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Context: these notes, page 7, last paragraph (between equations 2.7 and 2.8).

Given the $\mathcal{C}^r$ function with compact support $\varphi:\mathbb{R}^d\to \mathbb{R}$, we define $$\varphi_x^n:=2^{nd/2}\varphi(2^n(\cdot -x)).$$

Next, for some $\mathcal{C}^r$ function with compact support $\psi$, we try to bound

$$|\langle\varphi_x^n,\psi\rangle|.$$

Substituting in the definition of $\varphi_x^n$, we get

$$|\langle\varphi(2^n(\cdot -x)),\psi\rangle|$$ $$= 2^{nd/2}|\int \varphi(2^n(\cdot -x))\psi|$$ Since both $\varphi$ and $\psi$ are integrable and with compact support, I would expect the integral to be bounded. Like this, I would obtain

$$|\langle\varphi_x^n,\psi\rangle| \lesssim 2^{nd/2}.$$

However, the paper gives

$$|\langle\varphi_x^n,\psi\rangle| \lesssim 2^{-nd/2}.$$

Why might that be?

1 Answers 1

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Notice that the paper is a far better majoration. A change of variable will help us to get it.

$$\int_{\Bbb{R}^d } \varphi(2^n(y-x)) \psi(y) dy = \int_{\Bbb{R}^d } \varphi(y-2^nx) \psi(\frac{y}{2^n}) (2^{-n})^d dy$$

Now $\psi$ is bounded, and $\varphi$ is bounded with compact support $K$, so

$$ \left| \int_{\Bbb{R}^d } \varphi(2^n(y-x)) \psi(y) dy \right| = \left| \int_{\Bbb{R}^d } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right|$$

$$= \left| \int_{K+2^nx } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right|$$

$$\leq 2^{-nd} \int_{K+2^nx} \|\varphi\|_\infty \| \psi \|_\infty dy$$

$$\leq 2^{-nd} \mu(K) \|\varphi\|_\infty \| \psi \|_\infty$$

And if you multiply this by $2^{\frac{nd}{2}}$, you get the given coefficient

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    thanks for your answer. When you change the domain of the integral to $K+2^n x$, why are you allowed to shift it by $2^nx$? I get that $\varphi$ will only be nonzero in that subset, but what about $\psi$, whose argument doesn't get shifted? – 2017-01-30
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    @man_in_green_shirt : it doesn't matter what's "inside" $\psi$, as you majorate it by its supremum norm. – 2017-01-30
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    but what about before you take the supremum norm, when you write $\left| \int_{\Bbb{R}^d } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right| = \left| \int_{K+2^nx } \varphi(y-2^nx) \psi(2^{-n}y) 2^{-nd} dy \right|$? How do we know they're equal, when $\psi(2^{-n}y)$ may be nonzero outside of $K+2^nx$? – 2017-01-30
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    @man_in_green_shirt : but if $\varphi(y-2^nx)$ is zero outside $K+2^nx$, obviously, the product with anything else is zero too ! – 2017-01-30
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    Oh yeah, it's obvious. Thanks again for your answer (and patience)! – 2017-01-30