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I am reading the paper "Bounded Mean Oscillation and Sobolev Spaces" by Robert s. Strichartz, Indiana University Mathematics Journal , 1980, Vol. 29, pp. 539-558. In this paper he defines the space $I_{\alpha}(BMO)$ as the image of $BMO(\mathbb{R}^{n})$ under the Riesz potential $$ I_{\alpha}f=\mathcal{F}^{-1}(|\xi|^{-\alpha}\mathcal{F}(f)), $$ the Riesz potential being defined on tempered distributions modulo polynomials of degree $\leq\alpha$. For $\alpha=1$ he states that $f\in I_{1}(BMO)$ if and only if $\frac{\partial f}{\partial x_{i}}\in BMO(\mathbb{R}^{n})$ for every $i=1,\ldots,n$.

In Theorem 3.3 he proves that $f\in I_{1}(BMO)$ if and only if $f$ is of temperate growth and $$ \sup_{Q_{s}}\left( \frac{1}{|Q_{s}|}\int_{Q_{s}}\int_{|y|\leq s} \frac{|f(x+y)-2f(x)+f(x+y)|^{2}}{|x-y|^{n+2}}dydx\right) ^{1/2}<\infty. $$ In the proof he does not show that if $f\in I_{1}(BMO)$, then $f$ is of temperate growth, which I assume means that $f$ is bounded in absolute value by a polynomial. I don't see an easy way to prove this. Am I missing something trivial?

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  • $\begingroup$ This follows from the fact that for all $\alpha>0$, one can check that the chain of continuous inclusions $$BMO^{-\alpha}\subset \dot{\mathrm{B}}^{-\alpha}_{\infty,\infty}\subset\mathcal{S}'.$$ $\endgroup$
    – ToGle
    Commented Dec 8, 2023 at 11:42
  • $\begingroup$ Do you have a reference with a valid proof? In "Tempered homogeneous function spaces," Triebel defines all these spaces as spaces of tempered distributions. In "Theory of functions spaces", he takes all these spaces in $Z'$, which is defined as $\mathcal{S}'$ modulo polynomial. All I know is that $f$ is locally integrable, so it is a distribution but I don't see why it is a tempered distribution. $\endgroup$
    – Gio67
    Commented Dec 9, 2023 at 13:55
  • $\begingroup$ or the sum of a tempered distribution and a polynomial. This is what I would need to prove. I thought about the homogeneous Zygmund class, but all the inequalities (to prove that they are almost Lipschitz) I know only work for bounded functions. $\endgroup$
    – Gio67
    Commented Dec 9, 2023 at 14:27
  • $\begingroup$ As a complete normed vector space $BMO$ itself is no longer a space of distribution, and it cannot be. There is this paper by Bourdaud about it. $\endgroup$
    – ToGle
    Commented Dec 9, 2023 at 14:51
  • $\begingroup$ Proposition 1.31, from Dimitri Cobb's PhD Thesis, p.69. The same argument would apply for $BMO^{-\alpha}$ for any $\alpha>0$ ! $\endgroup$
    – ToGle
    Commented Dec 9, 2023 at 15:12

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