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$t\cdot \delta (t)$ Does it equal to zero? If so, how can we prove it?

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This answer is supposed to provide some more insight, even though I use "engineering math" that might not pass muster over at math.SE.

One defining property of the Dirac delta impulse is

$$\int_{-\infty}^{\infty}\delta(t)f(t)dt=f(0)\tag{1}$$

if $f(t)$ is continuous at $t=0$.

Note that the Dirac delta impulse is not an ordinary function but a generalized function or distribution. For that reason it is only meaningful under an integral. The product of a distribution (such as $\delta(t)$) and an ordinary function $f(t)$ is also a distribution, and it is defined by

$$\int_{-\infty}^{\infty}[\delta(t)f(t)]\phi(t)dt=\int_{-\infty}^{\infty}\delta(t)[f(t)\phi(t)]dt\tag{2}$$

where $\phi(t)$ is some test function.

And since (from $(1)$)

$$\int_{-\infty}^{\infty}\delta(t)[f(t)\phi(t)]dt=f(0)\phi(0)=\int_{-\infty}^{\infty}\delta(t)f(0)\phi(t)dt\tag{3}$$

we can conclude that

$$\delta(t)f(t)=\delta(t)f(0)\tag{4}$$

if $f(t)$ is continuous at $t=0$.

So for $f(t)=t$ we get

$$t\,\delta(t)=0\tag{5}$$

since $f(t)=t$ is continuous at $t=0$ and $f(0)=0$.

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  • $\begingroup$ Even when $f (0)=0$? $\endgroup$ – Codevan Dec 15 '17 at 9:13
  • $\begingroup$ @Codevan: Sure, as long as $f(t)$ is continuous at $t=0$. $\endgroup$ – Matt L. Dec 15 '17 at 9:38
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    $\begingroup$ i'm gonna let Jazz and Matt slug this one out. $\endgroup$ – robert bristow-johnson Dec 15 '17 at 11:57
  • $\begingroup$ @LaurentDuval: That's what I mean by engineering math; in engineering literature I've always come across the statement that $f(t)$ must be continuous for (1) to hold. I guess that $f\in C^{\infty}$ is another (sufficient ?) requirement, which is the case for $f(t)=t$. I'm not sure about compact support, because I believe to "know" that for $f(t)=t$ everything works out fine. $\endgroup$ – Matt L. Dec 15 '17 at 15:44
  • $\begingroup$ Related: this answer on math.SE. No mention of compact support. $\endgroup$ – Matt L. Dec 15 '17 at 15:47

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