How can $t \delta(t) = 0$ be proven using its defining propery? How can $t\delta(t)=0$ be proven using it's defining principle, $\int f(t) \delta(t-a) dt = f(a)$?
I have a "proof" so far but I know its not even close to rigorous or proper.
$\int f(t)\delta(t-a)dt = f(a)$
Let $f(t) = t, a = 0$:
$\int t\delta(t)dt=f(0) = 0$
Differentiate both sides:
$\frac{d}{dt} \int t\delta(t)dt = \frac{d}{dt}0$
$\Longrightarrow t\delta(t)=0$
What would be a more formal proof?
 A: Since
$$
\int t\delta(t) \, f(t) \, dt
= \int \delta(t) \, tf(t) \, dt
= \left. tf(t) \right|_{t=0}
= 0 f(0)
= 0
= \int 0 \, f(t) \, dt
$$
for every test function $f$ we can conclude that $t\delta(t)=0.$
That's really all we need.

Above I wrote things using an integral sign. But in the formal theory, distributions are not functions that can be integrated. Instead they are linear functionals acting on a space of test functions. Usually this is taken to be $C_c^\infty,$ the space of infinitely differentiable functions with compact support. The action of a distribution $u$ on a test function $\varphi$ is often written $\langle u, \varphi \rangle$ or $\langle u(t), \varphi(t) \rangle,$ where the variable $t$ is purely formal; it shouldn't be interpreted as $u$ having a value at $t$. This formal notation corresponds to the abuse of notation $\int_{-\infty}^{\infty} u(t)\,\varphi(t)\,dt.$
The definitions we need for proving $t\,\delta(t)=0$ are the following:

*

*If $f \in C^\infty$ and $u$ is a distribution, then $fu$ is the distribution defined by $\langle fu, \varphi \rangle = \langle u, f\varphi \rangle$ for every $\varphi\in C_c^\infty.$

*$\langle \delta, \varphi \rangle = \varphi(0)$ for every $\varphi\in C_c^\infty.$

*The zero distribution $0$ is defined by $\langle 0, \varphi \rangle = 0.$

*Two distributions $u$ and $v$ are equal if and only if $\langle u, \varphi \rangle = \langle v, \varphi \rangle$ for every $\varphi\in C_c^\infty.$
For any $\varphi \in C_c^\infty$ we have
$$
\langle t\,\delta(t), \varphi(t) \rangle
= \{ \text{ def. 1 } \}
= \langle \delta(t), t\,\varphi(t) \rangle
= \{ \text{ def. 2 } \}
\\
= \left. t\,\varphi(t) \right|_{t=0}
= 0
= \{ \text{ def. 3 } \}
= \langle 0, \varphi \rangle.
$$
Thus, by def. 4, we have $t\,\delta(t) = 0.$
A: This proof uses definition of convolution and Dirac delta and properties of integrals.
Define convolution of functions by
$$ h=f * g \quad \text{ iff } \quad h(x) = \int f(t)\,g(x-t)\,dt.\tag{1} $$
The identity property of the Dirac delta $\,\delta(x)\,$ is
that $\, f = f * \delta\,$ for all $\,f.$
Now given any function $\,f,\,$ define
$$ F(x) := x\,f(x),\quad
g(x) := x\,\delta(x)\quad \tag{2}$$  and $\, h=f*g \,$ as in
equation $(1)$.
By definition of convolution we have
$$ h(x) = \int f(t)\,(x-t)\,\delta(x-t)\,dt, \tag{3} $$
$$ h(x) = \int f(t)\,x\,\delta(x-t)\,dt - \int f(t)\,t\,\delta(x-t)\,dt, 
\tag{4} $$
$$ h(x) = x \int f(t)\,\delta(x-t)\,dt - \int F(t)\,\delta(x-t)\,
dt. \tag{5} $$
Thus $\, h(x) = x\,f(x) - F(x)\,$ because $\,\delta\,$ is the identity of
convolution. This implies $\,h(x)=0\,$ by definition of $\,F.\,$ Thus,
we have proved that $\,x\,\delta(x)=0.$
