Convolution by $\log$ maps $\mathrm{L}^1$ into $\mathrm{BMO}$

Question

It is stated in Stein's Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals, Ch. IV, 6.3(i) that $$ I_{n} f:=f\star\log|\cdot|\in\mathrm{BMO}(\mathbb{R}^n)\qquad\text{if }f\in\mathrm{L}^1(\mathbb{R}^n). $$ More precisely, $$ (I_{n}f)(x)=\int_{\mathbb{R}^n}f(y)\log|x-y|\mathrm{d}y\qquad\text{for }x\in\mathbb{R}^n, $$ and $f\in\mathrm{L}^1(\mathbb{R}^n)$ implies $I_{n}f\in\mathrm{BMO}(\mathbb{R}^n)$.

My questions revolve around obtaining a uniform bound for this inclusion:

1. How to prove it? (hopefully this settles 2. and, hence, 5.)
2. Is it the case that there exists a constant $C>0$ such that $$ \|I_{n}f\|_{\mathrm{BMO}(\mathbb{R}^n)}\leq C\|f\|_{\mathrm{L}^1(\mathbb{R}^n)}? $$
3. More generally, is it the case that the following generalization of Young's Convolution Inequality holds $$ \|f\star g\|_{\mathrm{BMO}(\mathbb{R}^n)}\leq \|f\|_{\mathrm{L}^1(\mathbb{R}^n)}\|g\|_{\mathrm{BMO}(\mathbb{R}^n)}? $$ For the classical statement, $\mathrm{BMO}$ is replaced by $\mathrm{L}^\infty$. Of course, this implies 2., but it is very interesting in itself!
4. Is a weaker estimate available? Perhaps 5.
5. Finally, for my purposes it suffices that we have a local estimate: $$ \|I_{n}f\|_{\mathrm{L}^1(B)}\leq C(B)\|f\|_{\mathrm{L}^1(\mathbb{R}^n)}, $$ for $f\in\mathrm{C}^\infty_c(\mathbb{R}^n)$. Here $B\subset\mathbb{R}^n$ is an open ball and $C(B)>0$ is, of course, allowed to depend on $B$.

Many thanks for your help!

Answer 1

Question 5 is not that bad. I'll provide a sketch: One takes the Fourier transform to get $$ \widehat{I_nf}(\xi)=|\xi|^{-n}\hat f(\xi)=|\xi|^{s-n}|\xi|^{s}\hat f(\xi) $$ for some fixed $0<s<n$. Actually, make $s=1$. The FT of $\log$ is computed in Samko, Hypersingular integrals and their applications, p. 44. It follows that $$I_nf=I_{1}I_{n-1}f.$$One then applies boundedness of Riesz potentials on domains (as in Lemma 7.12, Gilbarg-Trudinger, Elliptic partial differential equations of second order) twice to get $$ \begin{align} \|I_nf\|_{\mathrm{L}^q(B)}&=\|I_{1}I_{n-1}f\|_{\mathrm{L}^q(B)}\\ &\leq C(B)\|I_{n-1}f\|_{\mathrm{L}^p(B)}\\ &\leq C(B)\|f\|_{\mathrm{L}^1}, \end{align} $$ where $1\leq q < np/(n-p)$ and $1<p<n$. Sending $p\uparrow n$ allows to choose any $1\leq q<\infty$.

This, however, is just a fix for my problem and an indication that 2 is not necessarily wrong. The question of a limiting global embedding is still open for me ;)

Answer 2

I am sorry, but I think the estimate 5. you wrote is wrong. For a general $f\in L^1$, the function $I_nf$ is only defined up to a constant due to the fact that the logarithm is not bounded, so an estimate like 5. cannot hold, not even for test functions. Of course $I_n f$ is locally integrable, but the bound does not hold.

The problem with the proof in your answer is the first estimate. Call $g:=I_{n-1}f$. You cannot estimate the $L^q$ norm of $I_1 g$ on a ball $B$ with the $L^p$ norm of $g$ on the same ball $B$, since $I_1$ is a nonlocal operator, so you have to estimate with a norm that takes into account the values of $g$ on the whole $\mathbb R^n$. The boundedness of Riez potentials on domains cannot be applied here.

Estimate 2. has to be true, since any reasonable operator between Banach spaces (and defined for every vector of the domain) is bounded. I don't know id estimate 3. is true, it looks reasonable but I can't tell for sure.

Answer 3

I think 3. is true. We could use a duality argument since $BMO$ is the dual of $H ^1$. Since

$\int g h \;dx \le \|g\| _{BMO} \|h\| _{H ^1}$

and both norms are translation invariant, we have

$\| g * h \| _{L ^\infty} \le \|g\| _{BMO} \|h\| _{H ^1}$.

As a consequence, for any $h \in H ^1$, it holds that

$(f * g, h) = (f, g * h) \le \|f\| _{L ^1} \|g * h\| _{L ^\infty} \le \|f\| _{L ^1} \|g\| _{BMO} \|h\| _{H ^1}$.

This shows that $\|f * g\| _{BMO} \le \|f\| _{L ^1} \|g\| _{BMO}$.