-1
$\begingroup$

I have an elementary question on the definition of Hardy-Littlewood maximal function (which is a question on the definition of $\sup$ really).

Do we have that

$$Mf(x) = \sup_{r>0}\frac{1}{m(B_r(x))}\int_{B_r(x)} f(y) dy $$

is equal to $$Mf(x) = \sup_{R>0}\sup_{r \in (0,R)}\frac{1}{m(B_r(x))}\int_{B_r(x)} f(y) dy ? $$

$\endgroup$
  • $\begingroup$ @Dzoooks could you provide a formal proof? $\endgroup$ – Sambo Jun 29 '18 at 2:12
  • $\begingroup$ @Dzoooks This is intuitively obvious, but the proof is nontrivial. See my answer below. $\endgroup$ – Sambo Jun 29 '18 at 13:38
  • $\begingroup$ If you can make a cleaner proof than I did, then by all means leave an answer. But I don't find it obvious. You say LHS $\geq$ RHS is obvious, but individually, we don't immediately have $\sup \{ f(r) : 0 < r < R \} \geq f(R)$, in particular if $f$ is discontinuous. $\endgroup$ – Sambo Jun 29 '18 at 22:06
0
$\begingroup$

Let $f:(0,\infty) \rightarrow \mathbb{R}$. We want to show that the following two quantities are equal:

\begin{align} A &= \sup \{ f(r) : 0 < r < \infty \} \\ B &= \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} \end{align}

First remark that by definition, $A$ is an upper bound of $\{ f(r) : 0 < r < \infty \}$. Since this set includes $ \{ f(r) : 0 < r < R \}$ as a subset, $A$ is also an upper bound for this set for any $R$. But since the supremum of a set is less or equal than any upper bound, we have

$$ \sup \{ f(r) : 0 < r < R \} \leq A $$

for every $0 < R < \infty$. Hence $A$ is an upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$.

To show that $A$ is the supremum of this set, we must show that it is the least upper bound. Suppose that $D$ is another upper bound on $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$. Let $0 < r < \infty$, and pick $R=r+1$. Then we know that we must have $D \geq \sup \{ f(s) : 0 < s < R \}$. We know $f(r)$ is in this set, since $0 < r < r+1$, and since the supremum is an upper bound, it is greater than $f(r)$. Therefore, $D \geq f(r)$, and this holds for every $0 < r < \infty$.

Thus we conclude that $D$ is an upper bound on $\{ f(r) : 0 < r < \infty \}$. But since $A$ is the supremum of that set, we have that $A \leq D$. Hence we have shown that $A$ is less or equal to any upper bound of $\{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \}$, and since we have also shown that $A$ itself is an upper bound of this set, we get, by definition:

$$ A = \sup \{ \sup \{ f(r) : 0 < r < R \} : 0 < R < \infty \} = B $$

And this is what we wanted to show.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.