Find triple functions $ (g_0,g_1,g_2)$ such that $g_0+g_1'+g_2'' = \delta_0-\delta_1$

Question

I want to find a triple of compactly supported continuous functions $ (g_0,g_1,g_2)$ on $\mathbb{R}$ such that $$g_0+g_1'+g_2'' = \delta_0-\delta_1$$

This is seemingly not so hard but ive broken my over this, tried combinations of $x\cdot \mathbb{I}_{[0,1]} $ and other functions but I just dont get it..

Suggestions as to what might work are greatly appreciated.

Answer 1

Let's start with finding a continuous $f$ with $f'' = \delta_0 - \delta_1$. For some $a \in \mathbb R$, we have $$ f'(x) = a + \mathbb 1_{[0,1]}(x) = \begin{cases} a & x \not\in [0,1] \\ a + 1 & x \in [0,1] \end{cases} $$ (denoting by $\mathbb 1_E$ the indicator function of a set $E \subseteq \mathbb R$) and hence by integrating once more for some $b \in \mathbb R$ $$ f(x) = ax + x\mathbb 1_{[0,1]}(x) + \mathbb 1_{(1,\infty)}(x) + b $$ lets choose $a = b = 0$, leaving us with $$ f(x) = x\mathbb 1_{[0,1]}(x) + \mathbb 1_{(1,\infty)}(x) $$ Now let $\eta \in C^\infty_c(\mathbb R)$ be a test function with $\eta|_{[-1, 2]} = 1$ and $g_2 = f\eta$, then $$ g_2'' = f''\eta + 2f'\eta' + f\eta'' $$ Now $f''\eta = f''$, as $\eta$ equals 1 in an open neighbourhood of the support of $f''$. If we let now $g_1 = -2f\eta'$, we have $g_1' = -2f'\eta' - 2f\eta''$, giving $$ g_2'' + g_1' = f'' - f\eta'' $$ now let $g_0 = f\eta''$ and get $$ g_2'' + g_1' + g_0 = f'' = \delta_0 - \delta_1. $$

Answer 2

First set $$f(x) = \left\{\begin{array} 0 0 &:if\quad x\leq 0 \\ x &:if\quad 0\leq x\leq 1 \\ 1 &:if\quad 1\leq x\end{array}\right.$$

Then $f''(x) = \delta_0 - \delta_1$

Now suppose I can find some twice differentiable function $g$ with $g(x) = 0$ for sufficiently small $x$ and $g(x) = 1$ for sufficiently large $x$.

Then $$\begin{array}{rcl}g_1(x)&=&0 \\ g_2(x) &=& g'(x) \\ g_3(x) &=& f(x) - g(x) \end{array}$$ Will do nicely.

I assume you won't have much trouble finding a $g$.