I'm trying to prove that, for any function $f \in L^1(\mathbb{R}^n)$, we have that $$ \|f\|_1 \leq C \|Mf\|_{1,w} $$ for some constant $C >0$, where $\|\cdotp\|_{1,w}$ denotes the weak 1-norm, $$ \|g\|_{1,w} = \sup_{t > 0} \ t \ |\lbrace x \in \mathbb{R}^n : |g(x)|> t \rbrace | $$ and $|A|$ is the measure of a set $A$.
So far, my argument goes as follows:
If $Q_r$ denotes the cube of center zero and radius $r$, we can clearly write $$ \|f\|_1 = \sup_{r > 0} \int_{Q_r} |f| $$ Notice that, if $f \neq 0$, then $\int_{Q_r}|f| > 0$ for $r$ large enough. Also, if $x \in Q_r$, then
\begin{equation}\label{eq} \frac{1}{|Q_r|} \int_{Q_r} |f| \leq \sup_{Q \ni x} \frac{1}{|Q|} \int_Q |f| = Mf(x) \end{equation}
Now, set $$ t = \frac{1}{\beta} \frac{1}{|Q_r|} \int_{Q_r}|f| $$ where $\beta$ is any real number greater than 1, so that we have $$ t < \frac{1}{|Q_r|} \int_{Q_r}|f| < \beta^2 t $$ With this, we have that for any $x \in Q_r$, $$ t < \frac{1}{|Q_r|} \int_{Q_r}|f| \leq Mf(x) $$ so $Q_r \subset \lbrace x \in \mathbb{R}^n : Mf(x) > t \rbrace$. Therefore, $$ \int_{Q_r}|f| = |Q_r| \frac{1}{|Q_r|} \int_{Q_r}|f| < |\lbrace x \in \mathbb{R}^n : Mf(x) > t \rbrace| \beta^2 t \leq \beta^2 \sup_{t > 0} \ t \ |\lbrace x \in \mathbb{R}^n : Mf(x) > t \rbrace| = \beta^2 ||Mf||_{1,w} $$ and since $Q_r$ is any cube where $\int_{Q_r} |f| > 0$, then this inequality must also hold for the supremum over all those cubes. That is, $$ \|f\|_1 \leq \beta^2 \|Mf\|_{1,w} $$
I feel like there must be something wrong in here, because of the freedom I have for choosing the value of $\beta$, but I don't see where is my mistake. Can anybody tell me what I'm missing?
Thank you very much!
There's no reason to be concerned over that $\beta$. Look at it this way:
Suppose for a second that the weak-type "norm" was defined in terms of $Mf\ge t$ instead of $Mf>t$. Then we could just say this:
Let $t=\frac1{|Q|}\int_Q|f|$. Then $Mf\ge t$ on $Q$, so $$\int_Q|f|=t|Q|\le t|\{Mf\ge t\}|,$$and we're done.
You need the $\beta$ just to jiggle this a bit to get $Mf>t$ instead.