For every $f \in L^1(\mathbb{R})$, do we have $\sup_{n \in \mathbb{N}}|T_nf(x)| < \infty$ for a.e. $x$?

Question

Let $K: \mathbb{R} \to \mathbb{R}$ be given by $K(x) := \dfrac {\sqrt {|x|}} {1 + x^2}$ and for $n \in \mathbb{N}$, $K_n(x) := nK(nx)$. For $f \in L^1(\mathbb{R})$ define $T_n (f)(x) := \int \limits _\mathbb{R} K_n(x - y)\,f(y)\,dy$. Do we have that for every $f \in L^1(\mathbb{R})$, $\sup \limits _{n \in \mathbb{N}}|T_nf(x)| < \infty$ for a.e. $x$?

Answer 1

Yes. We just need to use the fact that the maximal function $$ (Mf)(x):=\sup_{r>0}\frac{\int_{B(x,r)}|f(y)|\,dy}{|B(x,r)|} $$ is finite for a.e. $x$ (for any $f\in L^1$). This is implied from the classical Hardy-Littlewood maximal inequality.

What you ask follows from this claim:

Claim: If $f\in L^1$, then for any $\Phi:\mathbb{R}\to\mathbb{R}$ satisfying

• $\Phi\in C_0(\mathbb{R})$, i.e. it is continuous and $\lim_{|x|\to\infty}\Phi(x)=0$
• $\Phi\ge 0$
• $\Phi$ is even, i.e. $\Phi(x)=\Phi(-x)$
• $\Phi$ is nonincreasing on $[0,+\infty)$

and any $x\in\mathbb{R}$ we have $|\Phi\ast f(x)|\le (\int\Phi(y)\,dy)(Mf)(x)$.

Indeed, if you compute the derivative of $K$ you will notice that $K$ is increasing on $\left[0,\frac{1}{\sqrt{3}}\right]$, while it decreases on $\left[\frac{1}{\sqrt{3}},+\infty\right)$. So it can be written as $K=\Phi-\Psi$, where $\Phi$ and $\Psi$ satisfy the hypotheses of the claim and are summable. Defining $\Phi_n$ and $\Psi_n$ in the same way as $K_n$, observe that $\int \Phi_n=\int \Phi<\infty$ is independent of $n$, and analogously for $\Psi_n$. So $$ |T_n(f)(x)|=|\Phi_n\ast f(x)-\Psi_n\ast f(x)|\le|\Phi_n\ast f(x)|+|\Psi_n\ast f(x)|\le ({\textstyle\int}\Phi+{\textstyle\int}\Psi)(Mf)(x). $$

Proof of the claim: The basic idea is that $\Phi$ is a superposition of characteristic functions of balls centered at $0$ (and the claim is clear if we replace $\Phi$ with one such characteristic function).
Now the details: choose some integer $n>0$ and real numbers $0<\alpha_1<\cdots<\alpha_n$, $\beta_1\ge\cdots\beta_n\ge 0$ and define a corresponding step function $S:\mathbb{R}\to\mathbb{R}$ as follows: $$ S(y)=\begin{cases}\beta_1 & y\in(-\alpha_1,\alpha_1) \\ \beta_2 & y\in(-\alpha_2,\alpha_2)\setminus(-\alpha_1,\alpha_1) \\ \cdots \\ \beta_n & y\in (-\alpha_n,\alpha_n)\setminus(-\alpha_{n-1},\alpha_{n-1}) \\ 0 & y\in\mathbb{R}\setminus(-\alpha_n,\alpha_n)\end{cases} $$ (drawing a picture could help). We have the inequality $$ |S\ast f(x)|\le ({\textstyle\int} S)(Mf)(x) $$ because we can write $S=\sum_{i=1}^n(\beta_i-\beta_{i+1})\mathbf{1}_{B(0,\alpha_i)}$, where $\beta_{n+1}:=0$, and (remember that $\beta_i-\beta_{i+1}\ge 0$) $$ |S\ast f(x)|\le S\ast |f|(x)=\sum_{i=1}^n(\beta_i-\beta_{i+1})\mathbf{1}_{B(0,\alpha_i)}\ast|f|(x) \\ =\sum_{i=1}^n(\beta_i-\beta_{i+1})\int_{B(x,\alpha_i)}|f| \le\sum_{i=1}^n(\beta_i-\beta_{i+1})|B(x,\alpha_i)|(Mf)(x) \\ =(Mf)(x)\sum_{i=1}^n(\beta_i-\beta_{i+1})\int\mathbf{1}_{B(0,\alpha_i)} =(Mf)(x)\int S. $$ Now for every $\epsilon>0$ we can approximate $\Phi$ with a step function $S$ of the above form, in such a way that $\Phi-\epsilon\le S\le\Phi$. So finally $$ |\Phi\ast f(x)|\le\Phi\ast|f|(x)\le (S+\epsilon)\ast|f|(x)=S\ast|f|(x)+\epsilon\|f\|_1 \\ \le ({\textstyle\int}S)(Mf)(x)+\epsilon\|f\|_1\le({\textstyle\int}\Phi)(Mf)(x)+\epsilon\|f\|_1 $$ and letting $\epsilon\to 0$ we get the thesis. $\blacksquare$