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Consider the initial value problem

$$ \begin{cases} \partial^{2}_{t} w - \Delta w = 0, \\[6pt] w(x,0) = 0\\[6pt] \partial_t w(x,0) = g(x) \end{cases} $$

$x \in R^3 , t \in R.$

if $g$ is a radial function, i know this

$$w(x,t) = w(\| x\|,t) = \frac{1}{2 \| x\|} \int_{ |\|x\| - t|}^{\| x\|+t} \rho g(\rho) \ d\rho.$$ The exercise is:

Use the Hardy-Littlewood maximal function to show that

$$ \left(\int_{-\infty}^{+\infty} \| w(\cdot,t)\|^2_\infty \ dt\right)^{1/2} \leq C \|g\|_2.$$

I have no idea to how to do that. someone can help me with this exercise ?

Thanks in advance!

Stephen Montgomery-Smith
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math student
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  • The Hardy-Littlewood maximal operator acts from $L^p$ to $L^p$ as follows: for $f\in L^p$: $Mf = \sup_{r>0} \frac{1}{|B(x,r)|}\int_{B(x,r)} |f(y)|dy.$ It is well-known that $|Mf|_p \leq C |f|_p$ that means the H-L operator is bounded from $L^p$ to $L^p$. Btw. revise your question, could it be that you have some typos? – Martingalo Feb 11 '14 at 23:17
  • the question is right ( i copied form the book ) .I believe that $||w( . , t )||_{\infty} = M(g)$. Do you know prove that ? Do you know if my affirmation is true ? – math student Feb 12 '14 at 03:06
  • Where did you get the formula for $w(x,t)$? See Proposition 4 in S. Klainerman and M. Machedon, Space time estimates for null forms and the local existence theorem, Communications of Pure and Applied Math., 46, (1993), 1221-1268. They have $||x|-t|$ for the lower limit. – Stephen Montgomery-Smith Feb 12 '14 at 03:10
  • @StephenMontgomery-Smith you're right. i did a mistake. i fixed the error . thanks to show to me the error ( my english is terrible , sorry) – math student Feb 12 '14 at 03:16

1 Answers1

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$$w(x,t) = w(\| x\|,t) = \frac{1}{2 \| x\|} \int_{ |\|x\| - t|}^{\| x\|+t} \rho g(\rho) \ d\rho.$$ So $$ |w(\| x\|,t)| \le \frac{1}{2 \| x\|} \int_{ t-\|x\|}^{t+\| x\|} |\rho g(\rho)| \ d\rho.$$ This is trivial if $t > \|x\|$, and if $t \le \|x\|$ it follows because $$ |w(\| x\|,t)| \le \frac{1}{2 \| x\|} \int_{ \|x\|-t}^{\| x\|+t} |\rho g(\rho)| \ d\rho \le \frac{1}{2 \| x\|} \int_{ t-\|x\|}^{t+\| x\|} |\rho g(\rho)| \ d\rho,$$ because the integrand is positive.

Thus by the definition of the one-dimensional Hardy-Littlewood maximal function, $\|w(\cdot,t)\| \le M h(t)$, where $h(s) = |sg(s)| $. Now use spherical coordinates to see that $\|g\|_{L^2(\mathbb R^3)} = \|h\|_{L^2(\mathbb R)}$.

Stephen Montgomery-Smith
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  • By the way, Klainerman and Machedon show the result isn't necessarily true if $g$ isn't radially symmetric. I extended this to show the result still isn't true if the $L^\infty$ norm is replaced by the BMO norm. http://www.math.missouri.edu/~stephen/preprints/schrod.html, Time decay for the bounded mean oscillation of solutions of the Schrödinger and wave equations. Duke Math J. 91 (1998), 393-408. – Stephen Montgomery-Smith Feb 12 '14 at 04:11
  • Hello Stephen Montgomery-Smith. It's so nice this solution. Thank you. And, just one more question, this solution above is just for the case $|x| - t <= 0$, and what if do and the other case? – math student Feb 13 '14 at 13:02
  • It still works if $|x|-t < 0$, it is just that the equality turns into an inequality. That is why I added the absolute value inside the integral. – Stephen Montgomery-Smith Feb 13 '14 at 13:04
  • I asked because in the formula of $w$ we have $| |x| - t |$ in the lower limit, and you analyzed the case $|x| - t \leq 0$. And the case $|x| - t \geq 0$? – math student Feb 13 '14 at 13:35
  • I meant to say it still works for $|x|-t > 0$. – Stephen Montgomery-Smith Feb 13 '14 at 16:00
  • In the case $|x| - t > 0$, the length is $2t$ and we have $2|x|$ in the denominator, therefore we don't have the maximal function $Mh$ as previously, right? Excuse me for thats questions, I really need to learn this. – math student Feb 13 '14 at 16:15
  • See my edit in the answer where I tackle your question explicitly. – Stephen Montgomery-Smith Feb 13 '14 at 16:24
  • Oh, I really understand now. Thank you VERY MUCH! You saved my life, rs. Thank you for explain me that. =D – math student Feb 13 '14 at 16:43