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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 $${\rm{sup}}_{k\ge1}|h\ast\varphi_k|(x)\le Mh(x)\in L^p(\mathbb{R}^n).$$

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  • $\begingroup$ 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
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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 $\mathbb{R}^n$.

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  • $\begingroup$ 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

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