How can $t \delta(t) = 0$ be proven using its defining propery?

Question

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?

Answer 1

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 $\,x\,\delta(x)=0.$

Answer 2

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.

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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:

1. 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 $\varphi\in C_c^\infty.$
2. $\langle \delta, \varphi \rangle = \varphi(0)$ for every $\varphi\in C_c^\infty.$
3. The zero distribution $0$ is defined by $\langle 0, \varphi \rangle = 0.$
4. 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 $t\,\delta(t) = 0.$