Proving $f(t)\delta(t-T)=f(T)\delta(t-T)$ where $f(t)$ has jump at $t=T$ I have the following function:
$$\frac{1}{5} t(H(t) - H(t - 5))$$
where $H(t)$ is the Heaviside function.
The derivative is:
$$\frac15 t (\delta(t)-\delta(t-5) ) + \frac{1}{5}(H(t)-H(t-5))=\frac{1}{5}(H(t)-H(t-5))-\delta(t-)$$
Note that 1 is the jump (see http://www.wolframalpha.com/input/?i=(t%2F5)+(HeavisideTheta%5Bt%5D+-+HeavisideTheta%5Bt+-+5%5D) ).
How can I prove that:
$$-\frac{1}{5} t \delta(t-5)=-\delta(t-5)?$$
Thank you for your time.
 A: First, if $\displaystyle f(t)=\frac15 t\,\left(H(t)-H(t-5) \right)$, then we have in distribution 
$$\begin{align}
f'(t)&=\frac15 \left(H(t)-H(t-5) \right)+\frac15 t \left(\delta(t)-\delta(t-5) \right)\\\\
&=\frac15 \left(H(t)-H(t-5) \right)+\frac15 \color{blue}{t\delta(t)}-\frac15\color{red}{ t\delta(t-5)}\\\\
&=\frac15 \left(H(t)-H(t-5) \right)+\frac15 \color{blue}{0}-\frac15 \color{red}{5\delta(t-5)}\\\\
&=\frac15 \left(H(t)-H(t-5) \right)-\delta(t-5)
\end{align}$$

We now show that $t\delta(t-a)=a\delta(t-a)$ in distribution.  In proceeding, we adopt the notation 
$$\langle \delta_a,\phi\rangle =\int_{-\infty}^\infty \delta(t-a)\phi(t)\,dt$$ 

Note that if $f\in C^\infty$ and $\phi(t)$ is a suitable test function, then  
$$\begin{align}
\langle f\delta_a, \phi\rangle&=\langle \delta_a,f\phi\rangle \\\\
&=f(a)\phi(a)\\\\
&=f(a)\langle \delta_a,\phi\rangle \\\\
&=\langle f(a)\delta_a,\phi\rangle
\end{align}$$
Therefore, $f(t)\delta(t-a) =f(a)\delta(t-a)$ in distribution.
Letting $f(t)=t$ and $a=0$, we see that in distribution 
$$t\delta(t)=0$$.
Letting $f(t)=t$ and $a=5$, we see that in distribution 
$$t\delta(t-5)=5\delta(t-5)$$
And we are done!
