# Is sampled absolutely integrable function absolutely summable?

Suppose I have function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that it's absolutely integrable: $\int_{\mathbb{R}}|f(x)|dx<\infty$.

I am sampling function $f(x)$ with some period $T_s$. I am interested whether

$$\sum_{k=-\infty}^{k=\infty}|f(kT_s)|<\infty$$

It seems to me that it's true, but I can't figure out how to prove that.

The reason I ask is that I know that if $f(x)$ is absolutely integrable, then its Fourier transform exists, and I am wondering if the sampled version is guaranteed to have a discrete Fourier transform. Sorry if this is a silly question.

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Let us consider, for example, $$f=\sum_{k\in\mathbb{Z}}g_k\in C^\infty(\mathbb{R})$$ obtained by taking $g_k=g(2^k(x-k)),\ \forall x\in\mathbb{R},k\in\mathbb{Z},$ for some $g\in C_c^\infty(]-1/2,1/2[),$ with $g(0)>0.$
Now taking $T_s=1,$ we get $\sum_{k\in\mathbb{Z}}f(kT_s)=\sum_{k\in\mathbb{Z}}g(0)=+\infty.$
 Thank you, @Giuseppe! However, I am sort of lost in your notation: what is the set $C_c^{\infty}(]-1/2,1/2[)$? – M.B.M. Apr 28 '12 at 7:18 $g$ is a smooth (indefinitely differentiable with continuity) function which vanishes outside $]-1/2,1/2[$. An example is $\exp[1/(x^2-1/4)]$ – Giuseppe Apr 28 '12 at 7:24
Not true, as Giuseppe said. However, if you put $$g(x)=\sum_{k=-\infty}^\infty f(kT_s+x)$$ then the sum defining $g$ will be absolutely convergent for almost every $x$, and $g$ will be periodic with period $T_s$. Also, $$\int_0^{T_s}g(x)\,dx=\int_{-\infty}^\infty f(x)\,dx.$$ This might be of some use, depending on your application.