-1
$\begingroup$

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

$\endgroup$

closed as unclear what you're asking by lennon310, Peter K. Dec 15 '17 at 16:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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 $f(0)=0$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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