Let $h\in L^1(\mathbb{R}^n)$. Let $\varphi\in S(\mathbb{R}^n)$, $\int_{\mathbb{R}^n}\varphi(x) dx=1$, where $S(\mathbb{R}^n)$ is the Schwartz function space and $\varphi$ is nonnegative, radial, and radially decreasing. Let $\varphi_k(x)=k^n\varphi(kx)$, $k=1,2,...$, which is a sequence of function approximations to the Dirac delta function $\delta_0$. Recall that $||h\ast \varphi_k||_1\le ||\varphi_k||_1||h||_1=||h||_1$ by Young's inequality. Then is there a function $g\in L^1(\mathbb{R}^n)$ such that $${\rm{sup}}_{k\ge1}|h\ast\varphi_k|(x)\le g(x),\ a.e. \ ?$$

Remark: (1)Using the Hardy-Littlewood maximal function we have ${\rm{sup}}_{k\ge1}|h\ast\varphi_k|(x)\le Mh(x)$, but unfortunately $Mh\notin L^1(\mathbb{R}^n)$ whenever $h\ne 0$ on a set with positive measure.

(2)By the Hardy-Littlewood maximal theorem, if $h\in L^p(\mathbb{R}^n)$, $1<p\le \infty$, then $${\rm{sup}}_{k\ge1}|h\ast\varphi_k|(x)\le Mh(x)\in L^p(\mathbb{R}^n).$$

  • $\begingroup$ What are you using this for? If you want to show that $h*\phi_k \to h$ in $L^1(\mathbb{R}^n)$, there are easier ways... $\endgroup$ – Jeff Sep 18 '16 at 2:20
  • $\begingroup$ @Jeff I am just curious about the existence of the dominated function for this convolution. It seems difficult to construct such function from $h$. But if $h\in L^p,\ 1<p\le \infty$, it is OK to choose $g=Mh$. $\endgroup$ – Right Sep 18 '16 at 2:25

Never for all functions $h$. Let $n=1$ for simplicity. Define the maximal operator $$ M_1 h = \sup_{k \geq 1} \lvert h \ast \varphi_k \rvert. $$ Define the function $h$ by $$ h(x) = \sum_{j = 0}^\infty c_j (x - j \delta)^{-1 + \varepsilon_j} \mathbb{1}_{(j\delta, (j+1)\delta]} $$ for some parameters $c_j, \delta, \varepsilon_j > 0$. Choose $\delta$ small such that $\varphi(x) > 1/2$ if $\lvert x \rvert \leq \delta$. Choose $\varepsilon_j = 2^{-j}$. Choose $c_j$ so that $$ c_j \int_{0}^{j\delta} x^{-1 + \varepsilon_j} = 2^{-j}. $$ Then $h \in L^1(\mathbb{R})$ with $\| h \|_{L^1} = \sum_{j=0}^\infty 2^{-j} = 2$.

If $x \in (j_0 \delta, (j_0 + 1)\delta)$, choose $k_0$ maximal such that $x - k_0^{-1}\delta < j_0 \delta$. Then $k_0 \geq \frac{1}{2} \delta (x - j_0 \delta)^{-1}$ by maximality. We have $$ \varphi_{k_0} \ast h(x) \geq \frac{1}{2} k_0 c_{j_0} \int_{0}^{x - j_0 \delta} y^{-1 + \varepsilon_{j_0}} = \frac{k_0 c_{j_0}}{2 \varepsilon_{j_0}}(x - j_0 \delta)^{\varepsilon_{j_0}} \geq \frac{c_{j_0} \delta}{4 \varepsilon_{j_0}} (x - j_0 \delta)^{-1 + \varepsilon_{j_0}}. $$ It follows that $$ \| M_1 h \|_{L^1} \geq \sum_{j=0}^\infty \frac{\delta}{4 \varepsilon_j} 2^{-j} = \frac{\delta}{4} \sum 1 = \infty. $$ And thus any dominating function is not in $L^1$. This example can be modified to work in $\mathbb{R}^n$.

  • $\begingroup$ Thanks! If we assume that $h\in L^1\cap L^\infty$, can we show that $M_1h\in L^1$? (Clearly, it is in $L^p, p>1$). $\endgroup$ – Right Sep 28 '16 at 2:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.