$L^{1}$ Boundedness of Hilbert Transform on $\left\{f\in L^{1}(\mathbb{R}) : \int_{\mathbb{R}}f=0\right\}$

Question

It is well-known that the Hilbert transform $H(f)$ of a bounded, compactly supported function $f:\mathbb{R}\rightarrow\mathbb{C}$ belongs to $L^{1}(\mathbb{R})$ precisely when $\int f=0$. One can relax the boundedness assumption to $f\in L^{p}$, for $p>1$, or even $L\log L$. More generally, we only have a hope that the Hilbert transform $H(f)$ of an integrable function $f$, not necessarily compactly supported, is itself integrable if $\int f=0$.

Problem. I am quite confident that it is true that we have the proper containment

$$H^{1}(\mathbb{R}):=\left\{f\in L^{1}(\mathbb{R}) : H(f)\in L^{1}(\mathbb{R})\right\}\subsetneq\left\{f\in L^{1}(\mathbb{R}) : \int_{\mathbb{R}}f=0\right\}$$

however, I am struggling to produce an example proving the properness of the containment. Can someone help me with such an example?

My motivation for this question is understanding how close the real Hardy space $\mathbb{R}^{n}$ is to integrable functions which have good cancellation (see also this question). In one dimension, an equivalent characterization of the Hardy space $H^{1}(\mathbb{R})$ is $L^{1}$ functions with $L^{1}$ bounded Hilbert transforms and in fact, $\left\|f\right\|_{L^{1}}+\left\|Hf\right\|_{L^{1}}$ defines an equivalent norm. In higher dimensions, the analogue of this equivalence is $L^{1}$ boundedness of the Riesz transforms.

Answer 1

This is actually much easier than I thought. In particular, the following example is compactly supported.

First, I claim that the Hilbert transform of the function $$f:x\mapsto\dfrac{1}{x(\log|x|)^{2}}\chi_{(0,1/2)}(x)$$ is not locally integrable. Indeed, for $x<0$, \begin{align*} \left|\int_{0}^{1/2}\dfrac{1}{y(\log|y|)^{2}}\dfrac{1}{x-y}dy\right|&\geq\int_{x}^{1/2}\dfrac{1}{y(\log|y|)^{2}}\dfrac{1}{y-x}dy\\ &\geq\int_{0}^{x}\dfrac{1}{y(\log|y|)^{2}}\dfrac{1}{y+|x|}dy\\ &\geq\dfrac{1}{2|x|}\int_{0}^{x}\dfrac{1}{y(\log|y|)^{2}}dy\\ &=\dfrac{1}{2|x|\log|x|}, \end{align*} Hence, the $H(f)$ is not integrable on the interval $(-1/4,1/4)$.

If $\varphi\in C_{c}^{\infty}(\mathbb{R})$, then it's not hard to show that $H(\varphi)$ is bounded and in particular, locally integrable. Take $\varphi$ to be supported in $(0,1/2)$ and satisfy $\int\varphi=1$. Note that $\int f=(\log 2)^{-1}$ and obtain a mean zero, compactly supported, integrable function $g:=f-(\log 2)^{-1}\varphi$. So by linearity, $$\left|\int_{-1/4}^{0}H(g)(x)dx\right|\geq\left|\int_{-1/4}^{0}H(f)(x)dx\right|-(\log 2)^{-1}\|H(\varphi)\|_{L^{\infty}}=\infty$$

Answer 2

I believe that I have a soft analysis argument that nonconstructively addresses my question in all dimensions. The idea of the argument is to use $H^{1}$-$BMO$ duality together with the fact that the space of essentially bounded functions is a proper subspace of $BMO$ to show that $H^{1}(\mathbb{R}^{n})$ is a proper subspace of $L^{1}(\mathbb{R}^{n})$ functions with vanishing integral.

In the interest of definiteness, we will take the $H^{1}$-norm $\|\cdot\|_{H^{1}}$ to be the atomic norm: $$\|f\|_{H^{1}}:=\inf\left\{\sum_{k}|\lambda_{k}| : f=\sum_{k=1}^{\infty}\lambda_{k}a_{k}\right\},$$ where the infimum is taken over all decompositions $f=\sum_{k}\lambda_{k}a_{k}$ as a sum of $(1,\infty)$-atoms $a_{k}$.

Denote the subspace of $L^{1}(\mathbb{R}^{n})$ consisting of functions with mean zero by $L_{0}^{1}(\mathbb{R}^{n})$. By endowing this subspace with the $L^{1}$-norm, we obtain a Banach space. Using the basic result, $(L^{1})^{*}\cong L^{\infty}$, we obtain the analogous statement for $(L_{0}^{1})^{*}$.

Lemma 1. Let $\sim$ denote the equivalence relation on measurable functions defined by $f\sim g$ if and only if $f-g$ is almost everywhere (a.e.) equal to a constant. Then $(L_{0}^{1})^{*}\cong L^{\infty}/\sim$. More precisely, for every every continuous linear functional $L$ on $L_{0}^{1}$, there exists a unique equivalence class $[g]_{\sim}\in L^{\infty}/\sim$, such that $$L(f)=\int_{\mathbb{R}^{n}}f(x)g(x)dx,\quad\forall f\in L_{0}^{1}(\mathbb{R}^{n})$$ and $\|L\|_{*}=\|g\|_{L^{\infty}}$.

