# On $L^p$ and $\ell^p$

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 $\ell^p$?

-
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 $L_p$. –  David Mitra Oct 30 '12 at 18:41
This is true, however, if the function $f$ is monotone. This is immediate from the integral test. –  JavaMan Oct 30 '12 at 18:43
@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 $\ell^p$? –  Mark Anderson Oct 30 '12 at 18:56
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 $|x - y| < \delta$. –  Robert Israel Oct 30 '12 at 19:58
In particular, this is the case when $f' \in L^p + L^1$. –  Robert Israel Oct 30 '12 at 20:09

I claimed:

1. A sufficient condition for $f \in L^p$ 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 $|x−y|<\delta$.
2. In particular, this is the case if $f' \in L^1 + L^p$.

[ EDIT: here we need to assume $f$ is absolutely continuous on bounded intervals, so that $f(y) - f(x) = \int_x^y f(t)\ dt$ ]

Suppose there exist $\delta \in (0, 1/2)$ and $g \in L^p$ such that $|f(x) - f(y)| \le g(x)$ whenever $|x - y| < \delta$, but $f(n) \notin \ell^p$. Then there is a nonnegative sequence $a_n \in \ell^q$ where $1/p + 1/q = 1$ such that $\sum_{n=1}^\infty a_n |f(n)| = \infty$.
Let $A(x) = a_n$ for $n-\delta < x < n+\delta$ (with $A(x) = 0$ if $x$ is in none of these intervals). Then $A \in L^q$. Now if $n- \delta < x < n + \delta$, $$a_n |f(n)| \le a_n |f(n) - f(x)| + a_n |f(x)| \le A(x) (g(x) + |f(x)|)$$ Thus

\eqalign{\sum_{n=1}^\infty a_n |f(n)| &\le \sum_{n=1}^\infty (2 \delta)^{-1} \int_{n-\delta}^{n+\delta} A(x) (g(x) + |f(x)|)\ dx\cr &\le (2\delta)^{-1} \int_0^\infty A(x) (g(x) + |f(x)|)\ dx < \infty}

contradiction. This proves the first claim.

Now suppose $f' \in L^1 + L^p$, i.e. $f' = u + v$ with $u \in L^1$ and $v \in L^p$. Note that $|f(x) - f(y)| = \left|\int_x^y f'(t)\ dt\right| \le \int_{x-\delta}^{x+\delta} |f'(t)|\ dt = (|f'| \star h)(x)$ (the convolution of $|f'|$ and $h$) where $h$ is the indicator function of the interval $(-\delta,\delta)$. We just have to show that $|f'| \star h \in L^p$ to apply the previous result. By Young's inequality, $|u| \star h$ and $|v| \star h$, and therefore $|f'| \star h$, are in $L^p$ with $\||u| \star h\|_p \le \|u\|_1 \|h\|_p$ and $\||v| \star h \|_p \le \|v\|_p \|h\|_1$.

-
Thanks, I think I understand slightly better now, but I'm still lost as to how to "arrive at" some $g(x)$ to show this. Let me simplify my problem with an example: Consider $f(x)=\text{sinc(x)}$, which is known to be in $L^2$ (or in realistic terms, let's say I know how to do the integration and show it is in $L^2$). Now how do I (or can I) show $f(n)=\text{sinc}(n),\ n\in\mathbb{Z}$ is in $\ell^2$ using only the earlier fact that $\text{sinc}(x)$ is in $L^2$ (let's assume I don't know how to do the summation or that for a different $f(x)$, it is hard to approach). –  Mark Anderson Oct 31 '12 at 2:51
Note that $f'(x) = O(1/|x|)$ as $|x| \to \infty$ and is continuous, therefore is in $L^2$. So this is covered by my second statement. –  Robert Israel Oct 31 '12 at 3:08
Of course here you have both $f(x) \in L^2$ and $f(n) \in \ell^2$ directly by the fact $f(x) = O(1/|x|)$. –  Robert Israel Oct 31 '12 at 18:06

We can give any particular values to a continuous integrable function at integers, smoothness doesn't improve the things. Here is an example.

Take $\phi>0$ a smooth function with compact support (say contained in $[-1/2,1/2]$) and $\phi(0)=1$. Then define $f(x)=\sum_{n=0}^{+\infty}n\phi\left(n^22^n(x-n)\right)$. Let $\phi_n(x):=\phi\left(n^22^n(x-n)\right)$. Then $$x\in \operatorname{supp}\phi_n\Leftrightarrow |n^22^n(x-n)|\leq 2^{—1}\Leftrightarrow |x-n|\leq 2^{-n-1}n^{—2},$$ so the supports of $\phi_n$ are disjoint and $f$ is smooth. We have \begin{align} \int_{\Bbb R}|f(x)|^pdx&=\sum_{n\geq 0}n^p\int_{n-2^{-n-1}n^{—2}}^{n+2^{-n-1}n^{—2}}\phi\left(n^22^n(x-n)\right)^pdx\\ &=\sum_{n\geq 1}n^p\int_{-2^{-n-1}n^{—2}}^{2^{-n-1}n^{—2}}\phi(n^22^nt)^pdt\\ &=\sum_{n\geq 1}n^p\int_{-1/2}^{1/2}n^pn^{-2}2^{-n}\phi(s)^pds\\ &=\int_{-1/2}^{1/2}\phi(s)^pds\sum_{n\geq 1}n^{p-2}2^{-n}<+\infty, \end{align} so $f\in L^p$ for all $1\leq p<\infty$. But $f(n)\geq n$.

-
Thanks. The following is the same comment I left for Robert, illustrating the crux of my confusion. Let me simplify my problem with an example: Consider $f(x)=\text{sinc(x)}$, which is known to be in $L^2$ (or in realistic terms, let's say I know how to do the integration and show it is in $L^2$). Now how do I (or can I) show $f(n)=\text{sinc}(n),\ n\in\mathbb{Z}$ is in $\ell^2$ using only the earlier fact that $\text{sinc}(x)$ is in $L^2$ (let's assume I don't know how to do the summation or that for a different $f(x)$, it is hard to approach). –  Mark Anderson Oct 31 '12 at 2:52