2
$\begingroup$

Let $f_j \in C_c^\infty$ and assume there exists $M>0$ such that $\text{supp}f_0 \subset \{ |x| \le M \}$ and $\text{supp} f \subset \{ 1/M \le |x| \le M \}$. Define $f_j$ by $$ f_j (y) = f(y/2^j) $$ for $j =1, 2, \cdots$. Suppose that $$ \sum_{j=0}^\infty f_j (x) = 1.$$ Then how can I show that there exist $C, C'>0$ such that$$ C' \sum_{j=0}^\infty \| f_j (g)\|_{L^2}^2 \le\| g \|_{L^2}^2 \le C \sum_{j=0}^\infty \| f_j (g)\|_{L^2}^2?$$ $C_c^\infty$ means $C^\infty$ functions with compact support and $\text{supp}$ means the support.

$\endgroup$

1 Answer 1

2
$\begingroup$

First of all, $f_j(g)$ should be the product $f_j\, g$, not composition. Otherwise the inequality is plainly false as it lacks homogeneity with respect to $g$.

It is easier to think about this estimate by writing out norms as integrals and introducing $F=\sum_j |f_j|^2$. Then the desired estimate simplifies to $$C'\int F|g|^2\le \int |g|^2\le C\int F|g|^2 \tag1$$ Since $g$ could be any $L^2$ function, the only way for (1) to hold is pointwise: that is, $$C' F|g|^2 \le |g|^2\le C F|g|^2 \tag2$$ In other words, we hope that $F$ is bounded between two positive constants: $$(C)^{-1} \le F \le (C')^{-1} \tag3$$ Once you have (3), the claim follows.

Both parts of (3) follow from $\sum f_j=1$ and the fact that the supports of $f_j$ have bounded overlap, shown in your earlier question A simple question of Littlewood-Paley decomposition. (You may also want to recall that all $\ell^p$ norms are equivalent on a finite-dimensional space.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .