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Consider $A=(-1;1)$ and the functional $\mu$ on $C_0^\infty (A)$ defined as $\mu(\phi(x))=\phi(0)$. Can $\mu$ be extended to a continuous linear functional on $H_0^0(A)$ (i.e. $L^2(A)$) or $H_0^1(A)$?

What i did

I proved that $\mu$ is bounded and linear. Is it true that $D(\mu)=C_0^\infty (A)$ is dense in both $L^2(A)$ and $H_0^1(A)$? In that case can we use the following theorem?

$X$,$Y$ Banach spaces, $D(\mu)$ dense in $X$, $\mu$ linear and continuous, then there exists a unique, continuous extension of $\mu$ to $X$.

2 Answers 2

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$\mu$ is already unbounded on $C^{\infty}$ with respect to the $L^2$ norm. For example: For $n\in \mathbb N$ let $f_n(x)=\exp \left(\frac {-1}{x-\frac {1}{n}}\right) $ for $x\in [0,1/n)$ and $f_n(x)=0$ for $x\in [1/n,1].$ Then $f$ is strictly decreasing on $[0,1/n]$ so $$\int_0^1f(x)^2dx=\int_0^{1/n}f(x)^2dx \sqrt n.$

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Well $\mu$ is just the dirac delta measure. If it were to be extended to $L^2$, by duality and the Riesz representation $\delta$ would have $L^2$ density (This is not true). For example, by CS-inequality, $|f(0)| \leq C\|f\|_2$. You can contradict this.

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    So is the theorem I wrote wrong? To prove that \mu can't be extended to L^2 is this way correct? If I consider the sequence of functions $f_n(x)=n$ if $x=0$ and $f_n(x)=0$ otherwise. $f_n(x) \in L^2(A)$ and the norm of $L^2$ is zero. We have that $|f(0)|=n$ goes to infinity. – 2017-01-08