# Delta function representation

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

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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
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

$$u(k) = \operatorname{comb}(t)u(t)$$ 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.