I am currently taking a course on test functions and distributions and my task is to prove that the Dirac delta is not a function.
Furthermore, I would also like to prove that it is continuous as a distribution in the sense that the inequality :
$|\delta_{a}(\phi)| \leq c\|{\phi}\|_{C^k}$, for $c>0$ and $\phi \in C_{0}^{\infty}$ holds.
Thank you!
Best Answer
Suppose that we have a function $\psi(x)$ so that $$ \delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1} $$ Define $$ \begin{align} \eta(x)&=\left\{\begin{array}{} 0&\text{if }|x|\le\frac12\\ \frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}}+e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\ 1&\text{if }|x|\ge1 \end{array}\right.\tag{2a}\\[6pt] \eta_k(x)&=\eta\!\left(2^{-k} x\right)\tag{2b} \end{align} $$ Then $\eta_k\in C^\infty$, $\eta_k(x)=0$ for $|x|\le2^{k-1}$, and $\eta_k(x)=1$ for $|x|\ge2^k$. Since $C^\infty$ is dense in $L_\text{loc}^1$, for any $\epsilon>0$, we can choose a $\psi_\epsilon\in C^\infty$ so that $$ \begin{align} \|(\psi-\psi_\epsilon)(1-\eta)\|_{L^1}&\le\epsilon/2\tag{3a}\\ \|(\psi-\psi_\epsilon)(\eta_k-\eta_{k+1})\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{3b} \end{align} $$ which means $$ \|\psi-\psi_\epsilon\|_{L^1}\le\epsilon\tag{3c} $$ Furthermore, let $\sigma$ be the signum function, then since $\psi_\epsilon\in L_\text{loc}^\infty$, we can choose a $\sigma_\epsilon\in C^\infty$ so that $|\sigma_\epsilon|\le1$ and $$ \begin{align} \|(\sigma\circ\psi-\sigma_\epsilon)(1-\eta)\psi_\epsilon\|_{L^1}&\le\epsilon/2\tag{4a}\\ \|(\sigma\circ\psi-\sigma_\epsilon)(\eta_k-\eta_{k+1})\psi_\epsilon\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{4b} \end{align} $$ which means $$ \|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}\le\epsilon\tag{4c} $$ Since $\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\le2$, we get $$ \begin{align} \|(\sigma\circ\psi-\sigma_\epsilon)\psi\|_{L^1} &\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|(\sigma\circ\psi-\sigma_\epsilon)(\psi-\psi_\epsilon)\|_{L^1}\tag{5a}\\ &\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\|\psi-\psi_\epsilon\|_{L^1}\tag{5b}\\ &\le3\epsilon\tag{5c} \end{align} $$
Define $\phi_k(x)=\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)$. Then we have $$ \begin{align} 0 &=\delta\left(\sum_{k=m}^n\phi_k\right)\tag{6a}\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\phi_k(x)\,\mathrm{d}x\tag{6b}\\ &=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)\mathrm{d}x\tag{6c}\\ &=\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6d}\\ &\ge\int_{\mathbb{R}}|\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\,\mathrm{d}x\\ &-\int_{\mathbb{R}}|(\sigma\circ\psi(x)-\sigma_\epsilon(x))\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6e}\\ &\ge\int_{2^m\lt|x|\lt2^n}|\psi(x)|\,\mathrm{d}x-3\epsilon\tag{6f} \end{align} $$ Explanation:
$\text{(6a)}$: $\phi_k\in C_c^\infty$ and is $0$ in a neighborhood of $0$
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: apply the definition of $\phi_k$
$\text{(6d)}$: compute the sum of the $\eta_k$
$\text{(6e)}$: $\psi\sigma_\epsilon=|\psi|-(\sigma\circ\psi-\sigma_\epsilon)\psi$
$\text{(6f)}$: apply $(5)$
Since $\epsilon$ was arbitrary, $(6)$ implies that on any annulus centered at $0$, $\psi$ is $0$. That is, $\psi(x)=0$ for all $x\ne0$. Since the value of a function at a single point does not affect the integral of that function, $(1)$ would imply that $$ \delta(\phi)=0\tag7 $$ for all $\phi$. Since $(7)$ is false, $(1)$ must be false.