Q:

Show existence of continuous functions $f$ with $f''=\delta_0-\delta_1$

Let $u$ be the distribution on $\mathbb{R}$ given by $$u=\delta_0-\delta_1 $$ 
(a) show there exists a continuous function $f$ such that $f''=u$ and indicate such one. I thought of doing this with Fourier-transforms: $$F(f'')=s^2F(f)=F(u) = 1-e^{-2\pi i s}$$ 
which would imply $f= F^{-1}(s^{-2}(1- e^{-2\pi i s}))$. My problem is that it is not continuous (by wolframalpha calculations). 
So how to approach this?
(b) Show there exists a triple of continuous functions $g_0,g_1,g_2$ on $\mathbb{R}$ with compact support such that $$u= g_0+g_1'+g_2''$$
and indicate such triple. 
Thanks for any help or suggestions! 

A:

Answer to (a) : Intuitively, 1-dimensional Dirac delta function is the derivative of a unit step function (also called the Heaviside step function). Thus $f'' = \delta_0 - \delta_1$ means that $f'$ has a step (jump) of size 1 at $x = 0$ and a step of size -1 at $x = 1$ as follows:
$$ f'(x) = \begin{cases} a & x < 0 \\ a+1 & 0 \leq x < 1 \\ a & 1 \leq x \end{cases} $$
Integrating, we have
$$ f(x) = ax + b + x \chi_{[0, 1)}(x) + \chi_{[1, \infty)}(x), \tag{1}$$
where $\chi_{E}$ is the characteristic (indicator) function of $E \subset \Bbb{R}$. Confirming that this is our desired function is also easy: For any test function $\varphi \in C_{c}^{\infty}(\Bbb{R})$,
\begin{align*}
\int_{\Bbb{R}} f\varphi'' \, dx
&= \int_{\Bbb{R}} (ax + b + x \chi_{[0, 1]} + \chi_{[1, \infty)}) \varphi'' \, dx \\
&= \int_{\Bbb{R}} (ax + b) \varphi'' \, dx + \int_{0}^{1} x \varphi''(x) \, dx + \int_{1}^{\infty} \varphi''(x) \, dx \\
&= 0 + \left[ x \varphi'(x) \right]_{0}^{1} - \int_{0}^{1} \varphi'(x) \, dx - \varphi'(1) \\
&= \varphi(0) - \varphi(1).
\end{align*}
and therefore (1) is proved.

