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 $L^1$.
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 $g_{j}=R_{j}g_{0}\in L^{1}(\mathbb{R}^{n})$.