Second derivative of a discontinuous function and the $\delta$ distribution

Question

Suppose the function $$ f(x)=\begin{cases} 0, \text{ if }x<0,\\x, \text{ if }x>0.\end{cases}$$

Its derivative is $$ f'(x)=\begin{cases} 0, \text{ if }x<0,\\1, \text{ if }x>0,\end{cases}$$ and its second derivative is the $\delta$ function, $f''(x)=\delta(x)$. Fine.

Now suppose the function $$ g(x)=\begin{cases} 0, \text{ if }x<0,\\1+x, \text{ if }x>0.\end{cases}$$ The derivative $g'$ equals $f'$ almost everywhere. They are both undefined at $x=0$.

Question: is it true that $g''=\delta(x)$?

Answer 1

The derivatives $g'$ and $f'$ are distributions. As such, the "almost everywhere" that you are trying to use doesn't apply.

The equation that does hold is $g'=f'+\delta$.

Therefore $g''=\delta + \delta'$.

Answer 2

No. Let $\phi$ be a smooth compactly-supported function. Then the distributional derivative of $g$ is $$ -\int_{-\infty}^{\infty} g(x) \phi'(x) \, dx = -\int_{0}^{\infty} (1+x)\phi'(x) \, dx = \phi(0) +\int_0^{\infty} \phi(x) \, dx, $$ i.e. $\delta+f'$. Thus the second derivative is $\delta'+f'' = \delta'+\delta$, which also follows from $$ (-1)^2\int_{-\infty}^{\infty} g(x)\phi''(x) \, dx = \dotsb = -\phi'(0) + \phi(0). $$