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Suppose I have a set of functions $(f_\epsilon)$ such that as $\epsilon\to 0$, $f_\epsilon\to F$ s.t.

$F(x)=0$ for $x\neq 0$ and $F(x)=\infty$ for $x=0$;

$\int_{-\infty}^\infty f_\epsilon(x) dx=1$ for all $\epsilon >0$

Then can I conclude that the limit is the delta function $\delta(x)$? (which has the sampling property too)

Thanks

J. M. ain't a mathematician
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Terry
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  • You say $f_{\varepsilon}\to F$, but in that sense? – Davide Giraudo Dec 25 '11 at 22:26
  • @DavideGiraudo: for example, say $f_\epsilon={\epsilon\over x^2 +\epsilon}$ – Terry Dec 25 '11 at 22:33
  • How do you define "the delta function $\delta (x)$"? – joriki Dec 25 '11 at 22:41
  • @joriki: Pretty much it having the properties of $F$ PLUS the sampling property. I guess I am mainly interested in whether a function $F$ being the limit of functions $f_\epsilon$ would naturally have the sampling property. If it is not generally true, when would it be true, perhaps in the example I posted as a response to Davide's comment? – Terry Dec 25 '11 at 22:45
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    It's not clear from your question what framework you're working in. You speak of the delta "function", and you write $F(x)=\infty$. This seems to imply that you're considering $F$ as a function from the reals to the extended reals (extended by adding infinity). In that case, a) the Lebesgue integral over your function $F$ is $0$, not $1$ as it should be for anything that deserves to be called "the delta function", and b) $F$ is already fully defined by the second line in the question, so it's not clear what you mean when you say you define the delta function as $F$ with additional properties. – joriki Dec 25 '11 at 23:25
  • I get the impression that you may not be aware that what's often called the delta "function" is usually rigorously defined either as a distribution or as a measure. See Wikipedia for these definitions. I don't see how to make sense of what you write in a framework dealing only with functions. – joriki Dec 25 '11 at 23:28
  • This related answer might help you approach @joriki's objections, which I fully share. – Did Dec 26 '11 at 00:01
  • @joriki: Thanks for the reference. I realize that my question is not a very good one, sorry about that. I am new to the Dirac delta "function" and was a bit confused... I shall read the refs. – Terry Dec 26 '11 at 00:19

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To be able to answer this question, you have to define a few important concepts such as distributions or generalized functions, the limiting process etc. and you have to be extremely careful when performing the limiting step. The usual intuitions generally (and only formally) fail to make sense. The main problem is that the Dirac Delta does not mean anything outside an integral,sometimes even inside, in particular it is not defined pointwise!! Therefore, sampling with the Dirac comb, which is a function(!) consisting Dirac Deltas repeated periodically is not a valid operation if we write

$$ u(k) = \operatorname{comb}(t)u(t) $$ where $k\in\mathbb{Z}$ and $t\in\mathbb{R}$.

In other words, there is no pointwise multiplication that gives you the sampled version of $u(t)$ but it is, without a doubt, a neat shortcut to state the operation.

For a wonderful storyline I strongly recommend you the Lecture 12 of Brad Osgood's Fourier Transform and Applications course. He defines almost all practical definitions and mention how they fail (roughly though). But the course material spends quite some effort on this issue.