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If a continuous and infinitely differentiable function $f(x): \mathbb{R}\to\mathbb{C}$ is in $L^p$, is it also true that $f(n),\ n\in \mathbb{Z}$ is in $\ell^p$?

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    No. For instance, in $L_1$, think of a function whose graph consists of spikes centered around the integers with the spike centered at $n$ having height $1/n$. The width of the spikes can be chosen so that the function is in $L_1$. Similar constructions can be made for other $L_p$.2012-10-30
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    This is true, however, if the function $f$ is monotone. This is immediate from the integral test.2012-10-30
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    @DavidMitra Does it hold for $1 then (or any set of $p$)? I must add that $f(x)$ is continuous and has derivatives of all order.2012-10-30
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    You certainly assumed that $f$ is *continuous*.2012-10-30
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    @MarkAnderson No. Take the height of the $n$'th spike to be $1/n^{1/p}$. You can still make the widths small enough so that the resulting function is in $L_p$. You can also arrange things so that $f$ is infinitely differentiable.2012-10-30
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    @DavidMitra Hmm... Ok, let me put it this way: What would it take for a continuous and infinitely differentiable function $f(x)$ in $L^p$ to be such that $f(n), n\in\mathbb{Z}$ is in $\ell^p$?2012-10-30
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    When you write $f(n)$ do you mean to write $f^{(n)}$?2012-10-30
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    @JacobSchlather No, I mean $f(x)$ evaluated at integers ($n\in\mathbb{Z}$)2012-10-30
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    A sufficient condition for $f(x) \in L^p$ to imply $f(n) \in \ell^p$ is that there exist $\delta > 0$ and $g \in L^p$ such that $|f(x) - f(y)| < g(x)$ whenever $|x - y| < \delta$.2012-10-30
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    In particular, this is the case when $f' \in L^p + L^1$.2012-10-30
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    @RobertIsrael Could you elaborate on that a bit more? Not sure I follow...2012-10-30
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    I'd better make that into an answer.2012-10-30

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