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I'm reading a one proof. This say

If $u$ is a test function (smooth function with compact support), then $$|\delta_0(u)|=|u(0)|=\left|\int_{—1}^0u'(t)dt\right|\leq \lVert u'\rVert_p\leq \lVert u\rVert_{W^{1,p}(-1,1)}.$$ By density of test function, we can extend $\delta_0$ to the functions of $W^{1,p}(-1,1)$.

I don't understand this "By density of test function, we can extend $\delta_0$ to the functions of $W^{1,p}(-1,1)$".

Please, Anybody will be able to explain me?

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It means that the test function space $C^{\infty}_c(I)$ is dense in $W^{1,p}_0(I)$ which is a closed subspace of $W^{1,p}(I)$. –  Shuhao Cao Nov 8 '12 at 19:14
    
but Why "we can extend $\delta_0$ to the functions of $W^{-1,p}(-1,1)$"? –  user46060 Nov 8 '12 at 19:35
    
$W^{-1,p}$ is the space of bounded linear functional on $W^{1,p}_0$, aka the dual space of $W^{1,p}_0$, please see my answer. –  Shuhao Cao Nov 8 '12 at 19:40

1 Answer 1

I think your interpretation on this topic sounds inaccurate to me, without context I am guessing the author meant to convey the idea

"Dirac delta, as a distribution, is a bounded linear functional in $W^{1,p}_0(I)$"

To establish this, simply using the definition $\delta_0[u] = u(0)$ doesn't make any sense for $u\in W^{1,p}(I)$ for $u$ can be any number on a measure zero set, think how we circumvent this to define the trace of function in a Sobolev space to make the boundary value problem meaningful.

Therefore we would like to establish this indirectly borrowing the test function space $C^{\infty}_c(I)$, which is dense in $W^{1,p}_0(I)$ under the norm $\|\cdot\|_{W^{1,p}}$. The boundedness of $\delta_0[\cdot]$ in $W^{1,p}_0(I)$ is established by the inequality you gave and so-called "density argument":

$\forall u\in W^{1,p}_0(I)$, choose $\phi_n\in C^{\infty}_c(I)$, and $\phi_n\to u$ under $\|\cdot\|_{W^{1,p}}$, define $\delta_0[u] = \lim\limits_{n\to \infty}\delta_0[\phi_n]$, then $\delta_0[\cdot]$ is a bounded linear functional in $W^{1,p}_0(I)$.

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the question is math.stackexchange.com/questions/208906/delta-dirac-function/…, sorry the Sobolev space is $W^{-1,p}$ –  user46060 Nov 8 '12 at 19:39
    
@user46060 this is same with what I wrote, $W^{-1,p}$ is the same with the space of bounded linear functional in $W^{1,p}_0$. –  Shuhao Cao Nov 8 '12 at 19:43

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