About a property of the Dirac delta function How can I show that there is no $u$ satisfying both (i) and (ii):$$(i) \; u \in L^p (\Bbb R^n )$$ and $$(ii) \int_{\Bbb R^n} \delta (x) \phi(x) dx= \int_{\Bbb R^n} u (x) \phi(x)dx\; ( \forall \phi  \in \mathcal S)\;?$$
Here $\mathcal S$ means Schwartz class, $\delta$ : dirac delta function, $p \geq 1$. 

For $p > 1$ we can argue via dimensional analysis. Denote by $p' = ( 1 - 1/p)^{-1}$ the dual exponent to $p$. 
Fix $\phi\in \mathcal{S}$. Let $\phi_k(x) = k^{n/p'}\phi(kx)$. Since $1 < p' < \infty$ we have that 
$$\lim_{k\to\infty} \left|\langle \delta,\phi_k\rangle\right| = \lim_{k\to \infty} |\phi_k(0)| = +\infty$$
while by Holder's inequality
$$ \left|\int_{\mathbb{R}^n} u(x) \phi_k(x) \mathrm{d}x\right| \leq \|u\|_{L^p} \|\phi_k\|_{L^{p'}} = \|u\|_{L^p} \|\phi\|_{L^{p'}} $$
using the scaling property of $\phi_k$. 

For $p = 1$ one has to use a bit more. In particular, you need to use that for any integrable function $u$
$$ \lim_{\delta\to 0} \int_{|x| < \delta} u(x) \mathrm{d}x = 0 $$
and now take $\phi_k$ as above but with $\phi$ of compact support. 
