Dirac distribution representation as a finite sum of derivatives of continuous functions with compact support

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I've been trying to express the Dirac measure (treated as a distribution) at the origin on the real line $\mathbb{R}$ as a finite sum of derivatives of continuous functions, all of which have their support on the interval $(-\epsilon,\epsilon)$ for arbitrary $\epsilon>0$.

Looking through existing answers on math.se, I've become aware of such representations in terms of:

  1. infinite sequences of compactly-supported continuous functions
  2. single continuous functions that are not compactly supported
  3. finite sequences of functions that are not locally integrable.

However, I have been unable to identify a finite sequence of derivatives of continuous functions (in the sense of distributions) which are locally integrable, compactly supported, and sum to the Dirac measure. By this, I mean that there exists a sequence of functions $f_i\in\mathcal{C}_c^0{((-\epsilon,\epsilon))}$ and an integer $M:2\le{M}<\infty$ such that $$\sum_{i=1}^M \int_\mathbb{R}(\partial/\partial x)^{p_i}f_i(x)\phi(x)=\phi(0)\hspace{0.25cm}\forall\phi\in\mathcal{C}_c^{\infty}(\mathbb{R})\text{, where }p_i\in\mathbb{N}\hspace{0.125cm}(1\le i\le M), f_i\in L_{1,loc}(\mathbb{R}).$$ Inspiration took me in the direction of Taylor's Theorem, and the best I've been able to do is to express $\delta_0(\phi)$ as a finite sum of derivatives of $\delta_a(\phi)$ for $a$ arbitrarily close to $0$. However, this is not a distribution expressible in the above form.

(Incidentally, I believe the answer to question 922708 shows there is no solution for $M=1$. This is anticipated by the exercise I'm trying to solve (see comment below).)

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Show that $f$ and $g$ below are both continuous and determine $f''-g.$ $$ f(x) = \begin{cases} 0 & (x \leq -1) \\ -\frac16 (x+1)^3 & (-1 \leq x \leq 0) \\ \frac16 (x-1)^3 & (0 \leq x \leq 1) \\ 0 & (1 \leq x) \end{cases} $$ $$ g(x) = \begin{cases} 0 & (x \leq -1) \\ -1-x & (-1 \leq x \leq 0) \\ -1+x & (0 \leq x \leq 1) \\ 0 & (1 \leq x) \end{cases} $$