2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

No, that is not true in general.
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 $\sum_{k\in\mathbb{Z}}f(kT_s)=\sum_{k\in\mathbb{Z}}g(0)=+\infty.$

$\endgroup$
  • $\begingroup$ Thank you, @Giuseppe! However, I am sort of lost in your notation: what is the set $C_c^{\infty}(]-1/2,1/2[)$? $\endgroup$ – M.B.M. Apr 28 '12 at 7:18
  • $\begingroup$ $g$ is a smooth (indefinitely differentiable with continuity) function which vanishes outside $]-1/2,1/2[$. An example is $\exp[1/(x^2-1/4)]$ $\endgroup$ – agtortorella Apr 28 '12 at 7:24
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.