Does $t\delta(t) = 0$?

Question

In middle of solving a problem I encountered terms looking like this: $$t\delta(t)$$ Or in a more general form: $$(t-t_{0})\delta(t-t_{0})$$ It appears that in the textbook I am reading "Linear Systems and Siganls by B. P. Lathi and Roger Green" this value is considered to equal zero although it is not explicitly mentioned (probably to avoid some complex math). Now from the properties of the impulse function: $$t\delta(t) = 0\times\delta(t)$$ Is this value $0$ or it is undefined? similar to $0\times\infty$? If the value is not $0$, does this mean that the unit step function $u(t)$ is not the derivative of the unit ramp function $tu(t)$ (as shown bellow)? $$\frac{d(tu(t))}{dt}=t\delta(t)+u(t)$$

Answer 1

As a distribution, yes this is true. Let your distribution be $T$, then for Schwartz $f$,

$$T(f) = \delta(tf) = 0\cdot f(0) = 0.$$

The shifted result also holds because any Schwartz function is a translation of another.