Proof. It is evident that $L^{\infty}/\sim$ embeds in $(L_{0}^{1})^{*}$. Let $L\in (L_{0}^{1})^{*}$. Fix a nonnegative $C_{c}^{\infty}$ function $\eta$ with $\int\eta=1$, and define a linear functional $L_{\eta}$ on $L^{1}$ by $$L_{\eta}(f):=L(f-(\int f)\eta),\quad f\in L^{1}(\mathbb{R}^{n})$$ I claim that $L_{\eta}\in (L^{1})^{*}$. Indeed, $$\|L_{\eta}(f)\|=\|L(f-(\int f)\eta)\|\leq\|f-(\int f)\eta\|_{L^{1}}\leq\|f\|_{L^{1}}+\|f\|_{L^{1}}\|\eta\|_{L^{1}}=2\|f\|_{L^{1}}$$ Thus, by $L^{p}$-$L^{p'}$ duality, there exists a $g\in L^{\infty}$ such that $$L_{\eta}(f)=\int_{\mathbb{R}^{n}}f(x)g(x)dx,\quad\forall f\in L^{1}(\mathbb{R}^{n})$$ If we instead choose a different $\gamma\in C_{c}^{\infty}$, nonnegative, and with $\int\gamma=1$, then $\eta-\gamma\in L_{0}^{1}$ and we have that there exists an $h\in L^{\infty}$ such that \begin{align*} \int_{\mathbb{R}^{n}}f(x)h(x)dx&=L_{\gamma}(f)=L\left((f-(\int f)\eta+(\int f)(\eta-\gamma)\right)\\ &=L_{\eta}(f)+\underbrace{L(\eta-\gamma)}_{:=c}\int f\\ &=\int_{\mathbb{R}^{n}}f(x)[g(x)+c]dx \end{align*} From Lebesgue differentiation theorem, we conclude that $f-h=c$ a.e. $\Box$

Lemma 2. Let $f\in L_{loc}^{2}(\mathbb{R}^{n})$ be supported in a cube $Q$, and suppose that $$\int_{Q}f(x)g(x)dx=0$$ for all compactly supported $g\in L_{0}^{2}(\mathbb{R}^{n})$ (square integrable zero mean functions). Then there is a constant $c\in\mathbb{C}$ such that $f=c$ a.e. in $Q$.

Proof. Let $\eta\in C_{c}^{\infty}$ be supported in $Q$, nonnegative, and satisfy $\int\eta=1$. Set $\eta_{\epsilon}:=\epsilon^{-n}\eta(\cdot/\epsilon)$. For a.e. $x\in Q^{o}$, we have that $$f(x)=\lim_{\epsilon\rightarrow 0}\int_{Q}f(y)\eta_{\epsilon}(x-y)dy$$ For such an $x$ and $\epsilon>0$ sufficiently small, we have that the function $\varphi_{\epsilon}(y):=\eta_{\epsilon}(x-y)-|Q|^{-1}\chi_{Q}(y)$ is compactly supported and in $L_{0}^{2}$. Whence, $$\int_{Q}f(y)\varphi_{\epsilon}(y)dy=0\Longrightarrow\int_{Q}f(y)\eta_{\epsilon}(x-y)dy=|Q|^{-1}\int_{Q}f(y),\quad\forall 0<\epsilon\ll 1$$ Letting $\epsilon\rightarrow 0$, we obtain that $$f(x)=|Q|^{-1}\int_{Q}f(y)dy,$$ and we see that this equality holds for a.e. $x\in Q$. $\Box$

Assume for the sake of a contradiction that $L_{0}^{1}(\mathbb{R}^{n})=H^{1}(\mathbb{R}^{n})$. Recall that we have the embedding $$(H^{1}(\mathbb{R}^{n}),\|\cdot\|_{H^{1}})\hookrightarrow (L_{0}^{1}(\mathbb{R}^{n}),\|\cdot\|_{L^{1}}),\quad \left\|f\right\|_{L^{1}}\leq\left\|f\right\|_{H^{1}}$$ By hypothesis, this embedding is in fact bijective. Whence by the Bounded Inverse Theorem, the inverse map is also bounded, showing that $\|\cdot\|_{H^{1}}$ and $\|\cdot\|_{L^{1}}$ are equivalent norms. Now recall that $BMO(\mathbb{R}^{n})$ is the dual of $H^{1}(\mathbb{R}^{n})$ in the sense that for any $L\in (H^{1})^{*}$, there exists a BMO representative function $g$ (whose equivalence class is unique), such that $$L(f)=\int_{\mathbb{R}^{n}}f(x)g(x)dx$$ for all compactly supported functions in $L_{0}^{2}(\mathbb{R}^{n})$ and $\left\|L\right\|_{*}\sim\|g\|_{BMO}$. Conversely, any BMO function extends to a bounded linear functional on $H^{1}(\mathbb{R}^{n})$ by being initially defined by integration against this subspace of $H^{1}$. Since $L$ also defines a continuous linear functional on $L_{0}^{1}$, by the first lemma, there exists an $h\in L^{\infty}$ such that $$L(f)=\int_{\mathbb{R}^{n}}f(x)h(x)dx,\quad\forall f\in L_{0}^{1}(\mathbb{R}^{n})$$ Since BMO functions are locally square integrable as a consequence of the John-Nirenberg inequality and $L^{\infty}$ functions are trivially locally square integrable, we can apply Lemma 2 to $g-h$ to conclude that $g-h=c$ is a.e. equal to a constant in a cube $Q$. By taking a sequence of increasing cubes, we obtain that $g-h=c\in\mathbb{C}$ a.e. in $\mathbb{R}^{n}$. But if we take $g(x):=\log|x|\in BMO$, we obtain a contradiction, as $g$ is clearly not essentially bounded.