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Let $f: \mathbb R \rightarrow \mathbb R$ be a bounded function that is smooth and also in $L^1(\mathbb R) \cap L^2(\mathbb R)$. I want to prove that the Cauchy transform of this function $Kf$ is in $H^{\infty}$ of the upper half plane and that $$ \lim_{y\rightarrow o} \|Kf(x+iy)-f(x)\|_{\infty} =0 $$ First of all, is this true? and if yes, can you give me some clues?

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  • $\begingroup$ Under the hypotheses of the question, wouldn't $Kf$ be a bounded analytic function in the upper half plane $\mathbb{R}_{+}^{2}$ which is continuous and bounded on the closed upper half plane $\overline{\mathbb{R}_{+}^{2}}$? Furthermore, $Kf$ has real boundary values, so by Schwarz reflection, $Kf$ extends to a bounded entire function. By Liouville's theorem $Kf$ must be constant, and since $f\in L^{1}$, $Kf=f=0$. Am I missing something? $\endgroup$ Commented Jun 6, 2015 at 20:16
  • $\begingroup$ Thanks, Matt. Your reasoning seems correct. I guess this proves that the Hilbert transform cannot take not take $L^\infty(\mathbb R)$ to $L^{\infty}(\mathbb R)$, except at the constant functions? $\endgroup$
    – RR-
    Commented Jun 7, 2015 at 22:04
  • $\begingroup$ Well, it is known that $L^{\infty}(\mathbb{R})$ doesn't map to $L^{\infty}(\mathbb{R})$ but rather $BMO(\mathbb{R})$ (the space of functions of bounded mean oscillation). Consider the characteristic function of an interval $[a,b]$. Regarding which functions the Hilbert transform $H$ maps to $L^{\infty}(\mathbb{R})$, I think $H(f)$, for any Schwartz function $f\in\mathcal{S}(\mathbb{R})$ will be bounded. See this question for more details. $\endgroup$ Commented Jun 8, 2015 at 0:01
  • $\begingroup$ Unless I am confused about the definition of the operator $K$ (I take it to be the Cauchy transform, not the Hilbert transform), I don't think you're asking the right question since, since $Kf(x+iy)\rightarrow f+iH(f)$ in $L^{2}$-norm, as $y\rightarrow 0^{+}$. Since $H$ is an isometry on $L^{2}(\mathbb{R})$, $H(f)\neq 0$ if $f\neq 0$. Whether $\left\|Kf(x+iy)-f(x)-iH(f)(x)\right\|_{\infty}\rightarrow 0$ as $y\rightarrow 0^{+}$, I do not know. $\endgroup$ Commented Jun 8, 2015 at 0:13
  • $\begingroup$ Well, functions in $H^p(\mathbb C_+)$ are identified with their boundary values. And $H^{p} = \mathcal{N^+} \cap L^p(\mathbb R)$, where $\mathcal N^+$ is the Nevanlinna-Smirnov class in $\mathbb C_+$. In our case, $Kf$ has boundary values $f+iHf$ (pointwise a.e.). Thus, if $Hf$ was also bounded, then $Kf$ would have to be bounded. I feel I am not doing something right, but don't know what. $\endgroup$
    – RR-
    Commented Jun 9, 2015 at 14:49

1 Answer 1

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After several exchanges with the author in the comments section and re-reading the original question, it seems that the author is really asking if the Hilbert transform $Hf$ of a nontrivial bounded function $f\in L^{\infty}(\mathbb{R})$ can also be bounded. The answer is yes. One can compute that the Hilbert transform of the functions $f(x)=1/(1+x^{2})$ is the function $g(x)=-x/(1+x^{2})$. Although the Hilbert transform does not map $L^{\infty}(\mathbb{R})$ to $L^{\infty}(\mathbb{R})$ (take the characteristic function of a compact interval for a counterexample), there is a class of functions for which $Hf\in L^{\infty}(\mathbb{R})$.

If $f\in\mathcal{S}(\mathbb{R})$, then $Hf\in C^{\infty}(\mathbb{R})$ and using the oddness of $1/y$ and the mean value theorem,

\begin{align*} \left|Hf(x)\right|=\lim_{\epsilon\rightarrow 0^{+}}\dfrac{1}{\pi}\left|\int_{\left|y\right|\geq\epsilon}\dfrac{f(x-y)}{y}\mathrm{d}y\right|&=\lim_{\epsilon\rightarrow 0^{+}}\dfrac{1}{\pi}\left|\int_{1\geq\left|y\right|\geq\epsilon}\dfrac{f(x-y)-f(x)}{y}\mathrm{d}y+\int_{\left|y\right|\geq 1}\dfrac{f(x-y)}{y}\mathrm{d}y\right|\\ &\leq\lim_{\epsilon\rightarrow 0^{+}}\dfrac{1}{\pi}\int_{1\geq\left|y\right|\geq\epsilon}\dfrac{\left|y\right|\left\|f^{'}\right\|_{\infty}}{\left|y\right|}\mathrm{d}y+\dfrac{\left\|f\right\|_{1}}{\pi}\\ &=\dfrac{2}{\pi}\left\|f'\right\|_{\infty}+\dfrac{1}{\pi}\left\|f\right\|_{1}, \end{align*} for all $x\in\mathbb{R}$. It's not necessary that $f$ be Schwartz; I think $f$ integrable and $f\in C^{1}(\mathbb{R})$ with bounded derivative would suffice.

