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A paper (penultimate line in page 7) I'm reading gives the following sum$$\sum_{x\in 2^{-n}\mathbb{Z}^d}|\langle \phi,\psi\rangle|$$ (where $n\in\mathbb{N}$) and says that, because $\psi$ is a function with compact support, the number of non-zero terms is bounded by $2^{nd}$.

First, I'd imagine what they mean is that the number of non-zero terms in the sum is bounded by some constant multiplied by $2^{nd}$.

But even so, where does the power of $d$ come from? I'd have said that the number of non-zero terms equals $2^{n}$ multiplied by some element of $\mathbb{Z}^d$, so that we get a constant multiplied by $2^{n}$.

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The number of terms is bounded by $2^{n}$ (multiplied by some constant), and for each term there are $d$ components to make a choice for, so the bound we get is $(2^{n})^d = 2^{nd}.$

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    I realize you are trying to answer your own Question, but this is not correct reasoning. Note that $2^{nd} = (2^n)^d$. The exponent $d$ arises from the fact that $x$ has $d$ components to choose. – 2017-02-03
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    Ah yes, I see! Thanks – 2017-02-03
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    Will you be editing your Answer to correct it? – 2017-02-03