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$1\leq p < \infty$. Space is $L^p(\mathbb{R}^n)$.

Let $\delta >0,\ R>0$ be constants. $Q$ is the open cube centered at origin such that $||y||<\frac{\delta}{2}, \forall y \in Q$.

Let $Q_1, \dots, Q_N$ be mutually non overlapping translates of $Q$ such that $B(0,R)\subset \bigcup_{i=1}^N Q_i $.

Define projection map, $P:L^p(\mathbb{R}^n) \rightarrow L^p(\mathbb{R}^n)\ $ as

$$Pf= \sum_{i=1}^N \bigg( \frac{1}{|Q_i|} \int_{Q_i}f(z)dz \bigg) \chi_{Q_i}$$

where, $|Q_i|=$ Lebesgue measure of $Q_i$ ; $\chi_{Q_i}$ is the characteristic function of $Q_i$.

I have to show that $||P||=1$.

Can this be done by the definition that, $||P||= Sup_{||f||_p \leq 1} ||Pf||_p$? I am getting stuck because there is a double integration involved. Any help is appreciated.

Epsilon
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  • Do you mean $|y_k|<\frac{\delta}{2}$ for each $k$, $y=(y_1,...,y_n)$? And by mutually non-overlapping translates, you mean mutually disjoint $\mathcal{Q}_i\cap\mathcal{Q}_j=\emptyset$ for $i\neq j$? Also what is the purpose of $B(0,R)\subset \bigcup_i\mathcal{Q}_i$? – charlestoncrabb Nov 16 '15 at 18:04
  • No. Definiton of $Q$ is such that any point $y \in Q$ satisfies $||y||< \frac{\delta}{2}$. And yes, they are mutually disjoint. $B(0,R)$ is there to give an idea how the $Q_i$ are arranged. It means $N$ cubes are covering the ball $B(0,R)$. I think $R$ is not involved in calculating the norm, though. – Epsilon Nov 16 '15 at 18:08
  • Perhaps then you mean $B(0,R)\subset\bigcup_i\overline{\mathcal{Q}_i}$? And by this definition then $\mathcal{Q}$ is just $B(0,\frac{\delta}{2})$? – charlestoncrabb Nov 16 '15 at 18:17
  • If you take $B(0,\frac{\delta}{2})$ then will it be a cube? Maybe you can look at the case of $\mathbb{R}^2, Q$ would be the diamond centered at origin with vertices at $(0,\frac{\delta}{2}),(\frac{\delta}{2},0),(0,-\frac{\delta}{2}), (-\frac{\delta}{2},0)$ – Epsilon Nov 16 '15 at 18:21
  • To make it the cube with width $\delta$, you should use the definition I asked about: with $y=(y_1,...,y_k,...,y_n)$, $|y_k|<\delta/2$ for each $k$. It doesn't really matter that much... – charlestoncrabb Nov 16 '15 at 18:25
  • You are right. I had misunderstood your first comment. Now it makes sense. – Epsilon Nov 16 '15 at 18:28

1 Answers1

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Look at what each term in the sum is: $\left(\frac{1}{|\mathcal{Q}_i|}\int_{\mathcal{Q}_i}fdx\right)\chi_{\mathcal{Q}_i}$ is a step function taking the average value of $f$ over $\mathcal{Q}_i$ in $\mathcal{Q}_i$ and zero everywhere else--in particular, zero on all other $\mathcal{Q}_j$s. Perhaps for ease of notation you should write $c_i=\int_{\mathcal{Q}_i}fdx$. The "double integral" as you say will only pick out a factor of $|\mathcal{Q}_i|$ in each term in the sum.

Think about (i'll leave the details to you, they're not hard) how the facts I mentioned yield the following equality: $$\|Pf\|^p_{L^p(\mathbb{R}^n)}=\sum_{i=1}^N\frac{|c_i|^p}{|\mathcal{Q}_i|^{p-1}}.$$

Now here's where it is less obvious where to proceed, but motivated by the fact that if $f$ is non-negative, the $|c_i|$'s are just the $L^1$ norms of $f$ restricted to the respective boxes, and using this nice embedding, we get

$$\|Pf\|^p_{L^p(\mathbb{R}^n)}\leq\sum_{i=1}^N\|f|_{\mathcal{Q}_i}\|^p_{L^p(\mathcal{Q}_i)}\leq\|f\|^p_{L^P(\mathbb{R}^n)}.$$ From here, think about how we might achieve equality?

charlestoncrabb
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