We return now to your original question as written. Let $P_{y}$, for $y>0$, denote the Poisson kernel \begin{align*} P_{y}(x)=c_{n}\dfrac{y}{\left(\left|x\right|^{2}+y^{2}\right)^{(n+1)/2}},\qquad (x,y)\in\mathbb{R}_{+}^{n+1} \end{align*}

We make use of the following theorems, proofs of which can be found in [Stein, E.M., and G.L. Weiss. Introduction to Fourier analysis on Euclidean spaces. Vol. 1. Princeton university press, 1971.].

Theorem 1. (Theorem 2.1) If $f\in L^{p}(\mathbb{R}^{n})$, $1\leq p\leq\infty$, and $u(x,y)=f\ast P_{y}(x)$ is the Poisson integral of $f$, then \begin{align*} \sup_{y>0}\left\|u(\cdot,y)\right\|_{L^{p}}\leq\left\|f\right\|_{L^{p}} \end{align*} and \begin{align*} \lim_{y\rightarrow 0^{+}}\left\|u(\cdot,y)-f\right\|_{L^{p}}=0 \end{align*} If $f$ is continuous and bounded, then $u(\cdot,y)$ converges to $f$ uniformly on compact subsets of $\mathbb{R}^{n}$.

and

Theorem 2. (Theorem 2.5) If $u$ is harmonic on $\mathbb{R}_{+}^{n}$ and there exists a constant $c>0$ and $p$, $1\leq p\leq\infty$, such that \begin{align*} \sup_{y>0}\left\|u(\cdot,y)\right\|_{L^{p}}\leq c<\infty, \end{align*} then $u(x,y)$ is the Poisson integral of a function $f\in L^{p}(\mathbb{R}$.

Let $Q_{y}$, for $y>0$, denote the conjugate Poisson kernel (on $\mathbb{R}_{+}^{2}$) \begin{align*} Q_{y}(x)=\dfrac{1}{\pi}\dfrac{x}{x^{2}+y^{2}},\qquad (x,y)\in\mathbb{R}_{+}^{2} \end{align*} The last result we will need is a consequence of the $L^{p}$ convergence of the truncated Hilbert transforms. A proof can be found in [Grafakos, L. Classical fourier analysis. Vol. 2. New York: Springer, 2008].

Theorem 3. If $f\in L^{p}(\mathbb{R})$, where $1<p<\infty$, then \begin{align*} \lim_{\epsilon\rightarrow 0^{+}}\left\|f\ast Q_{\epsilon}-Hf\right\|_{L^{p}}=0, \quad \lim_{\epsilon\rightarrow 0^{+}}f\ast Q_{\epsilon}(x)=Hf(x) \ \mathrm{a.e.} \end{align*}

If $f\in L^{p}(\mathbb{R})$, for $1<p<\infty$, then $Hf\in L^{p}(\mathbb{R})$ and so $f\ast Q_{y}(x)$ and $Hf\ast P_{y}(x)$ are harmonic functions on $\mathbb{R}_{+}^{2}$ with boundary value $Hf$. I claim that $f\ast Q_{y}=Hf\ast P_{y}$. Indeed, $\left\|f\ast Q_{y}-f\right\|_{L^{p}}\rightarrow 0$, as $y\rightarrow 0^{+}$, implies that $\sup_{y>0}\left\|f\ast Q_{y}\right\|_{L^{p}}<\infty$. By Theorem 2, $f\ast Q_{y}=g\ast P_{y}$ for some $g\in L^{p}(\mathbb{R})$. Letting $y\rightarrow 0^{+}$, we see that $Hf=g$ a.e. If $f$ above is such that $Hf$ is bounded, then Combining this result together with the Poisson integral of $f$, we have proven the following theorem.

Theorem 4. If $f\in L^{p}(\mathbb{R})$, for $1<p<\infty$, then the Cauchy transform $Kf(z=x+iy)=f\ast P_{y}(x)+iHf\ast P_{y}(x)\in H^{p}(\mathbb{R}_{+}^{2})$, and \begin{align*} \lim_{y\rightarrow0^{+}}\left\|Kf(\cdot+iy)-f-iHf\right\|_{L^{p}}=0, \quad\lim_{y\rightarrow 0^{+}}Kf(x+iy)=f(x)+iHf(x)\ \mathrm{a.e.} \end{align*} If $f\in L^{p}(\mathbb{R})\cap L^{\infty}(\mathbb{R})$ and $Hf\in L^{\infty}(\mathbb{R})$, then $f\in H^{\infty}(\mathbb{R}_{+}^{2})$. Moreover, if $Hf$ is also continuous, then for any compact subset $K\subset\mathbb{R}$, \begin{align*} \lim_{y\rightarrow 0^{+}}\sup_{x\in K}\left|Kf(x+iy)-f(x)-iHf(x)\right|=0 \end{align*}

Restrict attention to the case $f\in L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$ as in your question. Since the Hilbert transform is an isometry, in fact an invertible operator, on $L^{2}(\mathbb{R})$, we have that $Hf=0\Leftrightarrow f=0$. So the answer to your question as written is no, unless $f=0$.

At the moment, I don't know if your hypotheses on $f$ are strong enough to guarantee that $Hf\in L^{\infty}(\mathbb{R})$, let alone that the Cauchy transform converges to its boundary value uniformly on $\mathbb{R}$. My inclination is no, without some hypothesis on the growth of $f'$.

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  • $\begingroup$ Thank you, Matt. That is a really comprehensive reply. $\endgroup$
    – RR-
    Commented Jun 11, 2015 at 15:46

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