Distributional derivative of a function with jumps

Question

Let $f(x)$ be a real function of a single variable x that is continuously differentiable on the real line except at $x=a$ where $f$ has a jump discontinuity. Let's suppose also that left- and right-hand derivatives of $f$ at $x=a$ exist. In order to construct the distributional derivative of $f$ (wich exist since $f$ is locally integrable) some authors proceed like this.

Let denote $\Delta=f(a^+)-f(a^-)$ the amount of the jump of $f$ at $a$. Also, denote $[f']$ the function obtained by differentiating $f$ without regard to the jump. Finally, define $$g(x)=f(x)-\Delta H(x-a)$$ where $H$ is the Heaviside function. Now they observe that $g$ is continuous at $x=a$ and has a piecewise continuous derivative g' which coincides with its distributional derivative. Differentiating both sides of the previous in the sense of distribution we obtain $$g'=f'-\Delta \delta(x-a)$$ where $\delta(x-a)$ is the $\delta-$ Dirac centered at $x=a$. Since $g'=[f']$ we finally get $$f'=[f']+\Delta \delta(x-a).$$ I have a couple of question about this construction.

1) In order to say that left- and right-hand derivatives of $f$ at $x=a$ exist, it is necessary that $f$ is defined at $x=a$, or they just mean that the limits $\lim_{x\to a^-}f'(x)$, $\lim_{x\to a^+}f'(x)$ both exist?

2) How can I prove that $g$ is continuous at $x=a$?

3) Does $g'=[f']$ is intended in the sense of distribution?

For 2), it easily follow from the definition of $g$ that $\lim_{x\to a^-}g(x)=\lim_{x\to a^+}g(x)=f(a^-)$, but how can I be sure that this limit equals to $g(a)$?

Any help would be really appreciated.

Answer 1

1. They must mean that $\lim_{x\to a^-}f'(x)$ and $\lim_{x\to a^+}f'(x)$ both exist.

2. To prove that $g$ is continuous at $x=a$ we first must give $g(a)$ a value. To do this we can start with showing that $g(a-) = g(a+)$. But that follows directly from the definition: $$\begin{align} g(a+) - g(a-) &= (f(a+) - \Delta H((a+)-a)) - (f(a-) - \Delta H((a-)-a)) \\ &= (f(a+) - \Delta) - (f(a-) - 0) \\ &= (f(a+) - f(a-)) - \Delta \\ &= 0. \end{align}$$

3. No, $g' = [f']$ is intended in the sense of ordinary derivatives. Those derivatives are defined and coincide everywhere except at $x=a$.