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 $\ell^p$?
 A: I claimed:


*

*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 $|x−y|<\delta$.

*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 $\||v| \star h \|_p \le \|v\|_p \|h\|_1$.
A: 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 $f(n)\geq n$. 
