a question on the extension of an operator?.

Question

It is known that $C_0^{\infty}(\Omega)$ is dense in $W_0^{1,p}(\Omega)$ where $\Omega$ is a bounded domain of $\mathbb{R}^n$. Let $T:C_0^{\infty}(\Omega)\rightarrow\mathbb{R}$ be a continuous linear functional (a distribution). Can $T$ be continuously extended to the whole $W_0^{1,p}(\Omega)$?.

Answer 1

That's not true in general. The general theory says that if $A \subset B$ is dense in the Banach space $B$, then a linear functional $\phi$ on $A$ can be extended to $B$ if and only if there is $C$ so that

$$ |\phi (f)| \le C ||f||_B$$

for all $f\in A$.

For example, fix $x\in \Omega$. Then the functional $\phi_x (f) = f(x)$ defined on $C^\infty_0(\Omega)$ is not bounded in $W^{1,p}_0(\Omega)$-norm whenever $p<n$. To see clearly why, consider the function

$$f(x) = \log \left( \log \left(1+\frac 1{|x|}\right) \right).$$

Defined on $\Omega = B(0,\frac{1}{e-1}) \subset \mathbb R^3$. This is an unbounded sobolev functions in $W^{1, 2}_0(\Omega)$. Define $f_n = \min \{n, f\}$ and let $g_n \in C_0^\infty(\Omega)$ be a smooth function so that

$$||f_n - g_n||_{C^0}, ||f_n - g_n||_{W^{1,2}} <\epsilon/2^n,$$

Then as $f_n \to f$ in $W^{1, 2}$, we have $g_n \to f$ in $W^{1, 2}$ too. However, $g_n(0) \to \infty$ as $f$ is unbounded at $0$. Thus there isn't a $C$ so that

$$ |g_n(0)| \le C ||g_n||_{W^{1, 2}}$$

for all $n$. Thus $\phi (g) = g(0)$ cannot be extended to $W^{1, 2}(\Omega)$.

$p<n$ is used essentially because there are unbounded Sobolev function in this case. When $p>n$, we actually have the Sobolev embedding $W^{1, p}(\Omega) \to C(\overline\Omega)$ and indeed the functional $\phi (g) = g(x)$ can be extended.