Proper proof of a delta function

[tolove]

Prove:
tδ(t) = 0

The answer our TA has given isn't doing it for me:
$\int dt \delta(t)f(t) = (0)f(0) = 0$

I'm wanting to write:
$t \frac{d}{dt}\int \delta(t) dt = t \frac{d}{dt}(1) = t * 0 = 0$

Am I right here? This doesn't make use of a test function. I'm very sloppy with proofs!

Thanks for your time!

 

[member 428835]

do you mean $\delta (t)$ is the dirac delta function?

also, you are incorrect in stating this:

$t \frac{d}{dt}\int \delta(t) dt = t \frac{d}{dt}(1)$

note that $\frac{d}{dt}\int \delta (t) dt= const.$, thus and arbitrary $\delta (t)$ would work in the above "proof" (assuming it fits the criteria for the fundamental theorem). first, you misuse the fundamental theorem (no bounds). secondly, you assume a function of $t$ is not in fact a function of $t$. and lastly, if this question is a homework question, i think it belongs in that section.

 

[micromass]

Both $t\delta(t)$ and $0$ are distributions here. So you have to prove an equality of distributions. By definition, two distributions $F$ and $G$ are equal if

$$\int \varphi(t)F(t)dt = \int \varphi(t)G(t)dt$$

for all test functions $\varphi$. So this is what you need to check.

 

[pwsnafu]

Prove:
tδ(t) = 0

The answer our TA has given isn't doing it for me:
$\int dt \delta(t)f(t) = (0)f(0) = 0$

What don't you like about it? It's exactly the Schwartz product, that is if $f$ is a smooth function and $\mu$ a distribution and $g$ a test function then
$\langle f \, \mu , g \rangle := \langle \mu , fg \rangle$.