Can we extend Young's convolution inequality with $BMO$ instead of $L^\infty$ Obviously $\|f*g\|_{L^\infty}\leq\|f\|_{L^1}\|g\|_{L^\infty}$. Do we have the stronger bound $\|f*g\|_{L^\infty}\leq C\|f\|_{L^1}\|g\|_{BMO}$? Or almost as good, $\|f*g\|_{L^\infty}\leq C\|f\|_{H^1}\|g\|_{BMO}$? I think this might follow from the fact that interpolation still works when you replace $L^\infty$ with $BMO$.
Edit: It seems to me that the first statement is false for example if you take $f=1_{[0,1]}\in L^1$ and $g(x)=\log|x|\in BMO$. Then for $x>1$,
\begin{align*}
f*g(x)=x\log x-(x-1)\log(x-1)-1
\end{align*}
is not $L^\infty$.
For the second claim, I'm tempted to use the duality inequality for $H^1$ and $BMO$ to say something like
$$|\int f(t)g(x-t)dt|\leq\|f\|_{H^1}\|g\|_{BMO}$$
but I know that this is only really supposed to hold for $f\in H_0^1$.
 A: I don't believe the notation $H^1_0$ is absolutely standard. I'm going to assume that it denotes a dense subspace of $H^1$ such that $fg\in L^1$ for every $f\in H^1_0$ and $g\in BMO$; there are various subspaces with this property, but it won't matter below which one we're talking about.
The answer to your question is "yes or no, depending on how we define that convvolution".
No: As you sort of hint, $f\in H^1$ and $g\in BMO$ do not imply $fg\in L^1$, so the integral defining $f*g$ need not exist. So if we insist that the convolution is defined by that integral the the answer is clearly no; something that does't exist can't be in $L^\infty$.
Yes: On the other hand, it happens all the time that we extend the notion of convolution to situations where that integral does not converge (the convolution of a distribution with a test function, for example). If $X$ is any Banach space of functions on the line which is  invariant under translations and reflection then there is a perfectly natural notion of $f*g$ for $f\in X$ and $g\in X^*$. In the present situation, being just a tiny bit pedantic:
Of course when we say $(H^1)^*=BMO$ we mean this: If $g\in BMO$ there exists a unique $\Lambda_g\in (H^1)^*$ such that $$\Lambda_g f=\int fg\quad(f\in H^1_0).$$Although the integral need not converge, if $f\in H^1$ and $g\in BMO$ then people often write $$\int fg=\Lambda_g f,$$where strictly speaking the right side is the definition of the left side.
If we interpret the integral defining the convolution in that sense then yes, $f\in H^1$ and $g\in BMO$ imply $f*g\in L^\infty$. 
What about the inequality? Again, the answer is "it depends". There are various equivalent (standard) norms on $H^1$ and $BMO$.  If we choose our norms so that $||\Lambda_g||_{(H^1)^*}\le||g||_{BMO}$ then yes, it's clear that $$||f*g||_{L^\infty}\le||f||_{H^1}||g||_{BMO}.$$ Of course with any other choice of (equivalent) norms we get $||f*g||_{L^\infty}\le c||f||_{H^1}||g||_{BMO},$ good enough for applications.
You might also note that $f*g$ is continuous; this is clear because if $f\in H^1$ and $\tau_hf(x)=f(x-h)$ then $$||f-\tau_hf||_{H^1}\to0\quad(h\to 0).$$
So which is it, yes or no? If someone forced me to replace the above with a one-word answer I'd say "yes", simply because people do interpret the integral that way. (Or: Because if $X_0$ is a dense subspace of $X$ and $T:X_0\to Y$ is linear and bounded then people do interpret $Tf$ for $f\in X$ in the obvious way, regardless of whether the original definiton of $T$ makes sense for $f\in X$.)
(Otoh if we assume just $f\in L^1$ and $g\in BMO$ then as you suggest the answer is no, there is simply no such thing as $f*g$.)
