da \ts{(x, y) \in \RR^2 \st y > 0}\) denote the upper half plane. For \((x, y) \in \mathcal{U}\) define \[ u(x, y) \da f \convolve P_y(x) && \text{where } P_y(x) \da {1\over \pi}\qty{y \over t^2 + y^2} .\]
Prove that there exists a constant \(C\) independent of \(f\) such that for all \(x\in \RR\), \[ \sup_{y > 0} \abs{ u(x, y) } \leq C\cdot Hf(x) .\]
Hint: write the following and try to estimate each term: \[ u(x, y) = \int_{\abs t < y} f(x - t) P_y(t) \dt + \sum_{k=0}^{\infty } \int_{A_k} f(x-t) P_y(t)\dt && A_k \da \ts{2^ky \leq \abs t < 2^{k+1}y} .\]
Following the proof of the Lebesgue differentiation theorem, show that for \(f\in L^1(\RR)\) and for almost every \(x\in \RR\), \[ u(x, y) \converges{y\to 0} \to f(x) .\]
In the statement, the kernel is \(P_y(t) = \frac1\pi\frac{y}{t^2+y^2}\) as a function of \(t\), and \(u(x,y) = \int f(x-t)P_y(t)\,dt\).
Solution
Here \(P_y(t) = \frac{1}{\pi}\frac{y}{t^2 + y^2}\), \(u(x,y) = \int f(x-t)P_y(t)\,dt\), and \(Hf(x) = \sup_{r>0}\frac{1}{2r}\int_{x-r}^{x+r}|f(s)|\,ds\) is the Hardy–Littlewood maximal function.
1. \(\int_\RR P_y = 1\), \(P_y(t) \le \frac{1}{\pi y}\) for all \(t\), and \(P_y(t) \le \frac{1}{\pi 2^{2k}y}\) on \(A_k = \theset{2^k y \le |t| < 2^{k+1}y}\).
The substitution \(t = ys\) gives \(\int P_y = \frac1\pi\int\frac{ds}{1+s^2} = 1\). The bounds follow from \(t^2 + y^2 \ge y^2\) and, on \(A_k\), \(t^2 + y^2 \ge 2^{2k}y^2\).
2. For \(g \in L^1\), \(x \in \RR\) and \(y > 0\), \(\int |g(x-t)|P_y(t)\,dt \le \frac{10}{\pi}\sup_{0 < r \le R}\frac{1}{2r}\int_{x-r}^{x+r}|g|\) whenever \(g(x - t) = 0\) for \(|t| \ge R\) and \(R \geq y\); with \(R = \infty\) the right side is \(\frac{10}{\pi}Hg(x)\).
Let \(\alpha\) denote the supremum on the right. By step 1, \(\int_{|t| < y}|g(x-t)|P_y(t)\,dt \le \frac{1}{\pi y}\cdot 2y\,\alpha\). For each \(k\) with \(2^ky < R\), the set \(A_k\) lies in \(\theset{|t| < 2^{k+1}y}\), so \(\int_{A_k}|g(x-t)|P_y(t)\,dt \le \frac{1}{\pi 2^{2k} y}\int_{|t| < \min(2^{k+1}y, R)}|g(x-t)|\,dt \le \frac{1}{\pi 2^{2k}y}\cdot 2^{k+2}y\,\alpha = \frac{4}{\pi 2^k}\alpha\); terms with \(2^ky \ge R\) vanish. Summing, the total is at most \(\left(\frac2\pi + \frac4\pi\sum_{k \ge 0}2^{-k}\right)\alpha = \frac{10}{\pi}\alpha\).
3. \(\sup_{y>0}|u(x,y)| \le \frac{10}{\pi}Hf(x)\) for every \(x\).
\(|u(x,y)| \le \int|f(x-t)|P_y(t)\,dt\); apply step 2 with \(g = f\) and \(R = \infty\).
4. \(u(x,y) \to f(x)\) as \(y \to 0\) at every Lebesgue point \(x\) of \(f\), hence for a.e. \(x\).
Fix a Lebesgue point \(x\) and \(\eps > 0\), and choose \(\delta > 0\) with \(\frac{1}{2r}\int_{x-r}^{x+r}|f(s) - f(x)|\,ds < \eps\) for \(0 < r \le \delta\). By step 1, \(u(x,y) - f(x) = \int (f(x-t) - f(x))P_y(t)\,dt\). Let \(g(s) = (f(s) - f(x))\chi_{\theset{|s - x| < \delta}}\). For \(y \le \delta\), step 2 with \(R = \delta\) gives \(\int_{|t| < \delta}|f(x-t) - f(x)|P_y(t)\,dt \le \frac{10}{\pi}\eps\). On \(|t| \ge \delta\), \(P_y(t) \le \frac{y}{\pi\delta^2}\), so \[ \int_{|t| \ge \delta}|f(x-t) - f(x)|P_y(t)\,dt \le \frac{y}{\pi\delta^2}\|f\|_1 + |f(x)|\int_{|t|\ge\delta}P_y(t)\,dt, \] and \(\int_{|t|\ge\delta}P_y = 1 - \frac2\pi\arctan(\delta/y) \to 0\) as \(y \to 0\). So \(\limsup_{y\to0}|u(x,y) - f(x)| \le \frac{10}{\pi}\eps\) for every \(\eps > 0\). By the Lebesgue differentiation theorem a.e. \(x\) is a Lebesgue point.