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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, $p \geq 1$.

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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.

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For $p = 1$, the $p' = \infty$ and $\phi_k$ should be $\phi(kx)$, not what you wrote. And as I wrote, you cannot use Holder's inequality naively. You use the fact that $|\phi_k(x)| \leq A\chi_{|x| \leq \lambda / k}$ where $\chi$ is the characteristic function and $A,\lambda$ are some constants which exist based on the compact support and smoothness of $\phi$. Then you can use the limiting condition as I wrote under the break. – Willie Wong Jan 30 '13 at 8:05

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