Sublinearity of Hardy-Littlewood Maximal Function on Sobolev Spaces

Question

Define the (centered) Hardy-Littlewood maximal function by $$\mathcal{M}f(x)=\sup_{r>0}\dfrac{1}{m(B(x,r))}\int_{B(x,r)}\left|f(y)\right|dy,\ f\in L_{\mathrm{loc}}^{1}(\mathbb{R}^{d})$$

We say that an operator $T:X\rightarrow Y$ between function spaces is sublinear if $$\left|T(\lambda f)\right|=\left|\lambda\right|\left|T(f)\right| \text{ and }\left|T(f+g)\right|\leq \left|T(f)\right|+\left|T(g)\right|$$ for all $f,g\in X$ and $\lambda\in\mathbb{C}$. It is easy to see that if $T(f)\geq 0$ for all $f\in X$, then $$\left|T(f)-T(g)\right|\leq T(f-g)$$

It follows from the triangle inequality that $\mathcal{M}$ satisfies the above condition and therefore is sublinear on $L^{1}(\mathbb{R}^{d})$. Moreover, the Hardy-Littlewood-Wiener theorem tells us that $\mathcal{M}:L^{p}(\mathbb{R}^{d})\rightarrow L^{p}(\mathbb{R}^{d})$ is a bounded, sublinear operator for $1<p<\infty$. $\mathcal{M}$ is in fact continuous on $L^{p}(\mathbb{R}^{d})$. Indeed, it follows from $\left|\mathcal{M}f-\mathcal{M}g\right|\leq\mathcal{M}(f-g)$ that

$$\left\|\mathcal{M}f-\mathcal{M}g\right\|_{L^{p}}^{p}\leq\left\|\mathcal{M}(f-g)\right\|_{L^{p}}^{p}\leq C\left\|f-g\right\|_{L^{p}}^{p}$$

It is a well-known theorem of Kinnunen that $\mathcal{M}:W^{1,p}(\mathbb{R}^{d})\rightarrow W^{1,p}(\mathbb{R}^{d})$ is a bounded operator. In the paper [H. Luiro, "Continuity of the Maximal Operator in Sobolev Spaces"], the author claims that $\mathcal{M}$ is not sublinear. Is this a misprint, or am I missing something nontrivial? I understand why the argument for the $L^{p}$ case doesn't carry over for Sobolev space, since we also have to deal with the weak gradient, and therefore continuity is nontrivial. However, I don't see why restricting the domain affects sublinearity.

Answer 1

I think the author (somewhat unconventionally) uses the word "sublinearity" for the property $$\left\|\mathcal{M}f-\mathcal{M}g\right\| \leq\left\|\mathcal{M}(f-g)\right\| \tag{1}$$ which indeed depends on the norm used. As you noted, $(1)$ holds for $L^p$ norm. I suppose there is an example out there to show that $(1)$ fails for $W^{1,p}$ norm, but I do not have one.

Answer 2

After reading the Luiro's paper again more carefully, I realize that he answers why $\mathcal{M}$ is not sublinear in the sense that $$\left\|\mathcal{M}f-\mathcal{M}g\right\|_{W^{1,p}}\leq\left\|\mathcal{M}(f-g)\right\|_{W^{1,p}},\quad\forall f,g\in W^{1,p}(\mathbb{R}^{n})$$ in the concluding remark of the paper. The failure of sublinearity is a consequence of a more general result, which is that $\mathcal{M}$ has no modulus of continuity $\omega$ on $W^{1,p}(\mathbb{R}^{n})$. I.e., there does not exist a function $\omega:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ such that

$$\left\|\mathcal{M}f-\mathcal{M}g\right\|_{W^{1,p}}\leq\omega(\left\|f-g\right\|),\qquad\forall f,g\in W^{1,p}(\mathbb{R}^{n})$$

This follows from two properties: $\mathcal{M}$ is not Lipschitz continuous $W^{1,p}(\mathbb{R}^{n})\rightarrow W^{1,p}(\mathbb{R}^{n})$ and the scale invariance of $\mathcal{M}$.

If $\mathcal{M}$ were sublinear, then $\omega(t)=t$ would be a modulus of continuity for $\mathcal{M}$.

Theorem 1. [Korry 2002] There exists a $C_{c}^{\infty}(\mathbb{R}^{n})$ function $f$ such that $\mathcal{M}f$ does not belong to $F_{p,q}^{s}(\mathbb{R}^{n})$ for every $s>1+1/p$. In particular, $\mathcal{M}f\notin W^{2,p}(\mathbb{R}^{n})$.

The Triebel-Lizorkin space $F_{p,q}^{s}(\mathbb{R}^{n})$ is equivalent to the fractional Sobolev space $W^{s,p}$ when $q=2$. By Sobolev embedding, $f\notin W^{2,p}(\mathbb{R}^{n})$.

Lemma 2. [Gilbarg-Trudinger] Let $f\in L^{p}(\mathbb{R}^{n})$, $1<p<\infty$, and suppose that the difference quotient $\Delta_{i}^{h}f$ are uniformly bounded in $L^{p}$ norm by some constant $C>0$. Then $f\in W^{1,p}(\mathbb{R}^{n})$ and $\left\|\partial_{i}f\right\|_{L^{p}}\leq C$ for all $i$. Conversely, if $f\in W^{1,p}(\mathbb{R}^{n})$, then $\left\|\Delta_{i}^{h}f\right\|_{L^{p}}\leq\left\|\partial_{i}f\right\|_{L^{p}}$ for all $i$.

We now have the ingredients to prove the main result.

Proposition 3. $\mathcal{M}$ is not Lipschitz continuous.

Proof. Assume the contrary. For nonzero $h\in\mathbb{R}$, let $\Delta_{i}^{h}$ denote the difference quotient in the $i^{th}$ coordinate. Observe that by the translation invariance of $\mathcal{M}$, $$\left\|\Delta_{i}^{h}(\mathcal{M}f)\right\|_{1,p}=\left\|(\mathcal{M}(h^{-1}\tau^{-h}(f))-\mathcal{M}(h^{-1}f)\right\|_{1,p}\leq\left\|h^{-1}\tau^{-h}(f)-h^{-1}f\right\|_{1,p}=\left\|\Delta_{i}^{h}f\right\|_{1,p}$$ If $f\in W^{2,p}(\mathbb{R}^{n})$, then by Lemma 2, there exists $C>0$ such that $\left\|\Delta_{i}^{h}f\right\|_{1,p}\leq C$ for all $h\neq 0$ and $i$. Whence, $\mathcal{M}f\in W^{2,p}(\mathbb{R}^{n})$, which is false by Korry's result. $\Box$

Suppose that $\mathcal{M}$ has a modulus of continuity $\omega$ in the sense defined above. Then by scale invariance, we have that for all $f\neq g\in W^{1,p}$, \begin{align*} \dfrac{\left\|\mathcal{M}f-\mathcal{M}g\right\|_{1,p}}{\left\|f-g\right\|_{1,p}}&=\left\|\mathcal{M}(\left\|f-g\right\|_{1,p}^{-1}f)-\mathcal{M}(\left\|f-g\right\|^{-1}g)\right\|_{1,p}\\ &\leq\omega\left(\left\|\left\|f-g\right\|_{1,p}^{-1}f-\left\|f-g\right\|_{1,p}^{-1}g\right\|_{1,p}\right)\\ &=\omega(1), \end{align*} which is a contradiction.