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Let $B=\bigoplus_{j=0}^nL^1(\mathbb R^n)$ a Banach space with norm $\|(f_0,\ldots,f_n)\|=\|f_0\|_{L^1}+\cdots+\|f_n\|_{L^1}$. Define $$S=\{(f_0,f_1,\ldots,f_n):f_j=R_jf_0,\quad j=1,2,\ldots,n\}\subset B$$ where $R_j$ is the Riesz transform. Show that this is closed.

The claim by the author is that this is obvious so I'm quite embarrassed to be asking. It only seems obvious to me if you know that $R_j$ is continuous from the subspace $\{f\in L^1:R_jf\in L^1\}$ to $L^1$.

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  • $\begingroup$ Is it because $L^2\cap H^1$ is dense in $H^1$? $\endgroup$ – Funktorality May 2 '16 at 22:59
  • $\begingroup$ By $H^1$ I mean the Hardy space not Sobolev. $\endgroup$ – Funktorality May 3 '16 at 0:37
  • $\begingroup$ Given that this is an old (i.e. classical) characterization of the Hardy space $H^{1}(\mathbb{R}^{n})$, I take it this is from Stein's Singular Integrals? $\endgroup$ – Matt Rosenzweig May 7 '16 at 15:07
  • $\begingroup$ It's from Fefferman and Stein's paper on duality which uses Singular Integrals as a starting point. $\endgroup$ – Funktorality May 10 '16 at 18:41
  • $\begingroup$ If you're interested in Hardy spaces, there is a book called "Hardy Spaces on the Euclidean Space" by A. Uchiyama that you may find of interest. $\endgroup$ – Matt Rosenzweig May 11 '16 at 0:34
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Let $F_{k}=(f_{0,n},\ldots,f_{k,n})$ be a sequence in $S$, such that $F_{k}\rightarrow G$ in $B$. I assume you are familiar with the Fourier transform $$\widehat{\varphi}(\xi):=\int_{\mathbb{R}^{n}}\varphi(x)e^{-2\pi ix\cdot\xi}dx,\quad\xi\in\mathbb{R}^{n}$$ and the facts that the Fourier transform is bounded $L^{1}\rightarrow C_{0}$ and the Riesz transform $R_{j}$ has Fourier multiplier $$\widehat{R_{j}\varphi}(\xi)=-i\frac{\xi_{j}}{|\xi|}\widehat{\varphi}(\xi)$$ Taking the vector-valued Fourier transform (i.e. Fourier transform of each component function), we have that $\widehat{F}_{k}\rightarrow\widehat{G}$ in $C_{0}(\mathbb{R}^{n};\mathbb{C}^{n+1})$. In particular, $$\widehat{f_{0,n}}\rightarrow\widehat{g_{0}}\Rightarrow \widehat{f_{j,n}}=-i\frac{\xi_{j}}{|\xi|}\widehat{f_{0,n}}\rightarrow-i\frac{\xi_{j}}{|\xi|}\widehat{g_{0}}=\widehat{g_{j}}$$ By Fourier inversion (for tempered distributions), we have that $g_{j}=R_{j}g_{0}\in L^{1}(\mathbb{R}^{n})$.

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