Simplification of PDF using the unit step function

Question

\begin{align} f_T(y)&=\frac{d}{dy}\left[\left(\frac 13+\frac{2y}9\right)[u(y)-u(y-3)]+u(y+3)\right]\\ &=\frac 29[u(y)-u(y-3)]+\left(\frac 13+\frac{2y}{9}\right)[ \delta(y)-\delta(y-3)]+\delta(y-3) \\ &=\frac 13\delta(y)+ \frac 29[u(y)-u(y-3)] \end{align}

where $\delta(x)$ is the Dirac delta. I think $\frac 29[u(y)-u(y-3)]$ being unchanged in the calculation but how $(\frac 13+\frac{2y}{9})[ \delta(y)-\delta(y-3)]+\delta(y-3)$ simplify to $\frac 13\delta(y)$?

Answer 1

To see why it simplifies that way, notice that $y\delta(y) = 0$, and $y\delta(y-3) = 3\delta(y-3)$. So: \begin{align} \left(\frac 13+\frac{2y}{9}\right)[ \delta(y)-\delta(y-3)]+\delta(y-3) &= \frac{1}{3}\delta(y) +\frac{2y}{9}\delta(y) - \frac{1}{3}\delta(y-3) - \frac{2y}{9}\delta(y-3) +\delta(y-3) \\ &=\frac{1}{3}\delta(y) - \frac{1}{3}\delta(y-3) - \frac{2}{3}\delta(y-3) + \delta(y-3) \\ &=\frac{1}{3}\delta(y) \end{align}